(a+b)^{2}\geq (2 \sqrt{ab})^{2}$\Rightarrow \frac{(a+b)^{2}}{4}\geq 2ab$$VT=2ab+\frac{a+b}{4}=ab+\frac{a}{4}+ab+\frac{b}{4}áp dụng cauchy cho từng bộ 2 số\Rightarrow VT\geq 2\sqrt{ab\times \frac{a}{4}}+2\sqrt{ab\times \frac{b}{4}}hay VT\geq a\sqrt{b}+b\sqrt{a}$(đpcm)
(a+b)^{2}\geq (2 \sqrt{ab})^{2}$\Rightarrow \frac{(a+b)^{2}}{
2}\geq 2ab$$VT
\geq 2ab+\frac{a+b}{4}=ab+\frac{a}{4}+ab+\frac{b}{4}
áp dụng cauchy cho từng bộ 2 số\Rightarrow VT\geq 2\sqrt{ab\times \frac{a}{4}}+2\sqrt{ab\times \frac{b}{4}}
hay VT\geq a\sqrt{b}+b\sqrt{a}$(đpcm)