Track / Stage 6 / 338 of 400 #1338 of 1964
Problem 1338 National Olympiad, first round Difficulty 6.5 Prove it
Given that x , y , z x, y, z x , y , z x + y + z = 1 x+y+z=1 x + y + z = 1 x 4 y ( 1 − y 2 ) + y 4 z ( 1 − z 2 ) + \frac{x^{4}}{y\left(1-y^{2}\right)}+\frac{y^{4}}{z\left(1-z^{2}\right)}+ y ( 1 − y 2 ) x 4 + z ( 1 − z 2 ) y 4 + z 4 x ( 1 − x 2 ) ⩾ 1 8 \frac{z^{4}}{x\left(1-x^{2}\right)} \geqslant \frac{1}{8} x ( 1 − x 2 ) z 4 ⩾ 8 1
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Official solution 15. By the generalization of Cauchy's inequality, we get( y + x + z ) [ ( 1 + y ) + ( 1 + z ) + ( 1 + x ) ] [ ( 1 − y ) + ( 1 − z ) + ( 1 − x ) ] ⋅ [ x 4 y ( 1 − y 2 ) + y 4 z ( 1 − z 2 ) + z 4 x ( 1 − x 2 ) ] ⩾ ( x + y + z ) 4 \begin{array}{l}
(y+x+z)[(1+y)+(1+z)+(1+x)][(1-y)+(1-z)+(1-x)] \cdot \\
{\left[\frac{x^{4}}{y\left(1-y^{2}\right)}+\frac{y^{4}}{z\left(1-z^{2}\right)}+\frac{z^{4}}{x\left(1-x^{2}\right)}\right] \geqslant(x+y+z)^{4}}
\end{array} ( y + x + z ) [( 1 + y ) + ( 1 + z ) + ( 1 + x )] [( 1 − y ) + ( 1 − z ) + ( 1 − x )] ⋅ [ y ( 1 − y 2 ) x 4 + z ( 1 − z 2 ) y 4 + x ( 1 − x 2 ) z 4 ] ⩾ ( x + y + z ) 4
Since x + y + z = 1 x+y+z=1 x + y + z = 1 x 4 y ( 1 − y 2 ) + y 4 z ( 1 − z 2 ) + z 4 x ( 1 − x 2 ) ⩾ 1 8 \frac{x^{4}}{y\left(1-y^{2}\right)}+\frac{y^{4}}{z\left(1-z^{2}\right)}+\frac{z^{4}}{x\left(1-x^{2}\right)} \geqslant \frac{1}{8} y ( 1 − y 2 ) x 4 + z ( 1 − z 2 ) y 4 + x ( 1 − x 2 ) z 4 ⩾ 8 1
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Source: NuminaMath-1.5 ,
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