Question

Cho a,b > 0; \(a^2+b^2\le16\). Tìm giá trị lớn nhất của biểu thức:

\(M=a\sqrt{9b\left(a+8b\right)}+b\sqrt{9a\left(b+8a\right)}\)

Answer 1

By C-S and AM-GM's inequality

\(M=a\left(9b\left(a+8b\right)\right)^{\dfrac{1}{2}}+b\left(9a\left(b+8a\right)\right)^{\dfrac{1}{2}}\)

\(\le\left(\left(a^2+b^2\right)\left(9b\left(a+8b\right)+9a\left(b+8a\right)\right)\right)^{\dfrac{1}{2}}\)

\(=\left(\left(a^2+b^2\right)\left(18ab+72b^2+72a^2\right)\right)^{\dfrac{1}{2}}\)

\(=\left(\left(a^2+b^2\right)\left(18\cdot\dfrac{a^2+b^2}{2}+72b^2+72a^2\right)\right)^{\dfrac{1}{2}}\)

\(=\left(16\cdot\left(18\cdot\dfrac{16}{2}+72\cdot16\right)\right)^{\dfrac{1}{2}}=144\)

\("="\Leftrightarrow a=b=2\sqrt{2}\)