Question

cho a,b,c là các số dương và a+b+c = 6

tìm Max của \(A=\sqrt{a^2+4ab+b^2}+\sqrt{b^2+4bc+c^2}+\sqrt{c^2+4ac+a^2}\)

Answer 1

Vì \(a;b;c>0\Rightarrow2ab\le\frac{\left(a+b\right)^2}{2}\) thay vào \(\sqrt{a^2+4ab+b^2}\)ta có:

\(\sqrt{a^2+4ab+b^2}=\sqrt{\left(a+b\right)^2+2ab}\)

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

Tương tự: \(\sqrt{b^2+4bc+c^2}\le\sqrt{\frac{3}{2}}.\left(b+c\right)\)

\(\sqrt{c^2+4ca+a^2}\le\sqrt{\frac{3}{2}}.\left(c+a\right)\)

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

        \(\le\sqrt{\frac{3}{2}}.\left(2a+2b+2c\right)=\sqrt{\frac{3}{2}}.6=\sqrt{216}=6\sqrt{6}\)Vì a+b+c=6

Dấu = xảy ra khi a=b=c=2

Vây ......