cách 2
\(\dfrac{a^2}{x}+\dfrac{b^2}{y}\ge\dfrac{\left(a+b\right)^2}{x+y}< =>\left(\dfrac{a^2}{x}+\dfrac{b^2}{y}\right)\left(x+y\right)\ge\left(a+b\right)^2\)
\(< =>a^2+\dfrac{a^2y}{x}+\dfrac{b^2x}{y}+b^2\ge\left(a+b\right)^2\)(luôn đúng với mọi x,y dương)
thật vậy \(a^2+b^2+\dfrac{a^2y}{x}+\dfrac{b^2x}{y}\ge a^2+b^2+2ab=\left(a+b\right)^2\)
dấu"=" xảy ra<=>\(\dfrac{a}{x}=\dfrac{b}{y}\)