Ta có: $\dfrac{sin^4x}{a}+\dfrac{cos^4x}{b}\geq\dfrac{(sin^2x+cos^2x)^2}{a+b}=\dfrac{1}{a+b}$Dấu bằng xảy ra $<=>\dfrac{sin^2x}{a}=\dfrac{cos^2x}{b}=\dfrac{sin^2x+cos^2x}{a+b}=\dfrac{1}{a+b}$
$<=>\dfrac{sin^6x}{a^3}=\dfrac{cos^6x}{b^3}=\dfrac{1}{(a+b)^3}$
$<=>\dfrac{sin^8x}{a^3}=\dfrac{sin^2x}{(a+b)^3}$ và $\dfrac{cos^8x}{b^3}=\dfrac{cos^2}{(a+b)^3}$
$<=>\dfrac{sin^8x}{a^3}+\dfrac{cos^{8}x}{b^{3}}$
$=\dfrac{sin^2x+cos^2x}{(a+b)^3}$
$=\dfrac{1}{(a+b)^3}$