Đặt \(a=x\sqrt{y}\\ b=y\sqrt{x}\left(a,b>0\right)\)

hpt <=> \(\hept{\begin{cases}2\left(1+a\right)^2=9b\\2\left(1+b\right)^2=9a\end{cases}}\)

lấy 2 cái trừ nhau ta được

\(2\left(a-b\right)\left(a+b+2\right)=-9\left(a-b\right)\)

\(\left(a-b\right)\left(2a+2b+13\right)=0\)

Vì a,b >o

nên a=b

\(\hept{\begin{cases}2\left(1+x\sqrt{y}\right)^2=9y\sqrt{x}\\2\left(1+y\sqrt{x}\right)^2=9x\sqrt{y}\end{cases}\left(I\right)}\)

ĐK: x >=0; y >=0

Đặt \(a=x\sqrt{y};y=b\sqrt{x}\). ĐK a>=0; b>=0. Hệ (I) trở thành \(\hept{\begin{cases}2\left(1+a\right)^2=9b\left(1\right)\\2\left(1+b\right)^2=9a\left(2\right)\end{cases}}\)

Lấy (1) trừ đi (2) ta được: \(2\left(1+a\right)^2-2\left(1+b\right)^2=9\left(b-a\right)\)

<=> \(2\left(a-b\right)\left(a+b+2\right)+9\left(a-b\right)=0\)

<=> \(\left(a-b\right)\left(2a+2b+13\right)=0\)

<=> a=b (vì 2a+2b+13 >0 với mọi a,b>0)

Thay a=b vào (1) ta có:

\(2\left(1+a\right)^2=9a\Leftrightarrow\orbr{\begin{cases}a=2\Rightarrow b=2\left(tm\right)\left(3\right)\\a=\frac{1}{2}\Rightarrow b=\frac{1}{2}\left(tm\right)\left(4\right)\end{cases}}\)

(3) => \(\hept{\begin{cases}x\sqrt{y}=2\\y\sqrt{x}=2\end{cases}\Leftrightarrow x=y=\sqrt[3]{4}}\)

(4) => \(\hept{\begin{cases}x\sqrt{y}=\frac{1}{2}\\y\sqrt{x}=\frac{1}{2}\end{cases}\Leftrightarrow x=y=\sqrt[3]{\frac{1}{4}}}\)

Vậy hệ phương trình có nghiệm \(\left(\sqrt[3]{4};\sqrt[3]{4}\right);\left(\sqrt[3]{\frac{1}{4}};\sqrt[3]{\frac{1}{4}}\right)\)

Áp dụng tính chất : 1/a+b < = 1/4.(1/a+1/b) thì :

x/2x+y+z = x.(1/2x+y+z) = x.[1/(x+y)+(x+z)] < = x/4.(1/x+y + 1/x+z)

Tương tự : ..........

=> x/2x+y+z + y/x+2y+z + z/x+y+2z < = 1/4.(x/x+y + x/x+z + y/y+x + y/y+z + z/z+x + z/x+y )

                                                         = 1/4. [ ( x/x+y + y/x+y ) + ( y/y+z + z/z+y ) + ( z/z+x + x/x+z )

                                                         = 1/4.(1+1+1) = 3/4

Dấu "=" xảy ra <=> x=y=z

Vậy ..........

Tk mk nha

Đặt BT là P:

\(\text{P}=\frac{x}{\left(2x+y+z\right)}-1+\frac{y}{2y+z+x}-1+\frac{z}{\left(2z+x+y\right)}-1+3\)

\(\text{P}=-\frac{\left(x+y+z\right)}{\left(2x+y-z\right)}-\frac{\left(x+y+z\right)}{\left(2y+z+x\right)}-\frac{\left(x+y+z\right)}{\left(2z+x+y\right)}+3\)

\(\text{P}=-\left(x+y+z\right).\left[\frac{1}{\left(2x+y+z\right)}+\frac{1}{\left(2y+z+x\right)}+\frac{1}{\left(2z+x+y\right)}\right]+3\)

Co-si 3 số, ta có:

\(2x+y+z+2y+z+x+2z+x+y\ge3.\sqrt[3]{\left(2x+y+z\right)\left(2y+z+x\right)\left(2z+x+y\right)}\)

\(\Rightarrow4\left(x+y+z\right)\ge3.\sqrt[3]{\left(2x+y+z\right)\left(2y+z+x\right)\left(2z+x+y\right)}\)(1)

Co-si tiếp cho 3 số, ta có:

\(\frac{1}{\left(2x+y+z\right)}+\frac{1}{\left(2y+z+x\right)}+\frac{1}{\left(2z+x+y\right)}\ge3.\sqrt[3]{\frac{1}{\left(2x+y+z\right)}+\frac{1}{\left(2y+z+x\right)}+\frac{1}{\left(2z+x+y\right)}}\)(2)

Lấy (1) và (2) ta có: \(4\left(x+y+z\right)\left[\frac{1}{\left(2x+y+z\right)}+\frac{1}{\left(2y+z+x\right)}+\frac{1}{\left(2z+x+y\right)}\right]\ge9\)

\(\Rightarrow-\left(x+y+z\right).\left[\frac{1}{\left(2x+y+z\right)}+\frac{1}{\left(2y+z+x\right)}+\frac{1}{\left(2z+x+y\right)}\right]\le-\frac{9}{4}\)

Thay P, ta có:

\(\text{P}\le-\frac{9}{3}+3=\frac{3}{4}\left(ĐPCM\right)\)

Dấu "=" xảy ra khi x = y = z.

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