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zmudz
 one year ago
Show that for any two positive real numbers \(a\) and \(b\),
\(\frac{a+b}{2}  \sqrt{ab} \geq \sqrt{\frac{a^2+b^2}{2}}  \frac{a+b}{2}.\)
zmudz
 one year ago
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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anonymous
 one year ago
Best ResponseYou've already chosen the best response.0have you tried searching this on google?

Zarkon
 one year ago
Best ResponseYou've already chosen the best response.0I 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.

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0consider $$\frac{a+b}2\sqrt{ab}\ge\sqrt{\frac{a^2+b^2}2}\frac{a+b}2\\a+b\ge\sqrt{\frac{a^2+b^2}2}+\sqrt{\frac{2ab}2}\\(a+b)^2\ge \frac{a^2+b^2}2+\frac{2ab}2+\sqrt{a^3b+ab^3}=\frac12 (a+b)^2+\sqrt{ab(a^2+b^2)}\\(a+b)^2\ge2\sqrt{ab(a^2+b^2)}\\(a+b)^4\ge 4a^3b+4ab^3\\a^4+b^4+4a^3b+4ab^3+6a^2b^2\ge 4a^3b+4ab^3\\a^4+b^4+6a^2b^2\ge0$$ which is trivially true since \(a^2,b^2,a^4,b^4\ge 0\)

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0the steps are all reversible so that proves it

Loser66
 one year ago
Best ResponseYou've already chosen the best response.1I 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 \((a+b)^2\ge \frac{a^2+b^2}2+\frac{2ab}2+\sqrt{2a^3b+2ab^3}=\frac12 (a+b)^2+\sqrt{2ab(a^2+b^2)}\)

Loser66
 one year ago
Best ResponseYou've already chosen the best response.1\((a+b)^2\geq(\sqrt{\dfrac{a^2+b^2}{2}})^2+(\sqrt{\dfrac{2ab}{2}})^2+2\sqrt{\dfrac{a^2+b^2}{2}}*\sqrt{\dfrac{2ab}{2}}\) \((a+b)^2\geq\dfrac{a^2+b^2}{2}+\dfrac{2ab}{2}+\cancel{2}\sqrt{\dfrac{a^2+b^2}{\cancel{4}}*2ab}\)

Loser66
 one year ago
Best ResponseYou've already chosen the best response.1Hence the result is \((a+b)^4\geq 4(2ab(a^2+b^2))\\(a+b)^4\geq 8a^3b+8ab^3\)

Loser66
 one year ago
Best ResponseYou've already chosen the best response.1\(a^4+b^4 +4a^3b +4ab^3+6a^2b^2\geq 8a^3b+8ab^3\\a^4+b^4+6a^2b^2\geq 4a^3b4ab^3\) then???

Loser66
 one year ago
Best ResponseYou've already chosen the best response.1oh, I got it, it becomes \((ab)^4\geq 0\) and go back ward. Thank you so much.

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0good catch @Loser66 but as you showed it's salvageable
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