answer：Okay, so I have this problem here: I need to find the number of lines that are tangent to the circle (x-3)² + (y-4)² = 2 and have equal intercepts on both the x-axis and y-axis. The options are A:1, B:2, C:3, D:4. Hmm, let me think about how to approach this.First, I remember that the general equation of a circle is (x-h)² + (y-k)² = r², where (h,k) is the center and r is the radius. So, in this case, the center of the circle is at (3,4) and the radius is √2 because 2 is the right-hand side of the equation. Okay, that's straightforward.Now, the lines we're looking for have equal intercepts on both axes. That means if a line intercepts the x-axis at (a,0), it should also intercept the y-axis at (0,a). So, the equation of such a line can be written in intercept form as x/a + y/a = 1, which simplifies to x + y = a. Alternatively, it can also be written as y = -x + a, which is a line with a slope of -1 and y-intercept at a.But wait, there's another case where the intercepts are equal but negative. So, if a line intercepts the x-axis at (-a,0) and the y-axis at (0,-a), the equation would be x/(-a) + y/(-a) = 1, which simplifies to x + y = -a. So, in this case, the equation is x + y = -a, which is similar to the previous one but with a negative constant term.Therefore, the general form of the lines we're considering is x + y = c, where c can be any real number. So, the lines can be either x + y = c or x + y = -c. But actually, since c can be positive or negative, we can just consider x + y = c, where c is any real number, and it will cover both cases.Now, these lines need to be tangent to the given circle. So, the condition for a line to be tangent to a circle is that the distance from the center of the circle to the line is equal to the radius of the circle.The formula for the distance from a point (h,k) to the line Ax + By + C = 0 is |Ah + Bk + C| / sqrt(A² + B²). So, in our case, the line is x + y = c, which can be rewritten as x + y - c = 0. So, A=1, B=1, C=-c.The center of the circle is (3,4), so plugging into the distance formula, we get |1*3 + 1*4 - c| / sqrt(1² + 1²) = |7 - c| / sqrt(2). This distance should be equal to the radius, which is √2.So, setting up the equation: |7 - c| / sqrt(2) = sqrt(2). Multiplying both sides by sqrt(2), we get |7 - c| = 2.This absolute value equation will have two solutions: 7 - c = 2 and 7 - c = -2. Solving the first equation: 7 - c = 2 => c = 5. Solving the second equation: 7 - c = -2 => c = 9.So, the two lines are x + y = 5 and x + y = 9. Therefore, there are two lines with equal intercepts that are tangent to the circle.Wait, but hold on a second. I also need to consider lines that pass through the origin because if a line passes through the origin, its intercepts are both zero, which is equal. So, lines passing through the origin also have equal intercepts (both zero). So, I need to check if there are any such lines that are tangent to the circle.A line passing through the origin can be written as y = mx, where m is the slope. So, the equation is mx - y = 0. The distance from the center (3,4) to this line should be equal to the radius √2.Using the distance formula again: |m*3 - 4| / sqrt(m² + 1) = sqrt(2). So, |3m - 4| / sqrt(m² + 1) = sqrt(2). Let's square both sides to eliminate the square roots:(3m - 4)² / (m² + 1) = 2.Expanding the numerator: 9m² - 24m + 16.So, (9m² - 24m + 16) / (m² + 1) = 2.Multiply both sides by (m² + 1):9m² - 24m + 16 = 2m² + 2.Bring all terms to one side:9m² - 24m + 16 - 2m² - 2 = 0 => 7m² - 24m + 14 = 0.Now, solving this quadratic equation for m:Using the quadratic formula, m = [24 ± sqrt(24² - 4*7*14)] / (2*7).Calculating the discriminant:24² = 576, 4*7*14 = 392.So, sqrt(576 - 392) = sqrt(184) = sqrt(4*46) = 2*sqrt(46).Therefore, m = [24 ± 2sqrt(46)] / 14 = [12 ± sqrt(46)] / 7.So, there are two real solutions for m, meaning there are two lines passing through the origin that are tangent to the circle.Therefore, in total, we have two lines from the x + y = c case and two lines from the y = mx case, giving us four tangent lines in total that have equal intercepts on both axes.Wait, but hold on again. When I considered the lines passing through the origin, I assumed that their intercepts are both zero, which is equal. But in the problem statement, it says "equal intercepts on the two coordinate axes." So, does that include the origin? I think yes, because the intercepts are both zero, which is equal. So, those two lines should be counted as well.Therefore, combining both cases, we have four lines in total: two of the form x + y = c and two passing through the origin with slopes [12 ± sqrt(46)] / 7.Wait, but let me double-check. When I considered x + y = c, I found c = 5 and c = 9, giving two lines. Then, for lines through the origin, I found two more lines. So, in total, four lines. So, the answer should be D:4.But let me make sure I didn't miss anything. Is there any other case where the intercepts are equal? For example, lines where the intercepts are both positive or both negative, but not passing through the origin. Wait, but I already considered x + y = c, which includes both positive and negative c, so that should cover all cases except the origin.But actually, when c=0, x + y = 0 is a line passing through the origin, but in that case, the intercepts are both zero, which is equal. So, in my earlier calculation, when I solved |7 - c| = 2, I got c=5 and c=9, but c=0 wasn't a solution. So, x + y = 0 is another line passing through the origin, but in that case, it's a separate case because it's not covered by the x + y = c where c is non-zero.Wait, no. When I considered lines passing through the origin, I used the form y = mx, which is a different approach. So, actually, x + y = 0 is a specific case of y = mx where m = -1. So, in that case, when I solved for m, I found two slopes, which are [12 + sqrt(46)] / 7 and [12 - sqrt(46)] / 7. Neither of these is equal to -1, so x + y = 0 is not one of the tangent lines. Therefore, the two lines passing through the origin are distinct from the lines x + y = 5 and x + y = 9.So, in total, we have four tangent lines: two of the form x + y = c (c=5 and c=9) and two lines passing through the origin with slopes [12 ± sqrt(46)] / 7.Therefore, the correct answer should be D:4.Wait, but let me visualize this to make sure. The circle is centered at (3,4) with a radius of √2, which is approximately 1.414. So, it's a relatively small circle. The lines x + y = 5 and x + y = 9 are straight lines with a negative slope of -1. The distance from the center to these lines is equal to the radius, so they should be tangent.Similarly, the lines passing through the origin with slopes [12 ± sqrt(46)] / 7 should also be tangent to the circle. Since sqrt(46) is approximately 6.782, so [12 + 6.782]/7 ≈ 18.782/7 ≈ 2.683, and [12 - 6.782]/7 ≈ 5.218/7 ≈ 0.745. So, these are two distinct slopes, meaning two distinct lines passing through the origin that are tangent to the circle.Therefore, adding these up, we have four tangent lines in total that have equal intercepts on both axes. So, the answer is D:4.I think that's thorough. I considered both cases where the lines pass through the origin and where they don't, ensuring that all possibilities are covered. I also double-checked my calculations to make sure there were no errors. So, I feel confident that the answer is D:4.

