Question

[Phan Quang Thái]

cho a,b, c\(\ge\)0; a+b+c=1. Chứng minh rằng\(\sqrt{a+b}+\sqrt{b+c}+\sqrt{c+a}\le\sqrt{6}\)

Answer 1

[Hoàng Lê Bảo Ngọc]

Áp dụng BĐT Bunhiacopxki, ta có : 

\(\left(1.\sqrt{a+b}+1.\sqrt{b+c}+1.\sqrt{c+a}\right)^2\le\left(1^2+1^2+1^2\right)\left[\left(a+b\right)+\left(b+c\right)+\left(c+a\right)\right]\)

\(\Rightarrow\left(\sqrt{a+b}+\sqrt{b+c}+\sqrt{c+a}\right)^2\le6\left(a+b+c\right)\)

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