Ta có: $b + c = (b + c).(a + b + c)^2$ (vì $a + b + c = 1$)
Ta có $[ a + (b + c) ]^2 \geq 4(b+c)a$ (vì $(x + y)^2 \geq 4xy )$
$\Leftrightarrow (b + c).(a + b + c)^2\geq 4(b+c)^2.a$
lại có $(b+c)^2\geq 4bc \Rightarrow 4(b+c)^2.a \geq 16abc$ (đpcm)
b+c=(b+c).(a+b+c)2