Show that for any two positive real numbers \(a\) and \(b\), \(\frac{a+b}{2} - \sqrt{ab} \geq \sqrt{\frac{a^2+b^2}{2}} - \frac{a+b}{2}.\)

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Show that for any two positive real numbers \(a\) and \(b\), \(\frac{a+b}{2} - \sqrt{ab} \geq \sqrt{\frac{a^2+b^2}{2}} - \frac{a+b}{2}.\)

Mathematics
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holy mother of god
have you tried searching this on google?

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I can prove it using the fact that \[\left(\frac{a^2+b^2}{2}-ab\right)^2\ge0\] it is not the most elegant proof. If I find something elegant then I might post it.
@oldrin.bataku
consider $$\frac{a+b}2-\sqrt{ab}\ge\sqrt{\frac{a^2+b^2}2}-\frac{a+b}2\\a+b\ge\sqrt{\frac{a^2+b^2}2}+\sqrt{\frac{2ab}2}\\(a+b)^2\ge \frac{a^2+b^2}2+\frac{2ab}2+\sqrt{a^3b+ab^3}=\frac12 (a+b)^2+\sqrt{ab(a^2+b^2)}\\(a+b)^2\ge2\sqrt{ab(a^2+b^2)}\\(a+b)^4\ge 4a^3b+4ab^3\\a^4+b^4+4a^3b+4ab^3+6a^2b^2\ge 4a^3b+4ab^3\\a^4+b^4+6a^2b^2\ge0$$ which is trivially true since \(a^2,b^2,a^4,b^4\ge 0\)
the steps are all reversible so that proves it
I don't get line 3 from line 2 \( a+b\ge\sqrt{\frac{a^2+b^2}2}+\sqrt{\frac{2ab}2}\) square both sides, line 3 should be \((a+b)^2\ge \frac{a^2+b^2}2+\frac{2ab}2+\sqrt{2a^3b+2ab^3}=\frac12 (a+b)^2+\sqrt{2ab(a^2+b^2)}\)
\((a+b)^2\geq(\sqrt{\dfrac{a^2+b^2}{2}})^2+(\sqrt{\dfrac{2ab}{2}})^2+2\sqrt{\dfrac{a^2+b^2}{2}}*\sqrt{\dfrac{2ab}{2}}\) \((a+b)^2\geq\dfrac{a^2+b^2}{2}+\dfrac{2ab}{2}+\cancel{2}\sqrt{\dfrac{a^2+b^2}{\cancel{4}}*2ab}\)
Hence the result is \((a+b)^4\geq 4(2ab(a^2+b^2))\\(a+b)^4\geq 8a^3b+8ab^3\)
\(a^4+b^4 +4a^3b +4ab^3+6a^2b^2\geq 8a^3b+8ab^3\\a^4+b^4+6a^2b^2\geq 4a^3b-4ab^3\) then???
oh, I got it, it becomes \((a-b)^4\geq 0\) and go back ward. Thank you so much.
good catch @Loser66 but as you showed it's salvageable
:)

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