It's so great!
\(\frac{a^2}{b}+b+2b=\frac{a^2+b^2}{b}+2b\ge2\sqrt{2\left(a^2+b^2\right)}\)
\(\Rightarrow\frac{a^2}{b}\ge2\sqrt{2\left(a^2+b^2\right)}-3b\)
Tương tự hai BĐT còn lại và cộng theo vế thu được:
\(LHS=\frac{a^2}{b}+\frac{b^2}{c}+\frac{c^2}{a}\ge2\sqrt{2}\left(\sqrt{a^2+b^2}+\sqrt{b^2+c^2}+\sqrt{c^2+a^2}\right)-3\left(a+b+c\right)\)
\(=\frac{1}{\sqrt{2}}\left(\sqrt{a^2+b^2}+\sqrt{b^2+c^2}+\sqrt{c^2+a^2}\right)+\frac{3\sqrt{2}}{2}\left(\sqrt{a^2+b^2}+...\right)-3\left(a+b+c\right)\)
\(\ge\frac{1}{\sqrt{2}}\left(\sqrt{a^2+b^2}+\sqrt{b^2+c^2}+\sqrt{c^2+a^2}\right)=RHS\) (sử dụng Mincopxki)
Ta có đpcm.
P/s: Is that true?
