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

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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 $u={a+b \over 2}$ and $v={a-b \over 2}$, 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)