Áp dụng bất đẳng thức Bunhiacopxki ta có:
\(\left[\left(\sqrt{x}\right)^2+\left(\sqrt{y}\right)^2\right]\left[\left(\frac{a}{\sqrt{x}}\right)^2+\left(\frac{b}{\sqrt{y}}\right)^2\right]\ge\left(\frac{a}{\sqrt{x}}\cdot\sqrt{x}+\frac{b}{\sqrt{y}}\cdot\sqrt{y}\right)^2=\left(a+b\right)^2\)
\(\Rightarrow\frac{a^2}{x}+\frac{b^2}{y}\ge\frac{\left(a+b\right)^2}{x+y}\)
Dấu = khi \(\frac{a}{x}=\frac{b}{y}\)
Áp dụng ta có:
\(\frac{4}{a}+\frac{9}{b}\ge\frac{\left(2+3\right)^2}{a+b}=25\)
Dấu = khi a = 2/5, b = 3/5