Track / Stage 6 / 60 of 400 #1060 of 2000
Problem 1060 National olympiad, first round Difficulty 6.1 Prove it
6. Let non-negative real numbers x , y , z x, y, z x , y , z x 2 + y 2 + z 2 = 1 x^{2}+y^{2}+z^{2}=1 x 2 + y 2 + z 2 = 1 x 1 − x 2 + y 1 − y 2 + y 1 − z 2 ⩾ 3 3 2 \frac{x}{1-x^{2}}+\frac{y}{1-y^{2}}+\frac{y}{1-z^{2}} \geqslant \frac{3 \sqrt{3}}{2} 1 − x 2 x + 1 − y 2 y + 1 − z 2 y ⩾ 2 3 3
This one wants a proof. Work it on paper, read the official solution, then mark
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Official solution 6. ∑ cyc x 1 − x 2 ⩾ 3 3 2 ⇔ ∑ cyc ( x 1 − x 2 − 3 2 − 3 3 2 ( x 2 − 1 3 ) ) ⩾ 0 ⇔ ∑ cyc x ( 3 3 x 3 − 3 3 x + 2 ) 1 − x 2 ⩾ 0 \text { 6. } \begin{array}{l}
\sum_{\text {cyc }} \frac{x}{1-x^{2}} \geqslant \frac{3 \sqrt{3}}{2} \\
\Leftrightarrow \sum_{\text {cyc }}\left(\frac{x}{1-x^{2}}-\frac{\sqrt{3}}{2}-\frac{3 \sqrt{3}}{2}\left(x^{2}-\frac{1}{3}\right)\right) \geqslant 0 \\
\Leftrightarrow \sum_{\text {cyc }} \frac{x\left(3 \sqrt{3} x^{3}-3 \sqrt{3} x+2\right)}{1-x^{2}} \geqslant 0
\end{array} 6. ∑ cyc 1 − x 2 x ⩾ 2 3 3 ⇔ ∑ cyc ( 1 − x 2 x − 2 3 − 2 3 3 ( x 2 − 3 1 ) ) ⩾ 0 ⇔ ∑ cyc 1 − x 2 x ( 3 3 x 3 − 3 3 x + 2 ) ⩾ 0
In fact, by the Arithmetic Mean-Geometric Mean Inequality, we know3 3 x 3 − 3 3 x + 2 = 3 3 x 3 + 1 + 1 − 3 3 x ⩾ 3 3 3 x 3 ⋅ 1 2 3 − 3 3 x = 0 \begin{aligned}
3 \sqrt{3} x^{3}-3 \sqrt{3} x+2 & =3 \sqrt{3} x^{3}+1+1-3 \sqrt{3} x \\
& \geqslant 3 \sqrt[3]{3 \sqrt{3} x^{3} \cdot 1^{2}}-3 \sqrt{3} x=0
\end{aligned} 3 3 x 3 − 3 3 x + 2 = 3 3 x 3 + 1 + 1 − 3 3 x ⩾ 3 3 3 3 x 3 ⋅ 1 2 − 3 3 x = 0
Therefore, the original inequality holds.
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