# Violympic toán 9

cho x,y,z thỏa mãn $x+y+z\le\dfrac{3}{2}$ . tìm GTNN của $P=\dfrac{x\left(yz+1\right)^2}{z^2\left(xz+1\right)}+\dfrac{y\left(xz+1\right)^2}{y^2\left(xy+1\right)}+\dfrac{z\left(xy+1\right)^2}{x^2\left(yz+1\right)}$

Sigma CTV Hôm qua lúc 10:38

Áp dụng bất đẳng thức AM - GM:

$P\ge3\sqrt[3]{\dfrac{\left(xy+1\right)\left(yz+1\right)\left(zx+1\right)}{xyz}}$.

Áp dụng bất đẳng thức AM - GM ta có:

$xy+1=xy+\dfrac{1}{4}+\dfrac{1}{4}+\dfrac{1}{4}+\dfrac{1}{4}\ge5\sqrt[5]{\dfrac{xy}{4^4}}$.

Tương tự: $yz+1\ge5\sqrt[5]{\dfrac{yz}{4^4}};zx+1\ge5\sqrt[5]{\dfrac{zx}{4^4}}$.

Do đó $\left(xy+1\right)\left(yz+1\right)\left(zx+1\right)\ge125\sqrt[5]{\dfrac{\left(xyz\right)^2}{4^{12}}}$

$\Rightarrow\dfrac{\left(xy+1\right)\left(yz+1\right)\left(zx+1\right)}{xyz}\ge125\sqrt[5]{\dfrac{1}{4^{12}\left(xyz\right)^3}}$.

Mà $xyz\le\dfrac{\left(x+y+z\right)^3}{27}=\dfrac{1}{8}$

Nên $\dfrac{\left(xy+1\right)\left(yz+1\right)\left(zx+1\right)}{xyz}\ge125\sqrt[5]{\dfrac{8^3}{4^{12}}}=125\sqrt[5]{\dfrac{1}{2^{15}}}=\dfrac{125}{8}$

$\Rightarrow P\ge\dfrac{15}{2}$.

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