Question

cho a, b, c > 0 và \(a^2+b^2+c^2=3\) C/m \(\frac{a}{a^2+2b+3}+\frac{b}{b^2+2c+3}+\frac{c}{c^2+2a+3}\le\frac{1}{2}\)

Answer 1

\(\sum\frac{a}{a^2+1+2b+2}\le\sum\frac{a}{2a+2b+2}\)

Nên ta chỉ cần chứng minh: \(\sum\frac{a}{2a+2b+2}\le\frac{1}{2}\Leftrightarrow\sum\frac{a}{a+b+1}\le1\)

\(\Leftrightarrow\sum\frac{b+1}{a+b+1}\ge2\)

Đặt \(P=\sum\frac{b+1}{a+b+1}=\sum\frac{\left(b+1\right)^2}{\left(b+1\right)\left(a+b+1\right)}=\sum\frac{\left(b+1\right)^2}{ab+a+b^2+2b+1}\)

\(P\ge\frac{\left(a+b+c+3\right)^2}{ab+a+b^2+2b+1+bc+b+c^2+2c+1+ca+c+a^2+2a+1}\)

\(P\ge\frac{a^2+b^2+c^2+2\left(ab+bc+ca\right)+6\left(a+b+c\right)+9}{a^2+b^2+c^2+ab+bc+ca+3\left(a+b+c\right)+3}\)

\(P\ge\frac{2\left(ab+bc+ca\right)+6\left(a+b+c\right)+12}{ab+bc+ca+3\left(a+b+c\right)+6}=2\) (đpcm)

Dấu "=" xảy rakhi \(a=b=c=1\)