Question

Cho a+b+c = 3. Tìm MAX của :

M= \(\sqrt{2a+5\sqrt{ab}+2b}+\sqrt{2b+5\sqrt{bc}+2c}+\sqrt{2c+5\sqrt{ac}+2a}\)

Answer 1

Đề thiếu nhé, a,b,c >0

Áp dụng BĐT Bunhiacopxki, ta có:

\(M^2=\left(\sqrt{2a+5\sqrt{ab}+2b}+\sqrt{2b+5\sqrt{bc}+2c}+\sqrt{2c+5\sqrt{ca}+2a}\right)^2\)

\(\le3\left[4\left(a+b+c\right)+5\left(\sqrt{ab}+\sqrt{bc}+\sqrt{ca}\right)\right]\)

\(\le3\left[4\left(a+b+c\right)+5\left(a+b+c\right)\right]=81\)

\(\Rightarrow M\le9\)

\(MaxM=9\Leftrightarrow a=b=c=1\)

(\(\sqrt{ab}+\sqrt{bc}+\sqrt{ca}\le\sqrt{\left(a+b+c\right)\left(a+b+c\right)}=a+b+c\left(Bunhiacopxki\right)\))