Maths Olympiad Prep

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Algebra Difficulty 6.5 National olympiad Prove it

20. Prove the inequality sinn2x+(sinnxcosnx)21\sin ^{n} 2 x+\left(\sin ^{n} x-\cos ^{n} x\right)^{2} \leqslant 1. (Problem from the 26th Russian Mathematical Olympiad)

Solution

20. The inequality to be proved is
sin2nx+(2n2)sinnxcosnx+cos2nx1\sin ^{2 n} x+\left(2^{n}-2\right) \sin ^{n} x \cos ^{n} x+\cos ^{2 n} x \leqslant 1

Taking the nn-th power on both sides of the identity sin2x+cos2x=1\sin ^{2} x+\cos ^{2} x=1, we get
1=(sin2nx+cos2nx)+n(sin2xcosn2x+cos2xsinn2x)+Cn2(sin4xcosn4x+cos4xsinn4x)+(sin2nx+cos2nx)+(2n2)sinnxcosnx\begin{aligned} 1= & \left(\sin ^{2 n} x+\cos ^{2 n} x\right)+n\left(\sin ^{2} x \cos ^{n-2} x+\cos ^{2} x \sin ^{n-2} x\right)+ \\ & \mathrm{C}_{n}^{2}\left(\sin ^{4} x \cos ^{n-4} x+\cos ^{4} x \sin ^{n-4} x\right)+\cdots \geqslant \\ & \left(\sin ^{2 n} x+\cos ^{2 n} x\right)+\left(2^{n}-2\right) \sin ^{n} x \cos ^{n} x \end{aligned}

This is because the value in each parenthesis is no less than 2sinnxcosnx2 \sin ^{n} x \cos ^{n} x, and the sum of the coefficients equals 12(2n\frac{1}{2}\left(2^{n}-\right.

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Source: NuminaMath-1.5, licensed Apache-2.0. Statement and solution reproduced as published; topic and difficulty added by this site.