$(a+b)^{2}\geq (2\sqrt{ab})^{2}$$\Rightarrow \frac{(a+b)^{2}}{4}\geq 2ab$$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)
$(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}$$\Rightarrow \frac{(a+b)^{2}}{4}\geq 2ab$$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)