Question

cho a , b , c là 3 só thực dương thỏa mãn : a + 2b + 3c = 1 . Tìm max của \(P=\frac{6bc}{\sqrt{a+6bc}}+\frac{3ac}{\sqrt{2b+3ac}}+\frac{2ab}{\sqrt{3c+2ab}}\)

Answer 1

Dặt x=a, y=2b,z=3c

Khi đó

\(P=\frac{yz}{\sqrt{x+yz}}+\frac{xz}{\sqrt{y+xz}}+\frac{xy}{\sqrt{z+xy}}\)và x+y+z=1

Ta có \(\frac{yz}{\sqrt{x+yz}}=\frac{yz}{\sqrt{x\left(x+y+z\right)+yz}}=\frac{yz}{\sqrt{\left(x+y\right)\left(x+z\right)}}\le\frac{1}{2}yz\left(\frac{1}{x+y}+\frac{1}{x+z}\right)\)

=> \(P\le\frac{1}{2}\left(\frac{xz}{x+y}+\frac{yz}{x+y}\right)+\frac{1}{2}\left(\frac{xy}{y+z}+\frac{xz}{y+z}\right)+...=\frac{1}{2}\left(x+y+z\right)\)

                                                                                                                     \(=\frac{1}{2}\)

Vậy \(MaxP=\frac{1}{2}\)khi x=y=z=1/3 hay \(\hept{\begin{cases}a=\frac{1}{3}\\b=\frac{1}{6}\\c=\frac{1}{9}\end{cases}}\)