Question

cho \(a,b,c>0\)thỏa  mãn:  \(a+b+c=6\)

tìm GTLN: \(S=\sqrt{a^2+4ab+b^2}+\sqrt{b^2+4bc+c^2}+\sqrt{c^2+4ca+a^2}\)

Answer 1

Có \(\sqrt{a^2+4ab+b^2}=\sqrt{\left(\frac{3}{2}a^2+3ab+\frac{3}{2}b^2\right)-\left(\frac{1}{2}a^2-ab+\frac{1}{2}b^2\right)}\)

\(=\sqrt{\frac{3}{2}\left(a+b\right)^2-\frac{1}{2}\left(a-b\right)^2}\le\sqrt{\frac{3}{2}\left(a+b\right)^2}=\sqrt{\frac{3}{2}}\left(a+b\right)\)

Tương tự, ta có : \(\sqrt{b^2+4bc+c^2}\le\sqrt{\frac{3}{2}}\left(b+c\right);\sqrt{c^2+4ca+a^2}\le\sqrt{\frac{3}{2}}\left(c+a\right)\)

\(\Rightarrow\)\(S\le\sqrt{\frac{3}{2}}\left(a+b\right)+\sqrt{\frac{3}{2}}\left(b+c\right)+\sqrt{\frac{3}{2}}\left(c+a\right)=\sqrt{\frac{3}{2}}.2\left(a+b+c\right)=6\sqrt{6}\)

Dấu "=" xảy ra \(\Leftrightarrow\)\(a=b=c=2\)