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 \(\hept{\begin{cases}a=\frac{1}{3}\\b=\frac{1}{6}\\c=\frac{1}{9}\end{cases}}\)

Áp dụng Bat đẳng thức C.B.S dạng Angel

Dấu bằng xảy ra khi a=2;b=1;c=1

2ab + 6bc + 2ac = 7abc => \(\frac{2}{c}+\frac{6}{a}+\frac{2}{b}=7\)

đặt \(x=\frac{1}{a};y=\frac{1}{b};z=\frac{1}{c}\) => 6x + 2y + 2z = 7; x; y; z > 0

Khi đó, C = \(\frac{4}{\frac{1}{b}+\frac{2}{a}}+\frac{9}{\frac{1}{c}+\frac{4}{a}}+\frac{4}{\frac{1}{c}+\frac{1}{b}}=\frac{4}{2x+y}+\frac{9}{4x+z}+\frac{4}{y+z}\)

AD BĐT Cauchy ta có:

 \(\left(\frac{4}{2x+y}+\left(2x+y\right)\right)+\left(\frac{9}{4x+z}+\left(4x+z\right)\right)+\left(\frac{4}{y+z}+\left(y+z\right)\right)\)

\(\ge2\sqrt{4}+2.\sqrt{9}+2.\sqrt{4}=14\)

=>  \(\frac{4}{2x+y}+\frac{9}{4x+z}+\frac{4}{y+z}\)+ 7  > 14 =>  C > 7 

Dấu "=" xảy ra <=> a = 2; b = 1; c = 1

Vậy Min C = 7

2ab+6bc+2ac=7abc =>

Đặt => 6x + 2y + 2z = 7; x; y; z > 0

Khi đó C=

  TA CÓ:

 Dấu “=” xảy raóa=2;b=1;c=1

   Vậy c=7

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