Testwiki:Reference desk/Archives/Mathematics/2021 July 17

{| width = "100%"

|- ! colspan="3" align="center" | Mathematics desk |- ! width="20%" align="left" | < July 16 ! width="25%" align="center"|<< Jun | July | Aug >> ! width="20%" align="right" |Current desk > |}

Welcome to the Wikipedia Mathematics Reference Desk Archives
The page you are currently viewing is a transcluded archive page. While you can leave answers for any questions shown below, please ask new questions on one of the current reference desk pages.

July 17

Decomposition of convex combination

Please let me know if this is inappropriate. I don't know how to type TeX in wiki. So maybe it is better to paste the link of my question here: https://math.stackexchange.com/questions/4198466/is-this-decomposition-of-convex-combination-always-feasible Thank you very much!

Quotient problem

The square root of (a*b), divided by (a+b), equals 0.5 if and only if a=b. In any other case, the quotient is less than 0.5. I've established this experimentally, but I can't see why it's the case. I've only considered positive a and b, but what about negative a and/or b? -- Jack of Oz ^{[pleasantries]} 02:07, 17 July 2021 (UTC)

That's the inequality of arithmetic and geometric means. --116.86.4.41 (talk) 04:22, 17 July 2021 (UTC)

Perfect. Thanks. -- Jack of Oz ^{[pleasantries]} 08:47, 17 July 2021 (UTC)

If either a or b is negative, but not both, the numerator is the square root of a negative number, which is an imaginary number, and I doubt you want to go there. Dolphin (t) 07:14, 17 July 2021 (UTC)

Indeed not. -- Jack of Oz ^{[pleasantries]} 08:47, 17 July 2021 (UTC)

Other cases: If a and b are both zero, you end up with 0/0, so the result is undefined. If only one of the two equals 0, the result is also 0. If a and b are both negative, the quotient is also negative. It is then equal to −0.5 if a and b are equal, and otherwise a negative number anywhere between −0.5 and 0.  --Lambiam 08:58, 17 July 2021 (UTC)

My favorite proof is to let $u=\frac{a+b}{2}$ and $v=\frac{a-b}{2}$, so $a=u+v$ and $b=u-v$. Thus $ab=(u+v)(u-v)=u^2-v^2$ which, holding $a+b$ constant, reaches a maximum when $v^2=0$ which means $a=b$. 2601:648:8202:350:0:0:0:2B99 (talk) 09:56, 18 July 2021 (UTC)