Ta có:
$1+b+1+c\ge2\sqrt{(1+b)(1+c)}$
$\Leftrightarrow 4+2(b+c)\ge1+b+2\sqrt{(1+b)(1+c)}+1+c$
$\Leftrightarrow 4(1+\dfrac{b+c}{2})\ge(\sqrt{1+b}+\sqrt{1+c})^2$
$\Leftrightarrow 2\sqrt{1+\dfrac{b+c}{2}}\ge\sqrt{1+b}+\sqrt{1+c}$
$\Rightarrow 2\sqrt{1+\dfrac{b+c}{2}}\ge2\sqrt{1+a}$
$\Leftrightarrow b+c\ge2a$