Does $\frac{x+y}{2}>\frac{a+b}{2}$ hold? $a$ and $b$ are two real positive numbers. Given that $x=\sqrt{ab}$ and $y=\sqrt{\frac{a^2+b^2}{2}}$, which one has a higher value, $\frac{x+y}{2}$ or $\frac{a+b}{2}$?
We know that $y=\sqrt{\frac{a^2+b^2}{2}}>\frac{a+b}{2}>x=\sqrt{ab}$ by inequality, and at this point I'm stuck.
 A: we know $$\sqrt{\frac{x^2+y^2}{2}} \geq \frac{x+y}{2} $$ so plug x,y $$ \sqrt{\frac{x^2+y^2}{2}} \geq \frac{x+y}{2}\\\sqrt{\frac{(\sqrt{ab})^2+(\sqrt{\frac{a^2+b^2}{2}})^2}{2}} \geq \frac{x+y}{2}\\\sqrt{\frac{ab+\frac{a^2+b^2}{2}}{2}} \geq \frac{x+y}{2}\\\sqrt{\frac{\frac{a^2+b^2+2ab}{2}}{2}} \geq \frac{x+y}{2}\\\sqrt{\frac{(a+b)^2}{4}} \geq \frac{x+y}{2}\\ \frac{a+b}{2} \geq \frac{x+y}{2}$$
A: we will prove that $$\frac{x+y}{2}\le \frac{a+b}{2}$$ after squaring is this equivalent to $$2\sqrt{ab}\sqrt{\frac{a^2+b^2}{2}}\le \frac{(a+b)^2}{2}$$ squaring again we get $$ab\left(\frac{a^2+b^2}{2}\right)\le \frac{(a+b)^4}{16}$$ and this is equivalent to $$0\le (a-b)^4$$ and this is true.
A: We can drop the factor of $\frac{1}{2}$, so we have $\sqrt{ab}+\sqrt{\frac{a^2+b^2}{2}}$ and $a+b$. Squaring gives $ab + \frac{a^2+b^2}{2}+\sqrt{ab\frac{a^2+b^2}{2}}$ and $a^2+b^2+ab$. They both have a factor $ab$ and $\frac{a^2+b^2}{2}$, so we have to compare $\sqrt{ab\frac{a^2+b^2}{2}}$ and $\frac{a^2+b^2}{2}$. Rewriting we find $\sqrt{ab}\sqrt{\frac{a^2+b^2}{2}}$ and $\sqrt{\frac{a^2+b^2}{2}}\sqrt{\frac{a^2+b^2}{2}}$, so dropping out the common factor and squaring, we get $ab$ and $\frac{a^2+b^2}{2}$. We have that $\frac{a^2+b^2}{2}\geq ab$ implies that $\frac{a^2+b^2}{2} -ab = (a-b)^2 \geq 0$, which holds.
