<s>[INST] 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.
 [/INST] 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.
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