Question

Cho x,y,z > 0 thỏa mãn \(\frac{1}{x+1}+\frac{1}{y+1}+\frac{1}{z+1}=2\). CMR: \(xyz\le8\)

Answer 1

\(\frac{1}{x+1}=1-\frac{1}{y+1}+1-\frac{1}{z+1}=\frac{y}{y+1}+\frac{z}{z+1}\ge2\sqrt{\frac{yz}{\left(y+1\right)\left(z+1\right)}}\)

Tương tự: \(\frac{1}{y+1}\ge2\sqrt{\frac{zx}{\left(z+1\right)\left(x+1\right)}}\) ; \(\frac{1}{z+1}\ge2\sqrt{\frac{xy}{\left(x+1\right)\left(y+1\right)}}\)

Nhân vế với vế:

\(\frac{1}{\left(x+1\right)\left(y+1\right)\left(z+1\right)}\ge\frac{8xyz}{\left(x+1\right)\left(y+1\right)\left(z+1\right)}\)

\(\Rightarrow xyz\le\frac{1}{8}\)

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