Question

Cho a,b không âm và a²+b²=2. Tìm GTLN: M = \(a\sqrt{9b\left(4a+5b\right)}+b\sqrt{9a\left(4b+5a\right)}\)

Answer 1

Áp dụng BĐT Cô-si cho 2 số không âm

Ta có: \(\sqrt{9b\left(4a+b\right)}\)\(\le\) \(\dfrac{9b+4a+5b}{2}\)=\(\dfrac{14b+4a}{2}\)

\(\Rightarrow\) \(a\sqrt{9b\left(4a+5b\right)}\)\(\le\) \(\dfrac{14ab+4a^2}{2}\)=7ab+2a²

^{CMTT: \(b\sqrt{9a\left(4b+5a\right)}\)} \(\le\) 7ab+2b²

^{\(\Rightarrow\)} M\(\le\) 14ab + 2(a²+b²) \(\le\)7(a²+b²) + 2(a²+b²) = 9(a²+b²)=18

Vậy M_min=18

Dấu "=" xảy ra\(\Leftrightarrow\) a=b=1

Answer 2

\(M=a\sqrt{9b\left(4a+5b\right)}+b\sqrt{9a\left(4b+5a\right)}\le\dfrac{a\left(9b+4a+5b\right)}{2}+\dfrac{b\left(9a+4b+5a\right)}{2}=\dfrac{a\left(14b+4a\right)+b\left(14a+4b\right)}{2}=2a^2+7ab+7ab+2b^2=2\left(a^2+b^2\right)+14ab=4+14ab\le4+14\times\dfrac{a^2+b^2}{2}=4+14=18\)

Dấu "=" xảy ra <=> a = b = 1