Đặt vế trái của BĐT cần chứng minh là P
\(P=\sqrt{ab+bc+ca+a^2}+\sqrt{ab+bc+ca+b^2}+\sqrt{ab+bc+ca+c^2}\)
\(P=\sqrt{\left(a+b\right)\left(a+c\right)}+\sqrt{\left(a+b\right)\left(b+c\right)}+\sqrt{\left(a+c\right)\left(b+c\right)}\)
\(P\le\dfrac{1}{2}\left(a+b+a+c\right)+\dfrac{1}{2}\left(a+b+b+c\right)+\dfrac{1}{2}\left(a+c+b+c\right)=2\left(a+b+c\right)\)
Dấu "=" xảy ra khi \(a=b=c=\dfrac{2015}{3}\)