Question

Show that for any two positive real numbers \(a\) and \(b\),

\(\frac{a+b}{2} - \sqrt{ab} \geq \sqrt{\frac{a^2+b^2}{2}} - \frac{a+b}{2}.\)

Answer 1

.

Answer 2

holy mother of god

Answer 3

have you tried searching this on google?

Answer 4

.

Answer 5

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.

Answer 6

@oldrin.bataku

Answer 7

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 \(a^2,b^2,a^4,b^4\ge 0\)

Answer 8

the steps are all reversible so that proves it

Answer 9

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

\((a+b)^2\ge \frac{a^2+b^2}2+\frac{2ab}2+\sqrt{2a^3b+2ab^3}=\frac12 (a+b)^2+\sqrt{2ab(a^2+b^2)}\)

Answer 10

\((a+b)^2\geq(\sqrt{\dfrac{a^2+b^2}{2}})^2+(\sqrt{\dfrac{2ab}{2}})^2+2\sqrt{\dfrac{a^2+b^2}{2}}*\sqrt{\dfrac{2ab}{2}}\)

\((a+b)^2\geq\dfrac{a^2+b^2}{2}+\dfrac{2ab}{2}+\cancel{2}\sqrt{\dfrac{a^2+b^2}{\cancel{4}}*2ab}\)

Answer 11

Hence the result is
\((a+b)^4\geq 4(2ab(a^2+b^2))\\(a+b)^4\geq 8a^3b+8ab^3\)

Answer 12

\(a^4+b^4 +4a^3b +4ab^3+6a^2b^2\geq 8a^3b+8ab^3\\a^4+b^4+6a^2b^2\geq 4a^3b-4ab^3\)
then???

Answer 13

oh, I got it, it becomes
\((a-b)^4\geq 0\)
and go back ward.
Thank you so much.

Answer 14

good catch @Loser66 but as you showed it's salvageable

Answer 15

:)