$a, b, c$ are positives such that $a + b + c = 1$. Determine the maximal value of $\sum\limits_{cyc}\frac1{a(b + c)} - \frac{a^2 + b^2 + c^2}{2abc}$.

Question

$a, b, c$ 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.

Answer 1

We have that

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

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

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

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

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

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

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

Answer 2

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 $uvw$.