BĐT cần chứng minh tương đương với
\(\left(1-\frac{a^5-a^2}{a^5+b^2+c^2}\right)+\left(1-\frac{b^5-b^2}{b^5+c^2+a^2}\right)+\left(1-\frac{c^5-c^2}{c^5+a^2+b^2}\right)\le3\)
hay \(\frac{1}{a^5+b^2+c^2}+\frac{1}{b^5+c^2+a^2}+\frac{1}{c^5+a^2+b^2}\le\frac{3}{a^2+b^2+c^2}\)
Từ \(abc\ge1\) ta có:
\(\frac{1}{a^5+b^2+c^2}\le\frac{1}{\frac{a^5}{abc}+b^2+c^2}=\frac{1}{\frac{a^4}{bc}+b^2+c^2}\)
\(\le\frac{1}{\frac{2a^4}{b^2+c^2}+b^2+c^2}=\frac{b^2+c^2}{2a^4+\left(b^2+c^2\right)^2}\)
Do \(4u^2+v^2\ge4uv\Leftrightarrow4u^2+v^2\ge\frac{2}{3}\left(u+v\right)^2\)nên
\(2a^4+\left(b^2+c^2\right)^2\ge\frac{2}{3}\left(a^2+b^2+c^2\right)^2\)
Suy ra \(\frac{1}{a^5+b^2+c^2}\le\frac{3\left(b^2+c^2\right)}{2\left(a^2+b^2+c^2\right)^2}\)
Tương tự ta có \(\frac{1}{b^5+c^2+a^2}\le\frac{3\left(c^2+a^2\right)}{2\left(a^2+b^2+c^2\right)^2}\)
và \(\frac{1}{c^5+a^2+b^2}\le\frac{3\left(a^2+b^2\right)}{2\left(a^2+b^2+c^2\right)^2}\)
Cộng ba vế của các BĐT trên ta được
\(\frac{1}{a^5+b^2+c^2}+\frac{1}{b^5+c^2+a^2}+\frac{1}{c^5+a^2+b^2}\le\frac{3}{a^2+b^2+c^2}\)
Vậy \(\frac{a^5-a^2}{a^5+b^2+c^2}+\frac{b^5-b^2}{b^5+c^2+a^2}+\frac{c^5-c^2}{c^5+a^2+b^2}\ge0\)
(Dấu "="\(\Leftrightarrow a=b=c=1\))
