khó quá!!!!!!!!!!!

Áp dụng bđt AM-GM ta có:

\(\frac{x^2}{x+y}+\frac{x+y}{4}\ge2\sqrt{\frac{x^2}{x+y}.\frac{x+y}{4}}=x\)

\(\frac{y^2}{x+z}+\frac{x+z}{4}\ge2\sqrt{\frac{y^2}{x+z}.\frac{x+z}{4}}\ge y\)

\(\frac{z^2}{x+y}+\frac{x+y}{4}\ge2\sqrt{\frac{z^2}{x+y}.\frac{x+y}{4}}\ge z\)

Cộng từng vế các bđt trên ta được:

\(P+\frac{x+y+z}{2}\ge x+y+z\)

\(\Rightarrow P\ge\frac{x+y+z}{2}=1\)

Dấu"="xảy ra \(\Leftrightarrow x=y=z=1\)

Vậy Min P=1 \(\Leftrightarrow x=y=z=1\)

anh Châu ơi, 1+1+1 đâu có = 2 anh.

à anh xl nhầm x=y=z=\(\frac{2}{3}\)

Lời giải:
Áp dụng BĐT Cô-si:

$\frac{1}{x(x+1)}+\frac{x}{2}+\frac{x+1}{4}\geq 3\sqrt[3]{\frac{1}{x(x+1)}.\frac{x}{2}.\frac{x+1}{4}}=\frac{3}{2}$

Tương tự:

$\frac{1}{y(y+1)}+\frac{y}{2}+\frac{y+1}{4}\geq \frac{3}{2}$

$\frac{1}{z(z+1)}+\frac{z}{2}+\frac{z+1}{4}\geq \frac{3}{2}$

Cộng theo vế các BĐT trên:

$\frac{1}{x^2+x}+\frac{1}{y^2+y}+\frac{1}{z^2+z}+\frac{3}{4}(x+y+z)+\frac{3}{4}\geq \frac{9}{2}$

$\frac{1}{x^2+x}+\frac{1}{y^2+y}+\frac{1}{z^2+z}+\frac{9}{4}+\frac{3}{4}\geq \frac{9}{2}$

$\Rightarrow \frac{1}{x^2+x}+\frac{1}{y^2+y}+\frac{1}{z^2+z}\geq \frac{3}{2}$ 

Vậy gtnn của biểu thức là $\frac{3}{2}$ khi $x=y=z=1$

\(A=\dfrac{2x^2}{2x+2yz}+\dfrac{2y^2}{2y+2zx}+\dfrac{2z^2}{2z+2xy}+\dfrac{9}{8\left(x^2+y^2+z^2\right)}\)

\(A\ge\dfrac{2x^2}{x^2+1+y^2+z^2}+\dfrac{2y^2}{y^2+1+z^2+x^2}+\dfrac{2z^2}{z^2+1+x^2+y^2}+\dfrac{9}{8\left(x^2+y^2+z^2\right)}\)

\(A\ge\dfrac{2\left(x^2+y^2+z^2\right)}{x^2+y^2+z^2+1}+\dfrac{9}{8\left(x^2+y^2+z^2\right)}\)

Đặt \(x^2+y^2+z^2=a>0\)

\(\Rightarrow A\ge\dfrac{2a}{a+1}+\dfrac{9}{8a}=\dfrac{2a}{a+1}+\dfrac{9}{8a}-\dfrac{15}{8}+\dfrac{15}{8}\)

\(\Rightarrow A\ge\dfrac{\left(a-3\right)^2}{8a\left(a+1\right)}+\dfrac{15}{8}\ge\dfrac{15}{8}\)

\(A_{min}=\dfrac{15}{8}\) khi \(a=3\) hay \(x=y=z=1\)