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Let $a,b,c>0$:

Prove that: $S=\frac{a^2}{(2a+b)(2a+c)}+\frac{b^2}{(2b+c)(2b+a)}+\frac{c^2}{(2c+a)(2c+b)} \leq \frac{1}{3}$

My solution:

We have: $\left[\begin{matrix}\frac{1}{x}+\frac{1}{y} \geq \frac{4}{x+y} \\\frac{1}{x}+\frac{1}{y} +\frac{1}{z} \geq \frac{9}{x+y+z}\end{matrix}\right.$

$=>S=\frac{a^2}{(2a+b)(2a+c)}+\frac{b^2}{(2b+c)(2b+a)}+\frac{c^2}{(2c+a)(2c+b)}$

$\leq \frac{a^2}{4}.(\frac{1}{2a+b}+\frac{1}{2a+c})+\frac{b^2}{4}.(\frac{1}{2b+c}+\frac{1}{2b+a})+\frac{c^2}{4}.(\frac{1}{2c+a}+\frac{1}{2c+b})$

$=\frac{1}{4}.[a^2.(\frac{1}{2a+b}+\frac{1}{2a+c})+b^2.(\frac{1}{2b+a}+\frac{1}{2b+c})+c^2.(\frac{1}{2c+a}+\frac{1}{2c+b})]$

$\leq \frac{1}{4}.[\frac{a^2}{9}.(\frac{2}{a}+\frac{1}{b}+\frac{2}{a}+\frac{1}{c}) +\frac{b^2}{9}.(\frac{2}{b}+\frac{1}{c}+\frac{2}{b}+\frac{1}{a}) +\frac{c^2}{9}.(\frac{2}{c}+\frac{1}{a}+\frac{2}{c}+\frac{1}{b})]$

$=\frac{1}{36}.[a^2.(\frac{4}{a}+\frac{1}{b}+\frac{1}{c})+b^2.(\frac{1}{a}+\frac{4}{b}+\frac{1}{c})+c^2.(\frac{1}{a}+\frac{1}{b}+\frac{4}{c})]$

$= \frac{1}{36}.(4a+4b+4c+\frac{a^2}{b}+\frac{a^2}{c}+\frac{b^2}{a}+\frac{b^2}{c}+\frac{c^2}{a}+\frac{c^2}{b}) $  

We need prove that: $S \le \frac{1}{3}$  $=> S \le \frac{1}{12}.(4a+4b+4c+\frac{a^2}{b}+\frac{a^2}{c}+\frac{b^2}{a}+\frac{b^2}{c}+\frac{c^2}{a}+\frac{c^2}{b}) \le 1 $  <- in there, I don't know how to do that :<

I'm trying to find a solution to continue for me or another solution, Can you help me?

Oshawott
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Leomessi
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    The question makes me a bit recalling this one: https://math.stackexchange.com/questions/4230355/prove-the-inequality-sum-cyc-fraca2-sqrtbcb3c3-geq-frac-32-w –  Aug 27 '21 at 14:09
  • yep, thanks you, but that not prove $\le$ <-> i mean that not same :< – Leomessi Aug 27 '21 at 14:14
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    Your first inequality involving S is incorrect from what I can tell. You seem to be using $ 1/(2a+b) + 1/(2a+c) \geq 4/( (2a+b)(2a+c) )$, but the RHS based on your first stated inequlity should be $ 4/( (2a+b) \color{red}{+} (2a+c) )$. This helps explain why you added up with an additional degree at the very end. – Calvin Lin Aug 29 '21 at 15:16
  • @Calvin Lin: oh year, thanks you so much. – Leomessi Aug 29 '21 at 18:09
  • How about that? Let $a,b,c>0$ and $a+b+c=3$ , Find max:: $P= \frac{1}{3a^2+2b^2}+\frac{1}{3b^2+2c^2}+\frac{1}{3c^2+2a^2}$ Can we use that: $\left[\begin{matrix}\frac{1}{x}+\frac{1}{y} \geq \frac{4}{x+y} \\frac{1}{x}+\frac{1}{y} +\frac{1}{z} \geq \frac{9}{x+y+z}\end{matrix}\right.$ ? – Leomessi Aug 30 '21 at 01:10

4 Answers4

4

The inequality looks familiar. In fact, I solved the inequality a few weeks ago from an inequality handout. Here is the solution:

We have $$\begin{align}\frac{9a^2}{(2a+b)(2a+c)} &= \frac{(2a+a)^2}{2a(a+b+c)+2a^2+bc}\\ & \leq \frac{(2a)^2}{2a(a+b+c)}+\frac{a^2}{2a^2+bc} \\ &= \frac{2a}{a+b+c}+\frac{a^2}{2a^2+bc} \end{align}$$ from Cauchy-Schwarz inequality.