answer：Okay, so I have this inequality to prove: Given ( a, b, c > 0 ), show that[ frac{1}{a(1+b)} + frac{1}{b(1+c)} + frac{1}{c(1+a)} geqslant frac{3}{sqrt[3]{abc}(1+sqrt[3]{abc})}. ]Hmm, okay. Let me think about how to approach this. I remember that inequalities often involve techniques like AM-GM, Cauchy-Schwarz, or maybe even substitution methods. Since the right-hand side involves the cube root of ( abc ), maybe I can use some symmetric substitution or homogenization.First, let me note that the inequality is symmetric in ( a, b, c ), so perhaps assuming ( a = b = c ) might help me check if the inequality holds in that case, which could give me some insight.If ( a = b = c ), then ( sqrt[3]{abc} = a ). So the right-hand side becomes:[ frac{3}{a(1 + a)}. ]On the left-hand side, each term becomes ( frac{1}{a(1 + a)} ), so adding them up gives:[ frac{3}{a(1 + a)}. ]So, in the case where ( a = b = c ), both sides are equal. That suggests that the inequality holds with equality when ( a = b = c ). That's a good sign. Now, I need to see if I can generalize this.Maybe I can use the AM-GM inequality somewhere. Let's recall that AM-GM says that for positive real numbers, the arithmetic mean is at least the geometric mean. So, perhaps I can find some way to apply AM-GM to the terms in the denominators or numerators.Looking at the left-hand side, each term is of the form ( frac{1}{a(1 + b)} ). Maybe I can manipulate this expression. Let's consider each term separately.Let me rewrite ( frac{1}{a(1 + b)} ) as ( frac{1}{a} cdot frac{1}{1 + b} ). Hmm, not sure if that helps directly. Alternatively, maybe I can use substitution to simplify the expression.Let me set ( x = sqrt[3]{abc} ). Then, ( abc = x^3 ). Maybe I can express ( a, b, c ) in terms of ( x ) and some other variables. For example, I can let ( a = frac{x y}{z} ), ( b = frac{x z}{y} ), and ( c = frac{y z}{x} ). Wait, is that correct? Let me check:If ( a = frac{x y}{z} ), ( b = frac{x z}{y} ), and ( c = frac{y z}{x} ), then ( abc = frac{x y}{z} cdot frac{x z}{y} cdot frac{y z}{x} = x y z ). Hmm, but I wanted ( abc = x^3 ). So maybe I need a different substitution.Alternatively, perhaps I can set ( a = frac{x}{y} ), ( b = frac{y}{z} ), and ( c = frac{z}{x} ). Then, ( abc = frac{x}{y} cdot frac{y}{z} cdot frac{z}{x} = 1 ). But that's not helpful because I want ( abc = x^3 ). Maybe I need to include ( x ) in the substitution.Wait, maybe I can set ( a = k cdot frac{a_2}{a_1} ), ( b = k cdot frac{a_3}{a_2} ), and ( c = k cdot frac{a_1}{a_3} ), where ( k = sqrt[3]{abc} ). Let me check:Then, ( abc = k cdot frac{a_2}{a_1} cdot k cdot frac{a_3}{a_2} cdot k cdot frac{a_1}{a_3} = k^3 cdot frac{a_2 a_3 a_1}{a_1 a_2 a_3} = k^3 ). Perfect, that works.So, substituting ( a, b, c ) in terms of ( k ) and ( a_1, a_2, a_3 ), the left-hand side becomes:[ frac{1}{a(1 + b)} = frac{1}{k cdot frac{a_2}{a_1} left(1 + k cdot frac{a_3}{a_2}right)} = frac{1}{k cdot frac{a_2}{a_1} + k^2 cdot frac{a_3}{a_1}} = frac{a_1}{k a_2 + k^2 a_3}. ]Similarly, the other terms become:[ frac{1}{b(1 + c)} = frac{a_2}{k a_3 + k^2 a_1}, ][ frac{1}{c(1 + a)} = frac{a_3}{k a_1 + k^2 a_2}. ]So, the left-hand side of the inequality is:[ frac{a_1}{k a_2 + k^2 a_3} + frac{a_2}{k a_3 + k^2 a_1} + frac{a_3}{k a_1 + k^2 a_2}. ]Now, the right-hand side is:[ frac{3}{k(1 + k)}. ]So, the inequality becomes:[ frac{a_1}{k a_2 + k^2 a_3} + frac{a_2}{k a_3 + k^2 a_1} + frac{a_3}{k a_1 + k^2 a_2} geq frac{3}{k(1 + k)}. ]Hmm, this seems a bit complicated, but maybe I can factor out ( k ) from the denominators:[ frac{a_1}{k(a_2 + k a_3)} + frac{a_2}{k(a_3 + k a_1)} + frac{a_3}{k(a_1 + k a_2)} = frac{1}{k} left( frac{a_1}{a_2 + k a_3} + frac{a_2}{a_3 + k a_1} + frac{a_3}{a_1 + k a_2} right). ]So, the inequality simplifies to:[ frac{1}{k} left( frac{a_1}{a_2 + k a_3} + frac{a_2}{a_3 + k a_1} + frac{a_3}{a_1 + k a_2} right) geq frac{3}{k(1 + k)}. ]Multiplying both sides by ( k ), we get:[ frac{a_1}{a_2 + k a_3} + frac{a_2}{a_3 + k a_1} + frac{a_3}{a_1 + k a_2} geq frac{3}{1 + k}. ]Okay, so now I need to prove that:[ frac{a_1}{a_2 + k a_3} + frac{a_2}{a_3 + k a_1} + frac{a_3}{a_1 + k a_2} geq frac{3}{1 + k}. ]This looks similar to Nesbitt's inequality, which states that for positive real numbers ( a, b, c ):[ frac{a}{b + c} + frac{b}{a + c} + frac{c}{a + b} geq frac{3}{2}. ]But in our case, the denominators have an extra term with ( k ). Maybe I can use a similar approach as in Nesbitt's inequality, perhaps using the Cauchy-Schwarz inequality or some other technique.Let me recall that in Nesbitt's inequality, one common approach is to use the AM-HM inequality or to apply the Cauchy-Schwarz inequality in the form of:[ left( sum frac{a}{b + c} right) left( sum a(b + c) right) geq (a + b + c)^2. ]Maybe I can try something similar here. Let's denote the sum as ( S ):[ S = frac{a_1}{a_2 + k a_3} + frac{a_2}{a_3 + k a_1} + frac{a_3}{a_1 + k a_2}. ]I want to find a lower bound for ( S ). Let me consider applying the Cauchy-Schwarz inequality in the following form:[ left( sum frac{a_1}{a_2 + k a_3} right) left( sum a_1(a_2 + k a_3) right) geq (a_1 + a_2 + a_3)^2. ]Yes, that seems promising. Let me compute ( sum a_1(a_2 + k a_3) ):[ sum a_1(a_2 + k a_3) = a_1 a_2 + k a_1 a_3 + a_2 a_3 + k a_2 a_1 + a_3 a_1 + k a_3 a_2. ]Wait, that's:[ = a_1 a_2 + a_2 a_3 + a_3 a_1 + k(a_1 a_3 + a_2 a_1 + a_3 a_2) ][ = (1 + k)(a_1 a_2 + a_2 a_3 + a_3 a_1). ]So, ( sum a_1(a_2 + k a_3) = (1 + k)(a_1 a_2 + a_2 a_3 + a_3 a_1) ).Therefore, by Cauchy-Schwarz:[ S cdot (1 + k)(a_1 a_2 + a_2 a_3 + a_3 a_1) geq (a_1 + a_2 + a_3)^2. ]So,[ S geq frac{(a_1 + a_2 + a_3)^2}{(1 + k)(a_1 a_2 + a_2 a_3 + a_3 a_1)}. ]Now, I need to show that this is at least ( frac{3}{1 + k} ). So, I need:[ frac{(a_1 + a_2 + a_3)^2}{(1 + k)(a_1 a_2 + a_2 a_3 + a_3 a_1)} geq frac{3}{1 + k}. ]Multiplying both sides by ( (1 + k) ), we get:[ frac{(a_1 + a_2 + a_3)^2}{a_1 a_2 + a_2 a_3 + a_3 a_1} geq 3. ]This is a well-known inequality. In fact, it's a form of the AM-GM inequality. Specifically, it's known that:[ (a_1 + a_2 + a_3)^2 geq 3(a_1 a_2 + a_2 a_3 + a_3 a_1). ]Yes, this is true because expanding ( (a_1 + a_2 + a_3)^2 ) gives:[ a_1^2 + a_2^2 + a_3^2 + 2(a_1 a_2 + a_2 a_3 + a_3 a_1) geq 3(a_1 a_2 + a_2 a_3 + a_3 a_1), ]which simplifies to:[ a_1^2 + a_2^2 + a_3^2 geq a_1 a_2 + a_2 a_3 + a_3 a_1, ]and this is true by the rearrangement inequality or by noting that ( (a_1 - a_2)^2 + (a_2 - a_3)^2 + (a_3 - a_1)^2 geq 0 ).Therefore, we have:[ frac{(a_1 + a_2 + a_3)^2}{a_1 a_2 + a_2 a_3 + a_3 a_1} geq 3, ]which implies:[ S geq frac{3}{1 + k}. ]Thus, going back to our earlier substitution, we have:[ frac{a_1}{a_2 + k a_3} + frac{a_2}{a_3 + k a_1} + frac{a_3}{a_1 + k a_2} geq frac{3}{1 + k}, ]which means:[ frac{1}{a(1 + b)} + frac{1}{b(1 + c)} + frac{1}{c(1 + a)} geq frac{3}{sqrt[3]{abc}(1 + sqrt[3]{abc})}. ]So, the inequality is proven. I think this makes sense. I used substitution to simplify the variables, applied Cauchy-Schwarz to get a lower bound, and then used a known inequality to complete the proof. It seems a bit involved, but each step logically follows from the previous one. I don't see any mistakes in my reasoning, so I think this should be correct.

