Question

cho a,b,c>0 và a+b+c=1. Cmr:

\(\frac{a^2}{a^2+\left(b+c\right)^2}+\frac{b^2}{b^2+\left(a+c\right)^2}+\frac{c^2}{c^2+\left(a+b\right)^2}\ge\frac{3}{5}\)

Answer 1

Ta có \(\left(b+c\right)^2\le2\left(b^2+c^2\right)\)

=> \(\frac{a^2}{a^2+\left(b+c\right)^2}\ge\frac{a^2}{a^2+2b^2+c^2}\)

=> \(VT\ge\Sigma\frac{a^4}{a^4+2b^2a^2+2a^2c^2}\ge\frac{\left(a^2+b^2+c^2\right)^2}{a^4+b^4+c^4+4\left(a^2b^2+b^2c^2+c^2a^2\right)}\)

=> \(VT\ge\frac{\left(a^2+b^2+c^2\right)^2}{\left(a^2+b^2+c^2\right)^2+2\left(a^2b^2+b^2c^2+a^2c^2\right)}\ge\frac{\left(a^2+b^2+c^2\right)^2}{\left(a^2+b^2+c^2\right)^2+\frac{2}{3}\left(a^2+b^2+c^2\right)^2}=\frac{3}{5}\)(ĐPCM)

Dấu bằng xảy ra khi a=b=c