P= $\frac{a}{b+c}$ + $\frac{b}{a+2c}$ +$\frac{c}{a+2b}$ = $\frac{a}{b+c}$ + $\frac{b^2}{ab+2bc}$ +$\frac{c^2}{ac+2bc}$
≥ $\frac{a}{b+c}$ + $\frac{(b+c)²}{a(b+c) + 4bc}$
≥ $\frac{a}{b+c}$ + $\frac{(b+c)²}{a(b+c) + (b+c)²}$ (bđt 4xy≤ (x+y)² )
= $\frac{a}{b+c}$ + $\frac{1}{\frac{a}{b+c}+1}$
Do a≥ b+c >0 ∀ a,b,c >0 ⇒$\frac{a}{b+c}$ ≥ 1 Đặt $\frac{a}{b+c}$ =t ( t≥ 1)
⇒ P ≥ t + $\frac{1}{t+1}$ = $\frac{t+1}{4}$ + $\frac{1}{t+1}$ + $\frac{3t}{4}$ - $\frac{1}{4}$
≥ 2$\sqrt[]{\frac{t+1}{4}.\frac{1}{t+1}}$ + $\frac{3t}{4}$ - $\frac{1}{4}$ (cosi)
≥ 1 + $\frac{3}{4}$ -$\frac{1}{4}$ ( do t≥1)
= $\frac{3}{2}$
Dấu "=" xảy ra ⇔$\frac{a}{2}$ =b =c >0
