Ta có : \(\sqrt{2015a+bc}=\sqrt{\left(a+b+c\right)a+bc}=\sqrt{\left(a+b\right)\left(a+c\right)}\)
Áp dụng BĐT Bu-nhi-a-cốp-ski, ta có : \(\left(a+b\right)\left(a+c\right)=\left(\sqrt{a}^2+\sqrt{b}^2\right)\left(\sqrt{a}^2+\sqrt{c}^2\right)\ge\left(\sqrt{ac}+\sqrt{ab}\right)^2\)
\(\Rightarrow\sqrt{\left(a+b\right)\left(a+c\right)}\ge\sqrt{ac}+\sqrt{ab}\)
\(\Rightarrow\frac{a}{a+\sqrt{2015a+bc}}\le\frac{a}{a+\sqrt{ac}+\sqrt{ab}}=\frac{\sqrt{a}^2}{\sqrt{a}\left(\sqrt{a}+\sqrt{b}+\sqrt{c}\right)}=\frac{\sqrt{a}}{\sqrt{a}+\sqrt{b}+\sqrt{c}}\)
\(\Rightarrow\Sigma\frac{a}{a+\sqrt{2015a+bc}}\le\frac{\sqrt{a}+\sqrt{b}+\sqrt{c}}{\sqrt{a}+\sqrt{b}+\sqrt{c}}=1\)