answer：To calculate the value of the product (left(1 - frac{1}{2}right)left(1 - frac{1}{3}right)left(1 - frac{1}{4}right)left(1 - frac{1}{5}right) dotsm left(1 - frac{1}{100}right)), I start by simplifying each term individually.Each term (1 - frac{1}{n}) can be rewritten as (frac{n-1}{n}). This transforms the product into a sequence of fractions:[frac{1}{2} times frac{2}{3} times frac{3}{4} times frac{4}{5} times cdots times frac{99}{100}]Next, I observe that this sequence exhibits a telescoping pattern. In a telescoping product, most terms in the numerator and denominator cancel out sequentially. Specifically, the numerator of each fraction cancels with the denominator of the preceding fraction:[frac{1}{2} times frac{2}{3} = frac{1}{3}, quad frac{1}{3} times frac{3}{4} = frac{1}{4}, quad text{and so on.}]Continuing this cancellation process through all the terms, only the first numerator and the last denominator remain. Therefore, the entire product simplifies to:[frac{1}{100}]

answer：First, identify the weight lost by each person:- Seth lost 17.5 pounds.- Veronica lost 1.5 pounds more than Seth, which is 17.5 + 1.5 = 19 pounds.- Let the multiple by which Jerome lost weight compared to Seth be ( x ). Therefore, Jerome lost ( 17.5x ) pounds.Next, set up the equation for the total weight lost:[ 17.5 + 17.5x + 19 = 89 ]Combine like terms:[ 17.5x + 36.5 = 89 ]Solve for ( x ):[ 17.5x = 89 - 36.5 ][ 17.5x = 52.5 ][ x = frac{52.5}{17.5} ][ x = 3 ]Finally, determine the ratio of the pounds Jerome lost to the pounds Seth lost:[ frac{17.5 times 3}{17.5} = 3 ]Therefore, the ratio is ( 3:1 ).

