Cyclic sum of 1/((3-c)(4-c))

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Cyclic sum of 1/((3-c)(4-c))

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Source: ELMO Shortlist 2013: Problem A6, by David Stoner

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#1 h Jul 23, 2013, 1:24 AM • 5 Y

Y by tenplusten, HamstPan38825, megarnie, mathmax12, Adventure10

Let $a, b, c$ be positive reals such that $a+b+c=3$ . Prove that $18\sum_{\text{cyc}}\frac{1}{(3-c)(4-c)}+2(ab+bc+ca)\ge 15.$ Proposed by David Stoner

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arqady

#2 h Jul 23, 2013, 4:02 AM • 3 Y

Y by I_am_human, Adventure10, Mango247

v_Enhance wrote:

Let $a, b, c$ be positive reals such that $a+b+c=3$ . Prove that $18\sum_{\text{cyc}}\frac{1}{(3-c)(4-c)}+2(ab+bc+ca)\ge 15$

$18\sum_{\text{cyc}}\frac{1}{(3-c)(4-c)}+2(ab+bc+ca)-15=\sum_{cyc}\left(\frac{1}{(3-a)(4-a)}+3-a^2-5\right)=$
$=\sum_{cyc}\frac{(a-1)(-a^3+6a^2-8a+6)}{a^2-7a+12}=\sum_{cyc}\left(\frac{(a-1)(-a^3+6a^2-8a+6)}{a^2-7a+12}-\frac{a-1}{2}\right)=$
$=\sum_{cyc}\frac{a(9-2a)(a-1)^2}{2(3-a)(4-a)}\geq0$ .

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6870 posts

#3 h Jul 23, 2013, 1:40 PM • 6 Y

Y by anantmudgal09, Illuzion, HamstPan38825, myh2910, Adventure10, Ryan2010

Nice. Here was my solution: $\frac{1}{(3-c)(4-c)} \ge \frac{2c^2+c+3}{36} \iff c(c-1)^2(2c-9) \le 0$ which implies $2(ab+bc+ca) + 18\sum_{\text{cyc}} \left( \frac{2c^2+c+3}{36} \right) = (a+b+c)^2 + \frac{a+b+c+9}{2} = 15.$

Stronger version: $162\sum\frac{1}{(3-c)(4-c)}+19(ab+bc+ca)\ge 138$ .

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arqady

#4 h Jul 23, 2013, 3:10 PM • 2 Y

Y by Adventure10, Mango247

The following inequality is also true.
For positives $a$ , $b$ and $c$ such that $a+b+c=3$ prove that:
$\sum_{cyc}\frac{90}{(3-c)(4-c)}+14(ab+bc+ca)\ge 87$ No, it's wrong!
For $b=a$ and $c=3-2a$ we need to prove that $(a-1)^2(3-2a)(14a^3-70a^2+29a+60)\geq0,$ which is wrong for $0<a<1.5.$
Victoria_Discalceata1, thank you!

The following inequality is true already.
For positives $a$ , $b$ and $c$ such that $a+b+c=3$ prove that:
$\sum_{cyc}\frac{90}{(3-c)(4-c)}+13(ab+bc+ca)\ge 84$

This post has been edited 3 times. Last edited by arqady, Jan 10, 2022, 5:58 AM

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Booper

#5 h Dec 13, 2018, 6:07 PM • 4 Y

Y by Merlin233, MathQurious, Adventure10, Mango247

Nice solution v_Enhance. What was the motivation for your solution though?

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AMN300

#6 h Mar 28, 2021, 4:07 AM • 1 Y

Y by teomihai

Here is a motivated solution. Rewrite the left hand side as follows, so we want to show
$\sum 18 \frac{1}{(3-c)(4-c)}+c(3-c) \ge 15$ The idea is to show the inequality $\frac{18}{(3-x)(4-x)}+x(3-x) \ge \frac{7}{2}(x-1)+5$ on $(0,3)$ ; we find the right hand side by taking the linear approximation of the left hand side at the case of equality $a=b=c=1$ , known as the tangent line trick. This holds since
$\frac{18}{(3-x)(4-x)}+x(3-x) \ge \frac{7}{2}(x-1)+5 \iff 18 \ge (3-x)(4-x)(\frac{7x+3}{2}-x(3-x)) \iff 36 \ge (3-x)(4-x)(2x^2+x+3)$ Expand $(3-x)(4-x)(2x^2+x+3) = 2x^4 - 13x^3 + 20x^2 - 9x+36$ so the above holds
$\iff 2x^3 -13 x^2 + 20x-9 \le 0$ Which is true on $(0,3)$ since $2x^3 -13 x^2 + 20x-9 = (x-1)^2 (2x-9)$ . Now summing it follows that
$\sum 18 \frac{1}{(3-c)(4-c)}+c(3-c) \ge \frac{7}{2}((a+b+c)-3)+3 \cdot 5 = 15$

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#7 h Jun 15, 2021, 12:27 PM • 1 Y

