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Algebra Difficulty 5.9 AIME, harder Prove it

23. Let x,y,zx, y, z be positive real numbers, and xyz=1x y z=1, prove:
(1) (1+x+y)2+(1+y+z)2+(1+z+x)2⩾27(1+x+y)^{2}+(1+y+z)^{2}+(1+z+x)^{2} \geqslant 27;

Solution

23. (1) By Cauchy-Schwarz inequality and AM-GM inequality, we have
3[(1+x+y)2+(1+y+z)2+(1+z+x)2]⩾[(1+x+y)+(1+y+z)+(1+z+x)]2=[3+2(x+y+z)]2⩾[3+2×3(xyz)23]2=81\begin{array}{l} 3\left[(1+x+y)^{2}+(1+y+z)^{2}+(1+z+x)^{2}\right] \geqslant \\ {[(1+x+y)+(1+y+z)+(1+z+x)]^{2}=} \\ {[3+2(x+y+z)]^{2} \geqslant} \\ {\left[3+2 \times 3 \sqrt[3]{(x y z)^{2}}\right]^{2}=81} \end{array}

Therefore,
(1+x+y)2+(1+y+z)2+(1+z+x)2⩾27(1+x+y)^{2}+(1+y+z)^{2}+(1+z+x)^{2} \geqslant 27

Equality holds if and only if x=y=z=1x=y=z=1.

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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.