Đặt: $x=\dfrac{2a^2}{bc};y=\dfrac{2b^2}{ac};z=\dfrac{2c^2}{ab}$ với $a,b,c>0$.
Khi đó ta có:
$\dfrac{x^2}{x^2+2x+4}+\dfrac{y^2}{y^2+2y+4}+\dfrac{z^2}{z^2+2z+4}$
$=\dfrac{\dfrac{4a^4}{b^2c^2}}{\dfrac{4a^4}{b^2c^2}+\dfrac{4a^2}{bc}+4}+\dfrac{\dfrac{4b^4}{a^2c^2}}{\dfrac{4b^4}{a^2c^2}+\dfrac{4b^2}{ac}+4}+\dfrac{\dfrac{4c^4}{a^2b^2}}{\dfrac{4c^4}{a^2b^2}+\dfrac{4c^2}{ab}+4}$
$=\dfrac{a^4}{a^4+a^2bc+b^2c^2}+\dfrac{b^4}{b^4+b^2ac+a^2c^2}+\dfrac{c^4}{c^4+c^2ab+a^2b^2}$
$\ge\dfrac{(a^2+b^2+c^2)^2}{a^4+b^4+c^4+a^2bc+b^2ac+c^2ab+a^2b^2+b^2c^2+a^2c^2}\ge1$
Dấu bằng xảy ra khi: $x=y=z=2$