answer：Alright, so I've got this problem here that says if for any angle φ, the inequality a₁ cos φ + a₂ cos 2φ + ... + aₙ cos nφ is greater than or equal to -1, then the sum of all the a's, a₁ + a₂ + ... + aₙ, is less than or equal to n. Hmm, okay, let's try to unpack this.First off, I know that cosine functions oscillate between -1 and 1. So, each term aᵢ cos iφ is going to be somewhere between -aᵢ and aᵢ. The sum of these terms is supposed to always be at least -1, no matter what φ is. That seems like a pretty strict condition because φ can be any angle, which means the cosines can take on any value in their range.I wonder if there's a way to choose specific angles φ that make the expression a₁ cos φ + a₂ cos 2φ + ... + aₙ cos nφ as small as possible. If I can find such angles, then maybe I can derive some constraints on the a's.Wait, the problem mentions that this inequality holds for any φ. So, maybe I can pick specific φ's that simplify the expression or make it take on its minimum value. For example, if I set φ such that all the cosines are aligned in a certain way, maybe I can get the sum to reach its minimum.I recall that there's a result in trigonometry where the sum of cosines of equally spaced angles can be simplified. Maybe that's relevant here. Let me think... Oh, right! There's a formula for the sum of cosines of angles in arithmetic progression. Maybe I can use that.But before I dive into that, let me consider what happens if I set φ to specific values. For instance, if I set φ = 0, then cos 0 = 1 for all terms, so the expression becomes a₁ + a₂ + ... + aₙ. The inequality would then say that a₁ + a₂ + ... + aₙ ≥ -1. But that's not directly helpful because we need an upper bound, not a lower bound.Wait, but if I set φ = π, then cos iπ alternates between 1 and -1 depending on whether i is even or odd. So, the expression becomes a₁(-1) + a₂(1) + a₃(-1) + ... + aₙ(-1)^n. Hmm, that might not be directly useful either because it depends on the parity of n.Maybe I need a different approach. Since the inequality must hold for all φ, perhaps I can use some kind of optimization or consider the minimum value of the expression over all φ. If the minimum is at least -1, then maybe I can relate that to the sum of the a's.I remember that in Fourier series, the sum of cosines can be represented as a function, and maybe I can use some properties from there. But I'm not sure if that's the right path.Alternatively, maybe I can use the fact that the sum is bounded below by -1 and then relate that to the maximum value of the sum. Since the sum is bounded below, perhaps its maximum is related to the sum of the a's.Wait, another idea: what if I consider the sum when φ is such that all the cosines are equal to 1? That would make the expression equal to a₁ + a₂ + ... + aₙ. But earlier, I saw that setting φ = 0 gives me that the sum is greater than or equal to -1, but I need an upper bound.Hmm, maybe I need to use some kind of inequality, like Cauchy-Schwarz or something else, to relate the sum of the a's to the sum involving cosines.Let me think about Cauchy-Schwarz. If I have two vectors, say (a₁, a₂, ..., aₙ) and (cos φ, cos 2φ, ..., cos nφ), then their dot product is a₁ cos φ + a₂ cos 2φ + ... + aₙ cos nφ. The Cauchy-Schwarz inequality says that this dot product is less than or equal to the product of the norms of the two vectors.But in this case, the inequality is a lower bound, not an upper bound. So, maybe Cauchy-Schwarz isn't directly applicable here.Wait, but if I square both sides, maybe I can get something useful. Let's see:(a₁ cos φ + a₂ cos 2φ + ... + aₙ cos nφ)² ≤ (a₁² + a₂² + ... + aₙ²)(cos² φ + cos² 2φ + ... + cos² nφ)But again, this gives me an upper bound on the square of the sum, not a lower bound. So, I'm not sure if this helps.Maybe I need to think differently. Since the inequality holds for all φ, perhaps I can integrate both sides over φ from 0 to 2π. That might allow me to use orthogonality of the cosine functions.Let me try that. If I integrate the left side over φ from 0 to 2π, I get:∫₀²π (a₁ cos φ + a₂ cos 2φ + ... + aₙ cos nφ) dφAnd the integral of cos kφ over 0 to 2π is zero for any integer k ≠ 0. So, the integral of the left side is zero.On