Question

Cho a,b,c là các số thực dương thỏa \(a+b+c=1\). Tìm GTLN của :

\(P=\frac{ab}{c+1}+\frac{bc}{a+1}+\frac{ca}{b+1}\)

Answer 1

Áp dụng Bđt \(\frac{4}{x+y}\le\frac{1}{x}+\frac{1}{y}\) ta có:

\(\frac{ab}{c+1}=\frac{ab}{\left(a+c\right)+\left(b+c\right)}\le\frac{1}{4}\left(\frac{ab}{a+c}+\frac{ab}{b+c}\right)\)

Tương tự: 

\(\frac{bc}{a+1}\le\frac{1}{4}\left(\frac{bc}{b+a}+\frac{bc}{c+a}\right)\)\(;\)\(\frac{ac}{b+1}\le\frac{1}{4}\left(\frac{ac}{a+b}+\frac{ac}{c+b}\right)\)

Cộng theo vế ta được:

\(P\le\frac{1}{4}\left[\left(\frac{ab}{b+c}+\frac{ac}{c+b}\right)+\left(\frac{ab}{a+c}+\frac{bc}{c+a}\right)+\left(\frac{bc}{b+a}+\frac{ac}{a+b}\right)\right]\)

\(=\frac{1}{4}\cdot\left(a+b+c\right)=\frac{1}{4}\)

Dấu = khi \(a=b=c=\frac{1}{3}\)