## zmudz one year ago Show that for any two positive real numbers $$a$$ and $$b$$, $$\frac{a+b}{2} - \sqrt{ab} \geq \sqrt{\frac{a^2+b^2}{2}} - \frac{a+b}{2}.$$

1. IrishBoy123

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2. sepeario

holy mother of god

3. sepeario

have you tried searching this on google?

4. ganeshie8

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5. Zarkon

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.

6. Loser66

@oldrin.bataku

7. anonymous

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$$

8. anonymous

the steps are all reversible so that proves it

9. Loser66

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

10. Loser66

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

11. Loser66

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

12. Loser66

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

13. Loser66

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

14. anonymous

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

15. Loser66

:)