the right side, integrating -1 over 0 to 2π gives -2π. But the integral of the left side is zero, which is greater than or equal to -2π. That doesn't seem helpful because 0 ≥ -2π is always true, but it doesn't give me any new information about the a's.Hmm, maybe integrating isn't the way to go. Let me think again.I remember that in some problems involving trigonometric polynomials, we can use specific values of φ to extract information about the coefficients. Maybe I can choose φ such that the sum a₁ cos φ + a₂ cos 2φ + ... + aₙ cos nφ reaches its minimum value.Wait, there's a result that says that for certain roots of unity, the sum of cosines can be simplified. Maybe I can use that.Let me consider φ = 2πk/(n+1) for k = 1, 2, ..., n. These are the (n+1)th roots of unity. I think there's a property that the sum of cosines at these angles has a specific value.In fact, I recall that for φ_k = 2πk/(n+1), the sum cos φ_k + cos 2φ_k + ... + cos nφ_k equals -1. Is that right? Let me check for small n.For n=1, φ = 2π/2 = π. cos π = -1, which is indeed -1.For n=2, φ = 2π/3. cos(2π/3) = -1/2, cos(4π/3) = -1/2. So, the sum is -1/2 + (-1/2) = -1. Okay, that works.For n=3, φ = 2π/4 = π/2. cos(π/2) = 0, cos(π) = -1, cos(3π/2) = 0. So, the sum is 0 + (-1) + 0 = -1. Wait, but n=3, so we have cos φ + cos 2φ + cos 3φ. At φ=π/2, that's 0 + (-1) + 0 = -1. Okay, that works too.So, it seems that for φ_k = 2πk/(n+1), the sum of cosines from 1 to n is indeed -1. That's a useful result.Now, going back to the original inequality:a₁ cos φ + a₂ cos 2φ + ... + aₙ cos nφ ≥ -1If I set φ = φ_k = 2πk/(n+1), then the left side becomes:a₁ cos φ_k + a₂ cos 2φ_k + ... + aₙ cos nφ_kBut we know that for each k, the sum of cosines is -1. So, if I sum the inequalities over all k from 1 to n, I get:Sum_{k=1}^n [a₁ cos φ_k + a₂ cos 2φ_k + ... + aₙ cos nφ_k] ≥ Sum_{k=1}^n (-1)On the left side, I can factor out the a's:a₁ Sum_{k=1}^n cos φ_k + a₂ Sum_{k=1}^n cos 2φ_k + ... + aₙ Sum_{k=1}^n cos nφ_kBut from the result I just recalled, each Sum_{k=1}^n cos iφ_k = -1 for i = 1, 2, ..., n.So, the left side becomes:a₁(-1) + a₂(-1) + ... + aₙ(-1) = -(a₁ + a₂ + ... + aₙ)And the right side is Sum_{k=1}^n (-1) = -nSo, putting it together:-(a₁ + a₂ + ... + aₙ) ≥ -nMultiplying both sides by -1 (which reverses the inequality):a₁ + a₂ + ... + aₙ ≤ nAnd that's exactly what we wanted to prove!Wait, let me double-check that. I used the fact that for each k, the sum of cosines is -1, and then summed over all k. Then, I factored out the a's and used the result that each sum of cosines is -1. That seems correct.Also, when I multiplied both sides by -1, the inequality direction changed, which is important. So, the conclusion is that the sum of the a's is less than or equal to n.I think that makes sense. By choosing specific angles where the sum of cosines is minimized, we were able to derive a constraint on the sum of the coefficients a's. This approach leverages the given condition that the inequality holds for all φ, and by selecting specific φ's, we can extract the necessary information about the coefficients.So, in summary, by using the specific angles φ_k = 2πk/(n+1), we transformed the given inequality into a condition that directly relates to the sum of the a's, leading us to the desired result.

answer：First, I need to multiply the two polynomials ( ax^{2} + bx + 1 ) and ( 3x - 2 ). Multiplying each term of the first polynomial by each term of the second polynomial:[(ax^{2})(3x) = 3ax^{3}][(ax^{2})(-2) = -2ax^{2}][(bx)(3x) = 3bx^{2}][(bx)(-2) = -2bx][(1)(3x) = 3x][(1)(-2) = -2]Next, I combine like terms:[3ax^{3} + (-2ax^{2} + 3bx^{2}) + (-2bx + 3x) - 2]Simplifying further:[3ax^{3} + (3b - 2a)x^{2} + (-2b + 3)x - 2]Since the product does not contain the ( x^{2} ) or ( x ) terms, their coefficients must be zero:1. For the ( x^{2} ) term:[3b - 2a = 0]2. For the ( x ) term:[-2b + 3 = 0]Solving the second equation for ( b ):[-2b + 3 = 0 implies -2b = -3 implies b = frac{3}{2}]Substituting ( b = frac{3}{2} ) into the first equation:[3left(frac{3}{2}right) - 2a = 0 implies frac{9}{2} - 2a = 0 implies 2a = frac{9}{2} implies a = frac{9}{4}]Therefore, the values of the coefficients are ( a = frac{9}{4} ) and ( b = frac{3}{2} ).