Question

 Cho \(a,b>0\); \(a^2+b^2\le16\).

Tìm max của \(N=a\sqrt{9b\left(a+8b\right)}+b\sqrt{9a\left(b+8a\right)}\)

Answer 1

\(ab\le\frac{a^2+b^2}{2}\le\frac{16}{2}=8\)

Ta có: \(N^2=\left(a\sqrt{9b\left(a+8b\right)}+b\sqrt{9a\left(b+8a\right)}\right)^2\)

\(\le\left(a^2+b^2\right)\left[9b\left(a+8b\right)+9a\left(b+8a\right)\right]\)

\(\le16\left(18ab+72\left(a^2+b^2\right)\right)\le16\left(18.8+72.16\right)\)

\(=20736\)

=> \(N\le144\)

Dấu "=" xảy ra <=> a = b = \(\sqrt{8}\)

Vậy max N = 144 tại a = b = \(\sqrt{8}\)