By Cauchy-Schwarz we have: $$(x+1)^2+(y-1)^2 \ge \frac{1}{2}(x+1+y-1)^2=\frac{1}{2}(x+y)^2$$ Now, use AM-GM we have: $$x+y=x+\frac{y}{3}+\frac{y}{3}+\frac{y}{3}\ge 4.\sqrt[4]{\frac{xy^3}{27}}$$ so $$\frac{1}{2}(x+y)^2 \ge \frac{1}{2}.(4.\sqrt[4]{\frac{xy^3}{27}})^2=\frac{8y\sqrt{xy}}{3\sqrt{3}}$$ Done.

Trunglindan1995 wrote: By Cauchy-Schwarz we have: $$(x+1)^2+(y-1)^2 \ge \frac{1}{2}(x+1+y-1)^2=\frac{1}{2}(x+y)^2$$ Now, use AM-GM we have: $$x+y=x+\frac{y}{3}+\frac{y}{3}+\frac{y}{3}\ge 4.\sqrt[4]{\frac{xy^3}{27}}$$ so $$\frac{1}{2}(x+y)^2 \ge \frac{1}{2}.(4.\sqrt[4]{\frac{xy^3}{27}})^2=\frac{8y\sqrt{xy}}{3\sqrt{3}}$$ Done. $$(x+1)^2+(y-1)^2 \geq\frac{1}{2}(x+1+y-1)^2=\frac{1}{2}(x+\frac{y}{3}+\frac{y}{3}+\frac{y}{3})^2\geq\frac{8y\sqrt{xy}}{3\sqrt{3}}.$$Nice.

Trunglindan1995 wrote: By Cauchy-Schwarz we have: $$(x+1)^2+(y-1)^2 \ge \frac{1}{2}(x+1+y-1)^2=\frac{1}{2}(x+y)^2$$ Now, use AM-GM we have: $$x+y=x+\frac{y}{3}+\frac{y}{3}+\frac{y}{3}\ge 4.\sqrt[4]{\frac{xy^3}{27}}$$ so $$\frac{1}{2}(x+y)^2 \ge \frac{1}{2}.(4.\sqrt[4]{\frac{xy^3}{27}})^2=\frac{8y\sqrt{xy}}{3\sqrt{3}}$$ Done. I'm a fool when it comes to inequalities, but, isn't that first step through Power Means using powers 2 and 1 rather than Cauchy? How put Cauchy in there? Nice solution by the way!

GeometryIsMyWeakness wrote: Trunglindan1995 wrote: By Cauchy-Schwarz we have: $$(x+1)^2+(y-1)^2 \ge \frac{1}{2}(x+1+y-1)^2=\frac{1}{2}(x+y)^2$$ Now, use AM-GM we have: $$x+y=x+\frac{y}{3}+\frac{y}{3}+\frac{y}{3}\ge 4.\sqrt[4]{\frac{xy^3}{27}}$$ so $$\frac{1}{2}(x+y)^2 \ge \frac{1}{2}.(4.\sqrt[4]{\frac{xy^3}{27}})^2=\frac{8y\sqrt{xy}}{3\sqrt{3}}$$ Done. I'm a fool when it comes to inequalities, but, isn't that first step through Power Means using powers 2 and 1 rather than Cauchy? How put Cauchy in there? Nice solution by the way! Use AM-GM have:$$a^2+b^2 \ge \frac{1}{2}(a+b)^2$$

rogue wrote: For every $x,y\ge0$ prove that $(x+1)^2+(y-1)^2\ge\frac{8y\sqrt{xy}}{3\sqrt{3}}.$ Equality holds when $x=1,y=3.$

GeometryIsMyWeakness wrote: I'm a fool when it comes to inequalities, but, isn't that first step through Power Means using powers 2 and 1 rather than Cauchy? How put Cauchy in there? Nice solution by the way! Power Means using powers 2 and 1 is a conclusion from Cauchy. So doesn't matter which one you write, you're right.

WolfusA wrote: GeometryIsMyWeakness wrote: I'm a fool when it comes to inequalities, but, isn't that first step through Power Means using powers 2 and 1 rather than Cauchy? How put Cauchy in there? Nice solution by the way! Power Means using powers 2 and 1 is a conclusion from Cauchy. So doesn't matter which one you write, you're right. Thanks you!