Question

Cho a,b,c > 0 thỏa mãn a + b + c = 1.CMR: 

\(A=\sqrt{a+b}+\sqrt{b+c}+\sqrt{c+a}\le\sqrt{6}\)

Answer 1

Easy!

\(A=\sqrt{a+b}+\sqrt{b+c}+\sqrt{c+a}\)

\(=\sqrt{\frac{3}{2}}\left[\sqrt{\left(a+b\right).\frac{2}{3}}+\sqrt{\left(b+c\right).\frac{2}{3}}+\sqrt{\left(c+a\right).\frac{2}{3}}\right]\) (*)

Áp dụng BĐT Cô si ngược,ta có: 

(*) \(\le\sqrt{\frac{3}{2}}\left[\frac{a+b+\frac{2}{3}}{2}+\frac{b+c+\frac{2}{3}}{2}+\frac{c+a+\frac{2}{3}}{2}\right]\)

\(=\sqrt{\frac{3}{2}}\left(a+b+c+1\right)=\sqrt{\frac{3}{2}}.2=\sqrt{6}^{\left(đpcm\right)}\)

Dấu "=" xảy ra khi \(\hept{\begin{cases}a+b=b+c=c+a=\frac{2}{3}\\a+b+c=1\end{cases}\Leftrightarrow}a=b=c=\frac{1}{3}\)