rogue wrote:
For every $x,y\ge0$ prove that $(x+1)^2+(y-1)^2\ge2\sqrt{2xy}.$
Proof of Zhangyanzong:
$$(x+1)^2+(y-1)^2=2\sqrt{2xy}+(x-y+1)^2+2(\sqrt{xy}-\frac{1}{\sqrt{2}})^2\ge2\sqrt{2xy}.$$Equality holds when $x=\frac{\sqrt{3}-1}{2},y=\frac{\sqrt{3}+1}{2}.$
sqing wrote:
rogue wrote:
For every $x,y\ge0$ prove that $(x+1)^2+(y-1)^2\ge2\sqrt{2xy}.$
Proof of Zhangyanzong:
$$(x+1)^2+(y-1)^2=2\sqrt{2xy}+(x-y+1)^2+2(\sqrt{xy}-\frac{1}{\sqrt{2}})^2\ge2\sqrt{2xy}.$$Equality holds when $x=\frac{\sqrt{3}-1}{2},y=\frac{\sqrt{3}+1}{2}.$
who is Zhangyanzong?