Ta có :
\(\(a^2+b^2+c^2=3\ge\frac{1}{3}\left(a+b+c\right)^2\Rightarrow a+b+c\le3\)\)
+) \(\(\frac{2a^2}{a+b^2}+\frac{2b^2}{b+c^2}+\frac{2c^2}{c+a^2}=\frac{4a^4}{2a^3+2a^2b^2}+\frac{4b^4}{2b^3+2b^2c^2}+\frac{4c^4}{2c^3+2c^2a^2}\)\)
\(\(\ge\frac{4\left(a^2+b^2+c^2\right)^2}{2a^3+2b^3+2c^3+2a^2b^2+2b^2c^2+2c^2a^2}\)\)
\(\(\ge\frac{4.3^2}{a^4+a^2+b^4+b^2+c^4+c^2+2a^2b^2+2b^2c^2+2c^2a^2}\)\)
\(\(=\frac{36}{\left(a^2+b^2+c^2\right)^2+a^2+b^2+c^2}=\frac{36}{9+3}=3\ge a+b+c\left(dpcm\right)\)\)
_Minh ngụy_
Dễ thấy
\(3=a^2+b^2+c^2\ge\frac{1}{3}\left(a+b+c\right)^2\)
\(\Rightarrow a+b+c\le3\)
Do đó :
\(\frac{2a^2}{a+b^2}+\frac{2b^2}{b+c^2}+\frac{2c^2}{c+a^2}=\frac{4a^4}{2a^3+2a^2b^2}+\frac{4b^4}{2b^3+2b^2c^2}+\frac{4c^4}{2c^3+2c^2a^2}\)
\(\ge\frac{\left(2a^2+2b^2+2c^2\right)^2}{2a^3+2b^3+2c^3+2a^2b^2+2b^2c^2+2c^2a^2}\)
\(\ge\frac{36}{a^4+a^2+b^4+b^2+c^4+c^2+2a^2b^2+2b^2c^2+2c^2a^2}\)
\(=\frac{36}{\left(a^2+b^2+c^2\right)^2+a^2+b^2+c^2}=3\ge a+b+c\left(dpcm\right)\)
