Ta có:
\(\dfrac{1}{a^2}+\dfrac{1}{b^2}+\dfrac{1}{c^2}\ge\dfrac{1}{ab}+\dfrac{1}{bc}+\dfrac{1}{ca}=\dfrac{a+b+c}{abc}\) (1)
Đồng thời: \(3abc\left(a+b+c\right)\le\left(ab+bc+ca\right)^2\)
\(\Rightarrow abc\le\dfrac{3\left(ab+bc+ca\right)^2}{a+b+c}\) (2)
(1);(2) \(\Rightarrow\dfrac{1}{a^2}+\dfrac{1}{b^2}+\dfrac{1}{c^2}\ge\dfrac{3\left(a+b+c\right)^2}{\left(ab+bc+ca\right)^2}\) (3)
Lại có:
\(\left(a+b+c\right)^2=\left(a^2+b^2+c^2\right)+\left(ab+bc+ca\right)+\left(ab+bc+ca\right)\)
\(\Rightarrow\left(a+b+c\right)^2\ge3\sqrt[3]{\left(a^2+b^2+c^2\right)\left(ab+bc+ca\right)^2}\)
\(\Rightarrow\left(a+b+c\right)^6\ge27\left(a^2+b^2+c^2\right)\left(ab+bc+ca\right)^2\)
\(\Rightarrow\left(a+b+c\right)^4\left(a+b+c\right)^2\ge27\left(a^2+b^2+c^2\right)\left(ab+bc+ca\right)^2\)
\(\Rightarrow81\left(a+b+c\right)^2\ge27\left(a^2+b^2+c^2\right)\left(ab+bc+ca\right)^2\)
\(\Rightarrow3\left(a+b+c\right)^2\ge\left(a^2+b^2+c^2\right)\left(ab+bc+ca\right)^2\)
\(\Rightarrow\dfrac{3\left(a+b+c\right)^2}{\left(ab+bc+ca\right)^2}\ge a^2+b^2+c^2\) (4)
(3);(4) \(\Rightarrow\dfrac{1}{a^2}+\dfrac{1}{b^2}+\dfrac{1}{c^2}\ge a^2+b^2+c^2\)
Dấu "=" xảy ra khi \(a=b=c=1\)