July 17 Information

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!

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) reply

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

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

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) reply

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

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) reply

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