Now, $$\begin{align}9\sum_{cyc}\frac{a^2}{(2a+b)(2a+c)} &\leq 2\sum_{cyc}\frac{a}{a+b+c}+\sum_{cyc}\frac{a^2}{2a^2+bc}\\ &= 2+\sum_{cyc}\frac{a^2}{2a^2+bc}\leq 3 \end{align}$$ which is the desired result.

Here the last inequality follows because $$\sum_{cyc}\frac{a^2}{2a^2+bc}\leq1\iff \sum_{cyc}\frac{bc}{2a^2+bc}\geq1$$ is true by Cauchy-Schwarz.

Oshawott
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3

Let $a\geq b\geq c$.

Thus, $$\frac{1}{3}-\sum_{cyc}\frac{a^2}{(2a+b)(2a+c)}=\sum_{cyc}\left(\frac{a}{3(a+b+c)}-\frac{a^2}{(2a+b)(2a+c)}\right)=$$ $$=\frac{1}{3(a+b+c)}\sum_{cyc}\frac{a(a-b)(a-c)}{(2a+b)(2a+c)}\geq$$ $$\geq\frac{1}{3(a+b+c)}\left(\frac{a(a-b)(a-c)}{(2a+b)(2a+c)}-\frac{b(a-b)(b-c)}{(2b+a)(2b+c)}\right)\geq$$ $$\geq\frac{1}{3(a+b+c)}\left(\frac{a(a-b)\frac{a}{b}(b-c)}{(2a+b)(2a+c)}-\frac{b(a-b)(b-c)}{(2b+a)(2b+c)}\right)=$$ $$=\frac{(a-b)(b-c)}{3b(a+b+c)}\left(\frac{a^2}{(2a+b)(2a+c)}-\frac{b^2}{(2b+a)(2b+c)}\right)\geq0$$ because $$a(2b+c)\geq b(2a+c)$$ and $$a(2b+a)\geq b(2a+b).$$

Michael Rozenberg
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Another way.

After full expanding we need to prove that: $$\sum_{cyc}(2a^4b^2+2a^4c^2-4a^3b^3+5a^4bc+4a^3b^2c+4a^3c^2b-13a^2b^2c^2)\geq0,$$ which is true by Muirhead because

$(4,2,0)\succ(3,3,0),$ $(4,1,1)\succ(2,2,2)$ and $(3,2,1)\succ(2,2,2).$

Also, we can see it by AM-GM.

Michael Rozenberg
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2

Another way. $$\frac{1}{3}-\sum_{cyc}\frac{a^2}{(2a+b)(2a+c)}=\sum_{cyc}\left(\frac{1}{9}-\frac{a^2}{(2a+b)(2a+c)}\right)=$$ $$=\frac{1}{9}\sum_{cyc}\frac{2ab+2ac+bc-5a^2}{(2a+b)(2a+c)}=\frac{1}{18}\sum_{cyc}\frac{(c-a)(5a+b)-(a-b)(5a+c)}{(2a+b)(2a+c)}=$$ $$=\frac{1}{18}\sum_{cyc}(a-b)\left(\frac{5b+c}{(2b+c)(2b+a)}-\frac{5a+c}{(2a+b)(2a+c)}\right)=$$ $$=\frac{1}{18}\sum_{cyc}\frac{(a-b)^2(c^2+10ab-ac-bc)}{(2a+b)(2a+c)(2b+a)(2b+c)}\geq$$ $$\geq\frac{1}{18}\sum_{cyc}\frac{(a-b)^2(c^2-ac-bc+ab)}{(2a+b)(2a+c)(2b+a)(2b+c)}=$$ $$=\frac{1}{18\prod\limits_{cyc}((2a+b)(2a+c))}\sum_{cyc}(a-b)^2(c-a)(c-b)(2c+a)(2c+b)=$$ $$=\frac{(a-b)(b-c)(c-a)\sum\limits_{cyc}(b-a)(4c^2+2ac+2bc+ab)}{18\prod\limits_{cyc}((2a+b)(2a+c))}=$$ $$=\frac{(a-b)(b-c)(c-a)\sum\limits_{cyc}(4a^2c-4a^2b+2a^2b-2a^2c+a^2c-a^2b)}{18\prod\limits_{cyc}((2a+b)(2a+c))}=$$ $$=\frac{(a-b)(b-c)(c-a)\sum\limits_{cyc}(a^2c-a^2b)}{6\prod\limits_{cyc}((2a+b)(2a+c))}=\frac{\prod\limits_{cyc}(a-b)^2}{6\prod\limits_{cyc}((2a+b)(2a+c))}\geq0.$$

Michael Rozenberg
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