Cái đề thế này ah

\(\frac{xyz}{\left(x+y\right)\left(y+z\right)\left(z+x\right)}\)

Vì \(\hept{\begin{cases}x\ge0\\y\ge0\\z\ge0\end{cases}}\)

\(\Rightarrow\frac{xyz}{\left(x+y\right)\left(y+z\right)\left(z+x\right)}\ge0\)

-_- Làm như thế để chết nhắm :v
Dấu = xảy ra x=y=z=0 => Hỏng .
@Aliba...

\(\frac{xyz}{\left(x+y\right)\left(y+z\right)\left(z+x\right)}\)

áp dụng BĐT cô-si ta có :

\(x+y\ge2\sqrt{xy}\)

\(y+z\ge2\sqrt{yz}\)

\(z+x\ge2\sqrt{zx}\)

nhân vế với vế ta có 

\(\left(x+y\right)\left(y+z\right)\left(z+x\right)\ge8\sqrt{x^2y^2z^2}\)

\(\Leftrightarrow\left(x+y\right)\left(y+z\right)\left(z+x\right)\ge8xyz\)

\(\Leftrightarrow\frac{xyz}{\left(x+y\right)\left(y+z\right)\left(z+x\right)}\ge\frac{xyz}{8xyz}=\frac{1}{8}\)

vậy GTNN là \(\frac{1}{8}\) khi và chỉ khi \(x=y=z=1\)

:)

\(\frac{3}{2}\)

\(P=x^2+y^2+z^2+\dfrac{20}{x+y+z}\ge\dfrac{\left(x+y+z\right)^2}{3}+\dfrac{20}{x+y+z}\)

\(\Leftrightarrow P\ge\dfrac{\left(x+y+z\right)^2}{3}+\dfrac{9}{x+y+z}+\dfrac{9}{x+y+z}+\dfrac{2}{x+y+z}\)

\(\Leftrightarrow P\ge3\sqrt[3]{\dfrac{\left(x+y+z\right)^2}{3}.\dfrac{9}{x+y+z}.\dfrac{9}{x+y+z}}+\dfrac{2}{3}\)

 (theo AM-GM và do \(x+y+z\le3\Rightarrow\dfrac{2}{x+y+z}\ge\dfrac{2}{3}\))

\(\Leftrightarrow P\ge\dfrac{29}{3}\)

Dấu = xảy ra khi x=y=z=1

Vậy minP\(=\dfrac{29}{3}\)

kẻ lười biếng nạp card, đi ô tô

\(P^2=\dfrac{x^2}{y}+\dfrac{y^2}{z}+\dfrac{z^2}{x}+\dfrac{2xy}{\sqrt{yz}}+\dfrac{2yz}{\sqrt{zx}}+\dfrac{2zx}{\sqrt{xy}}\)

\(P^2=\left(\dfrac{x^2}{y}+\dfrac{xy}{\sqrt{yz}}+\dfrac{xy}{\sqrt{yz}}+z\right)+\left(\dfrac{y^2}{z}+\dfrac{yz}{\sqrt{zx}}+\dfrac{yz}{\sqrt{zx}}+x\right)+\left(\dfrac{z^2}{x}+\dfrac{zx}{\sqrt{xy}}+\dfrac{zx}{\sqrt{xy}}+y\right)-\left(x+y+z\right)\)

\(P^2\ge4\sqrt[4]{\dfrac{x^4y^2z}{y^2z}}+4\sqrt[4]{\dfrac{y^4z^2x}{z^2x}}+4\sqrt[4]{\dfrac{z^4x^2y}{x^2y}}-\left(x+y+z\right)=3\left(x+y+z\right)\ge36\)

\(\Rightarrow P\ge6\)

\(P_{min}=6\) khi \(x=y=z=4\)