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 $x=\frac{\sqrt{3}-1}{2},y=\frac{\sqrt{3}+1}{2}.$

By AM-GM we have: $$(x+1)^2+(y-1)^2=x^2+(y-1)^2+2x+1 \ge 2x(y-1)+2x+1=2xy+1 \ge 2\sqrt{2xy}$$ Done.

Trunglindan1995 wrote: By AM-GM we have: $$(x+1)^2+(y-1)^2=x^2+(y-1)^2+2x+1 \ge 2x(y-1)+2x+1=2xy+1 \ge 2\sqrt{2xy}$$ Done. Nice.

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?