Question

Cho a,b,c>0, abc=1 Cmr\(\frac{a}{\left(ab+a+1\right)^2}\)+\(\frac{b}{\left(bc+b+1\right)^2}\)+\(\frac{c}{\left(ca+c+1\right)^2}\)\(\ge\frac{1}{a+b+c}\)

Answer 1

Áp dụng BĐT Cauchy-Schwarz ta có: 

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

Mà \(\left(a+b+c\right)\left(a+a^2b+\frac{1}{c}\right)=\left(a+b+c\right)\left(a+a^2b+ab\right)\)

\(\Rightarrow\frac{a}{\left(ab+a+1\right)^2}\ge\frac{a}{\left(a+b+c\right)\left(a+a^2b+ab\right)}=\frac{1}{\left(a+b+c\right)\left(1+ab+b\right)}\)

Tương tự rồi cộng theo vế 3 BĐT ta có:

\(VT\ge\frac{1}{a+b+c}\left(Σ\frac{1}{1+ab+b}\right)=\frac{1}{a+b+c}\left(abc=1\right)\)

Đẳng thức xảy ra khi \(a=b=c=1\)