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Algebra Difficulty 7.3 National olympiad, round 2 Prove it

42. Given that x,y,zx, y, z are positive real numbers, and x2+y2+z2=1x^{2}+y^{2}+z^{2}=1, prove that x1+x2+y1+y2+z1+z2⩽334\frac{x}{1+x^{2}}+\frac{y}{1+y^{2}}+\frac{z}{1+z^{2}} \leqslant \frac{3 \sqrt{3}}{4}. (1998 Bosnia and Herzegovina Mathematical Olympiad Problem)

Solution

42. Let x=tan⁡α,y=tan⁡β,z=tan⁡γ(α,β,γx=\tan \alpha, y=\tan \beta, z=\tan \gamma (\alpha, \beta, \gamma are acute angles )), then tan⁡2α+\tan ^{2} \alpha+ tan⁡2β+tan⁡2γ=1\tan ^{2} \beta+\tan ^{2} \gamma=1, by the Cauchy-Schwarz inequality we have 3 (tan⁡2α+tan⁡2β+tan⁡2γ)⩾(tan⁡α+\left(\tan ^{2} \alpha+\tan ^{2} \beta+\tan ^{2} \gamma\right) \geqslant(\tan \alpha+ tan⁡β+tan⁡γ)2\tan \beta+\tan \gamma)^{2}, so tan⁡α+tan⁡β+tan⁡γ⩽3\tan \alpha+\tan \beta+\tan \gamma \leqslant \sqrt{3}, since α,β,γ\alpha, \beta, \gamma are acute angles, by Jensen's inequality we get
tan⁡α+β+γ3⩽tan⁡α+tan⁡β+tan⁡γ3=33\tan \frac{\alpha+\beta+\gamma}{3} \leqslant \frac{\tan \alpha+\tan \beta+\tan \gamma}{3}=\frac{\sqrt{3}}{3}

Thus, α+β+γ3⩽\frac{\alpha+\beta+\gamma}{3} \leqslant, and by Jensen's inequality we get
x1+x2+y1+y2+z1+z2=12(sin⁡2α+sin⁡2β+sin⁡2γ)⩽32sin⁡2(α+β+γ)3⩽334\begin{array}{r} \frac{x}{1+x^{2}}+\frac{y}{1+y^{2}}+\frac{z}{1+z^{2}}=\frac{1}{2}(\sin 2 \alpha+\sin 2 \beta+\sin 2 \gamma) \leqslant \\ \frac{3}{2} \sin \frac{2(\alpha+\beta+\gamma)}{3} \leqslant \frac{3 \sqrt{3}}{4} \end{array}

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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.