Áp dụng BĐT Holder:
\(\left(\dfrac{a^2}{b}+\dfrac{b^2}{c}+\dfrac{c^2}{a}\right)\left(\dfrac{a^2}{b}+\dfrac{b^2}{c}+\dfrac{c^2}{a}\right)\left(a^2b^2+b^2c^2+c^2a^2\right)\ge\left(a^2+b^2+c^2\right)^3\)
\(\Rightarrow\left(\dfrac{a^2}{b}+\dfrac{b^2}{c}+\dfrac{c^2}{a}\right)^2\ge\dfrac{\left(a^2+b^2+c^2\right)^3}{a^2b^2+b^2c^2+c^2a^2}\)
Mà:
\(\left(a^2+b^2+c^2\right)^2\ge3\left(a^2b^2+b^2c^2+c^2a^2\right)\)
\(\Rightarrow\left(\dfrac{a^2}{b}+\dfrac{b^2}{c}+\dfrac{c^2}{a}\right)^2\ge3\left(a^2+b^2+c^2\right)\)
\(\Rightarrow\dfrac{a^2}{b}+\dfrac{b^2}{c}+\dfrac{c^2}{a}\ge\sqrt{3\left(a^2+b^2+c^2\right)}\) (1)
Áp dụng BĐT Bunhiacopxki:
\(\sqrt{\dfrac{a^2+b^2}{2}}+\sqrt{\dfrac{b^2+c^2}{2}}+\sqrt{\dfrac{c^2+a^2}{2}}\le\sqrt{3\left(\dfrac{a^2+b^2}{2}+\dfrac{b^2+c^2}{2}+\dfrac{c^2+a^2}{2}\right)}\)
\(\Rightarrow\sqrt{\dfrac{a^2+b^2}{2}}+\sqrt{\dfrac{b^2+c^2}{2}}+\sqrt{\dfrac{c^2+a^2}{2}}\le\sqrt{3\left(a^2+b^2+c^2\right)}\) (2)
(1);(2) suy ra điều phải chứng minh
Dấu "=" xảy ra khi \(a=b=c\)
