Áp dụng bất đẳng thức AM - GM và kết hợp với giả thiết x + y + z = 3 ta có:

\(B=\sqrt{\dfrac{xy}{xy+z\left(x+y+z\right)}}+\sqrt{\dfrac{yz}{yz+x\left(x+y+z\right)}}+\sqrt{\dfrac{zx}{zx+y\left(x+y+z\right)}}\)

\(B=\sqrt{\dfrac{xy}{\left(x+z\right)\left(y+z\right)}}+\sqrt{\dfrac{yz}{\left(y+x\right)\left(z+x\right)}}+\sqrt{\dfrac{zx}{\left(z+y\right)\left(z+x\right)}}\le\dfrac{1}{2}\left(\dfrac{x}{x+z}+\dfrac{y}{y+z}+\dfrac{y}{y+x}+\dfrac{z}{z+x}+\dfrac{z}{z+y}+\dfrac{x}{z+x}\right)\)

\(B\le\dfrac{3}{2}\).

Đẳng thức xảy ra khi x = y = z = 1.

Vậy...

Bài làm:

Ta có: \(x+y+z=8\Leftrightarrow\left(x+y+z\right)^2=64\)

\(\Leftrightarrow x^2+y^2+z^2+2\left(xy+yz+zx\right)=64\)

Mà \(\hept{\begin{cases}x^2+y^2\ge2xy\\y^2+z^2\ge2yz\\z^2+x^2\ge2zx\end{cases}}\)\(\Rightarrow2\left(x^2+y^2+z^2\right)\ge2\left(xy+yz+zx\right)\)

\(\Rightarrow x^2+y^2+z^2\ge xy+yz+zx\)

Thay vào ta có: \(64\ge3\left(xy+yz+zx\right)\)

\(\Leftrightarrow xy+yz+zx\le\frac{64}{3}\)

Dấu "=" xảy ra khi: \(x=y=z=\frac{8}{3}\)

Vậy Max(B) = 64/3 khi x = y = z = 8/3

với mọi x, y, z ta có:
(x-y)^2 +(y-z)^2+ (z-x)^2>=0
<=>2x^2 +2y^2 + 2z^2 - 2xy -2yz - 2xz >=0
<=>x^2 + y^2 +z^2 - xy -yz -zx >=0
<=>(x+y+z)^2 >= 3(x+y+z)
<=>[(x+y+z)^2]/3 >= xy+yz+ zx
=>xy +yz + zx <=3
dấu = xảy ra khi x=y=z =1

Khi đó P=1.1+1.1+1.1=3

Quẩy lên các em êii

\(A=\frac{\sqrt{xy}}{z+2\sqrt{xy}}+\frac{\sqrt{yz}}{x+2\sqrt{yz}}+\frac{\sqrt{zx}}{y+2\sqrt{zx}}\)

\(2A=\frac{z+2\sqrt{xy}}{z+2\sqrt{xy}}-\frac{z}{z+2\sqrt{xy}}+\frac{x+2\sqrt{yz}}{x+2\sqrt{yz}}-\frac{x}{x+2\sqrt{yz}}+\frac{y+2\sqrt{zx}}{y+2\sqrt{zx}}-\frac{y}{y+2\sqrt{zx}}\)

\(=3-\left(\frac{x}{x+2\sqrt{yz}}+\frac{y}{y+2\sqrt{zx}}+\frac{z}{z+2\sqrt{xy}}\right)\le3-\left(\frac{x}{x+y+z}+\frac{y}{x+y+z}+\frac{z}{x+y+z}\right)\)

\(=3-\frac{x+y+z}{x+y+z}=3-1=2\)\(\Leftrightarrow\)\(A\le\frac{2}{2}=1\)

Dấu "=" xảy ra \(\Leftrightarrow\)\(x=y=z\)

...

Có: \(x+y+z=3\)

\(\Leftrightarrow x^2+y^2+z^2+2\left(xy+yz+zx\right)=9\)

Vì: \(\left(x-y\right)^2+\left(y-z\right)^2+\left(z-x\right)^2\ge0,\forall x,y,z\)

\(\Leftrightarrow x^2-2xy+y^2+y^2-2yz+z^2+z^2-2zx+x^2\ge0\)

\(\Leftrightarrow2\left(x^2+y^2+z^2\right)\ge2\left(xy+yz+zx\right)\)

\(\Leftrightarrow x^2+y^2+z^2\ge xy+yz+zx\)

\(\Leftrightarrow3\left(xy+yz+zx\right)\le x^2+y^2+z^2+2\left(xy+yz+zx\right)=9\)

\(\Leftrightarrow xy+yz+zx\le3\)

Vậu GTLN của P là 3 khi \(x=y=z=1\)