Giải thích các bước giải:

Ta có:

$P=(\dfrac{xy}z+\dfrac{yz}x+\dfrac{zx}y)^2+4(xy+yz+zx)$

$\to P=\dfrac{x^2y^2}{z^2}+\dfrac{y^2z^2}{x^2}+\dfrac{z^2x^2}{y^2}+2\cdot \dfrac{xy}z\cdot\dfrac{yz}x+2\cdot\dfrac{yz}x\cdot \dfrac{zx}y+2\cdot\dfrac{zx}y\cdot \dfrac{xy}z+4(xy+yz+zx)$

$\to P=\dfrac{x^2y^2}{z^2}+\dfrac{y^2z^2}{x^2}+\dfrac{z^2x^2}{y^2}+ 2x^2+y^2+z^2+4(xy+yz+zx)$

$\to P=\dfrac{x^2y^2}{z^2}+\dfrac{y^2z^2}{x^2}+\dfrac{z^2x^2}{y^2}+ 2(x+y+z)^2$

$\to P=\dfrac{x^2y^2}{z^2}+\dfrac{y^2z^2}{x^2}+\dfrac{z^2x^2}{y^2}+ 2$

$\to P\ge \dfrac{xy}z\cdot\dfrac{yz}x+\dfrac{yz}x\cdot \dfrac{zx}y+\dfrac{zx}y\cdot \dfrac{xy}z + 2$

$\to P\ge x^2+y^2+z^2+2$

$\to P\ge \dfrac13(x+y+z)^2+2$

$\to P\ge \dfrac73$

Dấu = xảy ra khi $x=y=z=\dfrac13$