Cho các số thực dương a,b,c thỏa mãn a+b+c=3, tìm GTLN của P= $\sqrt{a+b}$ + $\sqrt{b+c}$ $\sqrt{c+a}$

Theo Cauchy có:

\(\sqrt{\dfrac{1}{2}}.\sqrt{(a+b).2}\le \sqrt{\dfrac{1}{2}}.\dfrac{a+b+2}{2}\)

Tương tự: 

\(\sqrt{\dfrac{1}{2}}.\sqrt{(b+c).2}\le \sqrt{\dfrac{1}{2}}.\dfrac{b+c+2}{2}\\\sqrt{\dfrac{1}{2}}.\sqrt{(c+a).2}\le \sqrt{\dfrac{1}{2}}.\dfrac{a+c+2}{2}\\\to P\le \sqrt{\dfrac{1}{2}}.\dfrac{2(a+b+c)+6}{2}\\\le \sqrt{\dfrac{1}{2}}.6\\\le 3\sqrt{2}\)

Dấu "=" xảy ra khi: \(a=b=c=1\)