1,cho $\left\{\begin{matrix} x,y>0 & & \\ x^{2}+y^{2}=1& & \end{matrix}\right.$
tìm min:
$A=(1+x)(1+\frac{1}{y})+(1+y)(1+\frac{1}{x})$
2,cho a,b>0/ab=1
tìm min A=$(a+b+1)(a^{2}+b^{2})+\frac{4}{a+b}$
Ta có $A=2+(x+y)+(\frac{1}{x}+\frac{1}{y})+(\frac{x}{y}+\frac{y}{x})$
Áp dụng AM-GM ta có $A \geq 2+2\sqrt{xy}+\frac{2}{\sqrt{xy}}+2=4+2(\sqrt{xy}+\frac{1}{\sqrt{xy}})$
Xét $\sqrt{xy}+\frac{1}{\sqrt{xy}}=(\sqrt{xy}+\frac{1}{2\sqrt{xy}})+\frac{1}{2\sqrt{xy}} \geq \sqrt{2}+\frac{1}{x+y} \geq 2+\frac{1}{\sqrt{2(x^2+y^2)}} \geq \sqrt{2}+\frac{1}{\sqrt{2}}$
$A \geq 4+2(\sqrt{2}+\frac{1}{\sqrt{2}})=4+3\sqrt{2}$
Đẳng thức xảy ra khi $x=y=\frac{1}{\sqrt{2}}$