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$a, b, c$ are positives such that $a + b + c = 1$. Determine the maximal value of $$\large \sum_{cyc}\frac{1}{a(b + c)} - \frac{a^2 + b^2 + c^2}{2abc}$$

This is a problem in a recent exam, I got $3/20$ points (and also almost everybody did worse). I didn't know how to solve this problem, then our teacher went on our group chat and said that Sum Of Square works. Thanks.

Here was my attempt (during the time taking the exam). We have that

$$\sum_{cyc}\frac{1}{a(b + c)} - \frac{a^2 + b^2 + c^2}{2abc} = \sum_{cyc}\frac{1}{a(1 - a)} - \frac{1 - 2(bc + ca + ab)}{2abc}$$

$$ = \sum_{cyc}\frac{1}{a(1 - a)} - \left(\frac{1}{a} + \frac{1}{b} + \frac{1}{c}\right) - \dfrac{1}{2abc} = \left(\frac{1}{1 - a} + \frac{1}{1 - b} + \frac{1}{1 - c}\right) - \dfrac{1}{2abc}$$

$$ = 3 + \left(\frac{a}{1 - a} + \frac{a}{1 - b} + \frac{a}{1 - c}\right) - \dfrac{1}{2abc}$$

Below is the solution I can come up with after taking the exam. I'm disappointed in myself.

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We have that

$$\sum_{cyc}\frac{1}{a(b + c)} - \frac{a^2 + b^2 + c^2}{2abc} = \sum_{cyc}\frac{(a + b + c)^2}{a(b + c)} - \frac{a^2 + b^2 + c^2}{2abc}$$

$$ = \sum_{cyc}\frac{(b + c - a)^2 + 4a(b + c)}{a(b + c)} - \frac{a^2 + b^2 + c^2}{2abc}$$

$$ = \sum_{cyc}\frac{(b + c - a)^2}{a(b + c)} - \frac{a^2 + b^2 + c^2}{2abc} + 12$$

$$ \le \frac{1}{4}\sum_{cyc}\left[\frac{1}{a}\left(\frac{1}{b} + \frac{1}{c}\right) \cdot (b + c - a)^2\right] - \frac{a^2 + b^2 + c^2}{2abc} + 12$$

$$ = \frac{\displaystyle \sum_{cyc}\left[(b + c)(b + c - a)^2\right]}{4abc} - \frac{(a + b + c)(a^2 + b^2 + c^2)}{2abc} + 12$$

$$ = \frac{\displaystyle \sum_{cyc}\left[(b + c)(b + c - a)^2\right] - 2(a + b + c)(a^2 + b^2 + c^2)}{4abc} + 12$$

$$ = \frac{\displaystyle \sum_{cyc}\left[(b + c)(bc - ca - ab)\right]}{2abc} + 12 = \frac{\displaystyle \sum_{cyc}\left[bc(b + c - c - a - a - b)\right]}{2abc} + 12 = 9$$

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  • $\begingroup$ As always in this kind of problems, the extremum is attaint when the variables are all equal.($\frac13$ each). $\endgroup$ – Piquito Sep 10 at 16:17
  • $\begingroup$ @Piquito, unless you can characterize exactly what you mean by "this kind of problem," this is not particularly helpful advice, nor is it a proof. $\endgroup$ – Cheerful Parsnip Sep 10 at 18:56
  • $\begingroup$ It was a simple comment "non mathematical" if you want. You have misunderstood it.I didn't look for anything to say it. $\endgroup$ – Piquito Sep 10 at 22:25
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Let $a=b=c=\frac{1}{3}.$

Thus, $$\sum_{cyc}\frac{1}{a(b+c)}-\frac{a^2+b^2+c^2}{2abc}=9.$$ We'll prove that it's a maximal value.

Indeed, we need to prove that $$(a+b+c)^2\sum_{cyc}\frac{1}{a(b+c)}-\frac{(a+b+c)(a^2+b^2+c^2)}{2abc}\leq9$$ or $$2(a+b+c)^2\sum_{cyc}bc(a+b)(a+c)-(a^2+b^2+c^2)(a+b+c)\prod_{cyc}(a+b)\leq18abc\prod_{cyc}(a+b)$$ or $$\sum_{cyc}(a^3+a^2b+a^2c+6abc)\sum_{cyc}\left(a^2b+a^2c+\frac{2}{3}abc\right)\geq2\sum_{cyc}(a^2+2ab)\sum_{cyc}(a^2b^2+3a^2bc)$$ or $$\sum_{cyc}(a^5b+a^5c+a^4b^2+a^4c^2+a^3b^2c+a^3c^2b+2a^4bc+$$ $$+a^4b^2+a^4c^2+2a^4bc+2a^3b^3+2a^3b^2c+2a^3c^2b+2a^2b^2+2a^3b^2c+2a^3c^2b+$$ $$+18a^3b^2c+18a^3c^2b+12a^2b^2c^2-2a^4b^2-2a^4c^2-2a^2b^2c^2-6a^4bc-6a^3b^2c-6a^3c^2b-$$ $$-4a^3b^3-4a^3b^2c-4a^3c^2b-12a^3b^2c-12a^3c^2b-12a^2b^2c^2)\geq0$$ or $$\sum_{cyc}(a^5b+a^5c-2a^3b^3-2a^4bc+a^3b^2c+a^3c^2b)\geq0.$$ Now, let $a\geq b\geq c$.

Thus, $$\sum_{cyc}(a^5b+a^5c-2a^3b^3-2a^4bc+a^3b^2c+a^3c^2b)=$$ $$=\sum_{cyc}(ab(a^2-b^2)^2-abc(a+b)(a-b)^2)=$$ $$=\sum_{cyc}(a-b)^2ab(a+b)(a+b-c)\geq$$ $$\geq(a-c)^2ac(a+c)(a+c-b)+(b-c)^2bc(b+c)(b+c-a)\geq$$ $$\geq(b-c)^2ac(a+c)(a-b)+(b-c)^2bc(b+c)(b-a)=$$ $$=(b-c)^2(a-b)^2c(a+b+c)\geq0$$ and we are done!

Also, your inequality easy to prove by $uvw$.

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