Đề bài sai hoặc thiếu 1 điều kiện nào đó
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\(a^2+b^2+c^2\le abc\Rightarrow\frac{a^2+b^2+c^2}{abc}\le1\)
\(VT=\frac{a}{a^2+bc}+\frac{b}{b^2+ac}+\frac{c}{c^2+ab}\le\frac{a}{2\sqrt{a^2bc}}+\frac{b}{2\sqrt{b^2ac}}+\frac{c}{2\sqrt{c^2ab}}\)
\(VT\le\frac{1}{2}\left(\frac{1}{\sqrt{ab}}+\frac{1}{\sqrt{bc}}+\frac{1}{\sqrt{ac}}\right)\le\frac{1}{4}\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{b}+\frac{1}{c}+\frac{1}{a}+\frac{1}{c}\right)\)
\(VT\le\frac{1}{2}\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)=\frac{1}{2}\left(\frac{ab+bc+ca}{abc}\right)\le\frac{1}{2}\left(\frac{a^2+b^2+c^2}{abc}\right)\le\frac{1}{2}\)
Dấu "=" xảy ra khi \(a=b=c=3\)
