Đáp án: $P\ge \sqrt{3}+2$
Giải thích các bước giải:
Ta có:
$x+y+z=xyz\to \dfrac{1}{y}\cdot \dfrac{1}{z}+\dfrac{1}{z}\cdot \dfrac{1}{x}+\dfrac{1}{x}\cdot \dfrac{1}{y}=1$
$\to \dfrac{1}{x}+\dfrac{1}{y}+\dfrac{1}{z}\ge \sqrt{3\left(\dfrac{1}{y}\cdot \dfrac{1}{z}+\dfrac{1}{z}\cdot \dfrac{1}{x}+\dfrac{1}{x}\cdot \dfrac{1}{y}\right)}=\sqrt{3}$
Lại có:
$P=\dfrac{y+2}{x^2}+\dfrac{z+2}{y^2}+\dfrac{x+2}{z^2}$
$\to P=\left(\dfrac{y}{x^2}+\dfrac{z}{y^2}+\dfrac{x}{z^2}\right)+\left(\dfrac{2}{x^2}+\dfrac{2}{y^2}+\dfrac{2}{z^2}\right)$
$\to P=\left(\dfrac{y}{x^2}+\dfrac{z}{y^2}+\dfrac{x}{z^2}\right)+\left(\dfrac{2}{x^2}+\dfrac{2}{y^2}+\dfrac{2}{z^2}\right)$
$\to P=\left(\dfrac{\left(\dfrac1x\right)^2}{\dfrac1y}+\dfrac{\left(\dfrac1y\right)^2}{\dfrac1z}+\dfrac{\left(\dfrac1z\right)^2}{\dfrac1x}\right)+2\left(\dfrac{1}{x^2}+\dfrac{1}{y^2}+\dfrac{1}{z^2}\right)$
$\to P\ge \dfrac{\left(\dfrac1x+\dfrac1y+\dfrac1z\right)^2}{\dfrac1y+\dfrac1z+\dfrac1x}+2\left(\dfrac{1}{x}\cdot\dfrac1y+\dfrac{1}{y}\cdot\dfrac1z+\dfrac{1}{z}\cdot \dfrac1x\right)$
$\to P\ge \left(\dfrac1x+\dfrac1y+\dfrac1z\right)+2\cdot 1$
$\to P\ge \sqrt{3}+2$
Dấu = xảy ra khi $x=y=z=\sqrt{3}$
