<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 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.
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