Đặt \(x=\sqrt{a^2+b^2+c^2}\)
Có: \(x=\sqrt{a^2+b^2+c^2}\ge\sqrt{\frac{1}{3}\left(a+b+c\right)^2}=\sqrt{3}\)
\(x=\sqrt{a^2+b^2+c^2}=\sqrt{\left(a+b+c\right)^2-2\left(ab+bc+ca\right)}\le\sqrt{\left(a+b+c\right)^2}=3\)
\(\Rightarrow\sqrt{3}\le x\le3\)
Khi đó, có: \(P=\sqrt{a^2+b^2+c^2}+\frac{1}{a^2+b^2+c^2}=x+\frac{1}{x^2}\)
Ta chứng minh \(P=x+\frac{1}{x^2}\le\frac{28}{9}\)
BĐT \(\Leftrightarrow9x^3-28x^2+9\le0\)
\(\Leftrightarrow\left(x-3\right)\left(9x^2-x-3\right)\le0\)(Luôn đúng vì \(\sqrt{3}\le x\le3\))
Vậy \(maxP=\frac{28}{9}\Leftrightarrow x=3\Leftrightarrow\left(a,b,c\right)\in\left\{\left(0;0;3\right)\right\}\)và các hoán vị.
