Áp dụng BĐT Bunhia cho 4 số [tex]\frac{x}{\sqrt{a}},\frac{y}{\sqrt{b}},\sqrt{a},\sqrt{b}[/tex], Ta có:
[tex](\frac{x^2}{a}+\frac{y^2}{b})(a+b)\geq (x+y)^2[/tex]
[tex]\Leftrightarrow \frac{x^2}{a}+\frac{y^2}{b}\geq \frac{(x+y)^2}{a+b}[/tex] (đpcm)
dấu "=" xảy ra [tex]\Leftrightarrow \frac{x}{a}=\frac{y}{b}[/tex]