Track / Stage 6 / 104 of 400 #1104 of 1964
Problem 1104 National olympiad, first round Difficulty 6.2 Prove it
13. Given x , y , z > 0 x, y, z>0 x , y , z > 0 x 2 1 + x 2 + y 2 1 + y 2 + z 2 1 + z 2 = 2 \frac{x^{2}}{1+x^{2}}+\frac{y^{2}}{1+y^{2}}+\frac{z^{2}}{1+z^{2}}=2 1 + x 2 x 2 + 1 + y 2 y 2 + 1 + z 2 z 2 = 2 x 1 + x 2 + y 1 + y 2 + z 1 + z 2 ⩽ 2 \frac{x}{1+x^{2}}+\frac{y}{1+y^{2}}+\frac{z}{1+z^{2}} \leqslant \sqrt{2} 1 + x 2 x + 1 + y 2 y + 1 + z 2 z ⩽ 2
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Official solution 13. 1 + x 2 1 + x 2 + 1 + y 2 1 + y 2 + 1 + z 2 1 + z 2 = 3 , x 2 1 + x 2 + y 2 1 + y 2 + z 2 1 + z 2 = 2 \frac{1+x^{2}}{1+x^{2}}+\frac{1+y^{2}}{1+y^{2}}+\frac{1+z^{2}}{1+z^{2}}=3, \frac{x^{2}}{1+x^{2}}+\frac{y^{2}}{1+y^{2}}+\frac{z^{2}}{1+z^{2}}=2 1 + x 2 1 + x 2 + 1 + y 2 1 + y 2 + 1 + z 2 1 + z 2 = 3 , 1 + x 2 x 2 + 1 + y 2 y 2 + 1 + z 2 z 2 = 2
Therefore 1 1 + x 2 + 1 1 + y 2 + 1 1 + z 2 = 1 \frac{1}{1+x^{2}}+\frac{1}{1+y^{2}}+\frac{1}{1+z^{2}}=1 1 + x 2 1 + 1 + y 2 1 + 1 + z 2 1 = 1 x 1 + x 2 + y 1 + y 2 + z 1 + z 2 \frac{x}{1+x^{2}}+\frac{y}{1+y^{2}}+\frac{z}{1+z^{2}} 1 + x 2 x + 1 + y 2 y + 1 + z 2 z = x 1 + x 2 ⋅ 1 1 + x 2 + y 1 + y 2 ⋅ 1 1 + y 2 + z 1 + z 2 ⋅ 1 1 + z 2 ⩽ x 2 1 + x 2 + y 2 1 + y 2 + z 2 1 + z 2 ⋅ 1 1 + x 2 + 1 1 + y 2 + 1 1 + z 2 = 2.
\begin{array}{l}
=\frac{x}{\sqrt{1+x^{2}}} \cdot \frac{1}{\sqrt{1+x^{2}}}+\frac{y}{\sqrt{1+y^{2}}} \cdot \frac{1}{\sqrt{1+y^{2}}}+\frac{z}{\sqrt{1+z^{2}}} \cdot \frac{1}{\sqrt{1+z^{2}}} \\
\leqslant \sqrt{\frac{x^{2}}{1+x^{2}}+\frac{y^{2}}{1+y^{2}}+\frac{z^{2}}{1+z^{2}}} \cdot \sqrt{\frac{1}{1+x^{2}}+\frac{1}{1+y^{2}}+\frac{1}{1+z^{2}}}=\sqrt{2 .}
\end{array}
= 1 + x 2 x ⋅ 1 + x 2 1 + 1 + y 2 y ⋅ 1 + y 2 1 + 1 + z 2 z ⋅ 1 + z 2 1 ⩽ 1 + x 2 x 2 + 1 + y 2 y 2 + 1 + z 2 z 2 ⋅ 1 + x 2 1 + 1 + y 2 1 + 1 + z 2 1 = 2.
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Source: NuminaMath-1.5 ,
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