Question

\(\frac{a^2\left(b+1\right)}{a+b+ab}+\frac{b^2\left(c+1\right)}{b+c+bc}+\frac{c^2\left(a+1\right)}{c+a+ca}\)cho a,b,c > 0 và a+b+c =3 CMR : 

Answer 1

Ta có \(\frac{a^2\left(b+1\right)}{a+b+ab}=\frac{a\left(ab+a+b\right)-ab}{ab+a+b}=a-\frac{ab}{ab+a+b}\)

Mà \(\frac{1}{ab+b+a}\le\frac{1}{9}\left(\frac{1}{ab}+\frac{1}{b}+\frac{1}{a}\right)\)

=> \(\frac{a^2\left(b+1\right)}{a+b+ab}\ge a-\frac{1}{9}ab\left(\frac{1}{ab}+\frac{1}{a}+\frac{1}{b}\right)=\frac{8}{9}a-\frac{1}{9}b-\frac{1}{9}\)

=> \(VT\ge\frac{7}{9}\left(a+b+c\right)-\frac{1}{3}=\frac{7}{3}-\frac{1}{3}=2\)

MinVT=2  khi a=b=c=1