Áp dụng bất đẳng thức Schwarz và AM - GM ta có:
\(VT=\dfrac{a^2}{ab}+\dfrac{b^2}{bc}+\dfrac{c^2}{ca}+\dfrac{a+b+c}{\sqrt{3\left(a^2+b^2+c^2\right)}}\)
\(\ge\dfrac{\left(a+b+c\right)^2}{ab+bc+ca}+\dfrac{3\left(a+b+c\right)}{\sqrt{3\left(a^2+b^2+c^2\right)}}-\dfrac{2\left(a+b+c\right)}{\sqrt{3\left(a^2+b^2+c^2\right)}}\)
\(\ge2\sqrt{\dfrac{3\left(a+b+c\right)^3}{\left(ab+bc+ca\right)\sqrt{3\left(a^2+b^2+c^2\right)}}}-\dfrac{2\left(a+b+c\right)}{a+b+c}\)
\(=2\sqrt[4]{\dfrac{3\left(a+b+c\right)^6}{\left(a^2+b^2+c^2\right)\left(ab+bc+ca\right)^2}}-2\)
\(\ge2\sqrt[4]{\dfrac{3\left(a+b+c\right)^6}{\dfrac{\left(ab+bc+ca+ab+bc+ca+a^2+b^2+c^2\right)^3}{27}}}-2\)
\(=6-2=4=VP\left(đpcm\right)\).
Đặt vế trái của biểu thức là P
\(P=\dfrac{a^2}{ab}+\dfrac{b^2}{bc}+\dfrac{c^2}{ca}+\dfrac{a+b+c}{\sqrt{3\left(a^2+b^2+c^2\right)}}\)
\(P\ge\dfrac{\left(a+b+c\right)^2}{ab+bc+ca}+\dfrac{a+b+c}{\sqrt{3\left(a^2+b^2+c^2\right)}}\)
\(P\ge\dfrac{2\left(a+b+c\right)^2}{3\left(ab+bc+ca\right)}+\dfrac{\left(a+b+c\right)^2}{6\left(ab+bc+ca\right)}+\dfrac{\left(a+b+c\right)^2}{6\left(ab+bc+ca\right)}+\dfrac{a+b+c}{\sqrt{12\left(a^2+b^2+c^2\right)}}+\dfrac{a+b+c}{\sqrt{12\left(a^2+b^2+c^2\right)}}\)
\(P\ge\dfrac{2\left(a+b+c\right)^2}{\left(a+b+c\right)^2}+4\sqrt[4]{\dfrac{\left(a+b+c\right)^6}{432\left(ab+bc+ca\right)\left(ab+bc+ca\right)\left(a^2+b^2+c^2\right)}}\)
\(P\ge2+4\sqrt[4]{\dfrac{\left(a+b+c\right)^6}{432\left(\dfrac{2ab+2bc+2ca+a^2+b^2+c^2}{3}\right)^3}}\)
\(P\ge2+4\sqrt[4]{\dfrac{\left(a+b+c\right)^6}{16\left(a+b+c\right)^6}}=4\)
Dấu "=" xảy ra khi \(a=b=c\)
