\(P=\sum\frac{x^2\left(y+z\right)}{yz}\ge\sum\frac{4x^2\left(y+z\right)}{\left(y+z\right)^2}=\sum\frac{4x^2}{y+z}\ge\frac{4\left(x+y+z\right)^2}{y+z+z+x+x+y}=2\left(x+y+z\right)=2\)
\(P_{min}=2\) khi \(x=y=z=\frac{1}{3}\)
Câu 2 có dương không nhỉ? Không dương thì không làm được
\(A=\frac{1}{x^2+y^2}+\frac{1}{2xy}+\frac{1}{2xy}\ge\frac{4}{x^2+y^2+2xy}+\frac{2}{\left(x+y\right)^2}=\frac{6}{\left(x+y\right)^2}\ge6\)
\(A_{min}=6\) khi \(x=y=\frac{1}{2}\)
1) \(P\ge\frac{x^2.2\sqrt{yz}}{yz}+\frac{y^2.2\sqrt{zx}}{zx}+\frac{z^2.2\sqrt{xy}}{xy}=\frac{2x^2}{\sqrt{yz}}+\frac{2y^2}{\sqrt{zx}}+\frac{2z^2}{\sqrt{xy}}\ge4\left(\frac{x^2}{y+z}+\frac{y^2}{z+x}+\frac{z^2}{x+y}\right)=4\left\{\left[\frac{x^2}{y+z}+\frac{1}{4}\left(y+z\right)\right]+\left[\frac{y^2}{z+x}+\frac{1}{4}\left(z+x\right)\right]+\left[\frac{z^2}{x+y}+\frac{1}{4}\left(x+y\right)\right]\right\}-2\left(x+y+z\right)\ge4\left(x+y+z\right)-2\left(x+y+z\right)=2\)
Dấu "=" xảy ra <=> \(x=y=z=\frac{1}{3}\)
2) \(A=\left[\frac{1}{x^2+y^2}+4\left(x^2+y^2\right)\right]+\left(\frac{1}{xy}+16xy\right)-4\left(x+y\right)^2-8xy\ge4+8-4-2.\left(x+y\right)^2=8-2.\left(x+y\right)^2\ge8-2=6\)
Dấu "=" xảy ra <=> \(x=y=\frac{1}{2}\)
