Question

\(cho\)\(a,b,c>0\)\(CM\)

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

Answer 1

Áp dụng BĐT Bu-nhi-a-cốp-ski, ta có: 

\(\left(a+b+c\right)\left[\frac{a}{\left(ac+a+1\right)^2}+\frac{b}{\left(bc+b+1\right)^2}+\frac{c}{\left(ca+c+1\right)^2}\right]\)

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

Lại có: \(\frac{a}{ac+a+1}+\frac{b}{bc+b+1}+\frac{c}{ca+c+1}\)

\(=\frac{a}{ac+a+abc}+\frac{b}{bc+b+1}+\frac{bc}{abc+bc+b}\)                             ( Do abc=1 )

\(=\frac{1}{bc+b+1}+\frac{b}{bc+b+1}+\frac{bc}{bc+b+1}\)

\(=1\)                                                                                              \(\left(2\right)\)

Từ (1) và (2) suy ra \(\left(a+b+c\right)\left[\frac{a}{\left(ac+a+1\right)^2}+\frac{b}{\left(bc+b+1\right)^2}+\frac{c}{\left(ca+c+1\right)^2}\right]\ge1\)

Mà \(a;b;c>0\Rightarrow a+b+c>0\)

\(\Rightarrow\frac{a}{\left(ac+a+1\right)^2}+\frac{b}{\left(bc+b+1\right)^2}+\frac{c}{\left(ca+c+1\right)^2}\ge\frac{1}{a+b+c}\)                (đpcm)