BĐT cần chứng minh tương đương:
\(\sum\dfrac{a}{\sqrt{\left(a+2b\right)\left(a+b+c\right)}}\ge1\)
Đặt vế trái là P, ta có:
\(P=\sum\dfrac{a}{\sqrt{\left(a+2b\right)\left(a+b+c\right)}}\ge\sum\dfrac{2a}{2a+3b+c}=2\sum\dfrac{a^2}{2a^2+3ab+ac}\)
\(P\ge\dfrac{2\left(a+b+c\right)^2}{2a^2+2b^2+2c^2+4ab+4bc+4ca}=1\) (đpcm)