Có: $\frac{a}{b+c}$ + $\frac{b}{c+a}$ = $\frac{a+b+c}{b+c}$ + $\frac{a+b+c}{c+a}$ -2
= ( a+b+c). ($\frac{1}{b+c}$ + $\frac{1}{c+a}$ ) -2
≥ ( a+b+c). $\frac{4}{a+b+2c}$ -2 ( bđt $\frac{1}{x}$ +$\frac{1}{y}$ ≥ $\frac{4}{x+y}$
+, $\sqrt[]{\frac{2c}{a+b}}$ = $\sqrt[]{\frac{2c.2c}{(a+b).2c}}$ = $\frac{2c}{\sqrt[]{(a+b).2c}}$ ≥ $\frac{4c}{a+b+2c}$ ( cosi)
⇒ $\frac{a}{b+c}$ + $\frac{b}{c+a}$ + $\sqrt[]{\frac{2c}{a+b}}$ ≥ ( a+b+c). $\frac{4}{a+b+2c}$ + $\frac{4c}{a+b+2c}$ -2
= $\frac{4}{a+b+2c}$. ( a+b+2c) -2
= 4-2 =2
Dấu "=" xra ⇔ a=b=c >0