Y by maths_arka

Here is my solution i know it can be solved in much simpler way but for curiosity of Tangent line Trick i solved in this way if something wrong in my solution please tell.
My solution:
Proof:
Lets solve this by Tangent line trick so we need to do is use $a+b+c=3$ so
$\begin{align*} (a+b+c)^2=a^2+b^2+c^2+2ab+2bc+2ca & \Leftrightarrow 9=a^2+b^2+c^2=2(ab+bc+ca) &\\ \Leftrightarrow9-(a^2+b^2+c^2)=2(ab+bc+ca) \end{align*}$ so we use this in orignal inequality as $\sum_{\text{cyc}}\frac{18}{(3-c)(4-c)}-c^2\ge 6.$ so let suppose $f(x)=\frac{18}{(3-x)(4-x)}-x^2$ we have to prove that $f(a)+f(b)+f(c)\geq 6$ so lets apply \textbf{\textit{Tangent line trick}} in $f(x)\geq f(\alpha)+f'(\alpha)(x-\alpha)$ here $\alpha=\frac{a+b+c}{3}$ and we know that $a+b+c=3$ so actually we need to find $f(1)$ and $f"(1)$
$f(\alpha)=f(1)=2$ and $f"(\alpha)=f"(1)=\frac{1}{2}$ so now have new equality we have is $\frac{18}{(3-x)(4-x)}-x^2\geq 2+\frac12(x-1)$ so its suffice to show $\implies\frac{18}{(3-x)(4-x)}-x^2\geq \frac{x+3}{2}$ Taking cyclic sum on both side on $f(a)$ $\frac{18}{(3-a)(4-a)}-x^2\geq \frac{(a+3)}{2}$ $\implies\sum_{cyc}\left(\frac{18}{(3-a)(4-a)}-x^2\right)\geq \sum_{cyc}\frac{(a+3)}{2}$ Also we can write it as $\implies\sum_{cyc}\left(\frac{18}{(3-a)(4-a)}-a^2\right)\geq \left(\frac{(a+3)}{2}+\frac{(b+3)}{2}+\frac{(c+3)}{2}\right)$ $\implies\sum_{cyc}\left(\frac{18}{(3-c)(4-c)}-c^2\right)\geq \left(\frac{3+9}{2}\right)$ $\implies\sum_{cyc}\left(\frac{18}{(3-c)(4-c)}-c^2\right)\geq 6$ and we are done. :-D

:-D

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#8 h Jun 15, 2021, 4:41 PM • 1 Y

Y by Abdullahil_Kafi

Note that $9=(a+b+c)^2=a^2+b^2+c^2+2(ab+bc+ca)$ , so we can rewrite the original inequality as:
$\sum_{\text{cyc}}\left(\frac{18}{(3-a)(4-a)}-a^2\right)\ge6$ Denote $f(x)=\frac{18}{(3-x)(4-x)}-x^2$ . Using the tangent line trick, we note the inequality
$f(x)\ge\frac{x+3}2\Leftrightarrow x(2x-9)(x-1)^2\le0,$ which is true for all $x\in(0,3)$ . Then:
$\sum_{\text{cyc}}\left(\frac{18}{(3-a)(4-a)}-a^2\right)\ge\frac{a+b+c+9}2=6$ as desired. $\square$

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#9 h Jun 15, 2021, 4:47 PM

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thanks its seems simple :-D

:-D

This post has been edited 1 time. Last edited by MathThm, Jun 15, 2021, 4:48 PM

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#10 h Jan 5, 2022, 8:33 PM

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Really nice solution. Specially for the part of computing what the tangent line is.

This post has been edited 1 time. Last edited by Tafi_ak, Jan 5, 2022, 8:35 PM
Reason: hide

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mihaig

#11 h Jan 6, 2022, 7:54 AM

Y by

arqady wrote:

The following inequality is also true.
For positives $a$ , $b$ and $c$ such that $a+b+c=3$ prove that:
$\sum_{cyc}\frac{90}{(3-c)(4-c)}+14(ab+bc+ca)\ge 87$

Good one

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Victoria_Discalceata1

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Victoria_Discalceata1

#12 h Jan 10, 2022, 4:59 AM • 1 Y

Y by arqady

arqady wrote:

The following inequality is also true.
For positives $a$ , $b$ and $c$ such that $a+b+c=3$ prove that:
$\sum_{cyc}\frac{90}{(3-c)(4-c)}+14(ab+bc+ca)\ge 87$

I think that taking $b=a,\ c=3-2a$ we can find a counterexample.

This post has been edited 2 times. Last edited by Victoria_Discalceata1, Jan 11, 2022, 3:15 AM

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Victoria_Discalceata1

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Victoria_Discalceata1

#14 h Jan 11, 2022, 3:46 AM

Y by

arqady wrote:

The following inequality is true already.
For positives $a$ , $b$ and $c$ such that $a+b+c=3$ prove that:
$\sum_{cyc}\frac{90}{(3-c)(4-c)}+13(ab+bc+ca)\ge 84$

Yes. It can be transformed equivalently into $\sum\frac{180}{(4-a)(3-a)}-13\sum a^2\ge 51$ , then assuming WLOG $a\ge b\ge c$ we have three cases considering $f(x)=\frac{180}{(4-x)(3-x)}-13x^2$ on $[0,3)$ .
$\bullet$ If $c\ge\frac{9}{13}$ then we use tangent lines.
$\bullet$ If $\frac{9}{13}>c$ and $\frac{9}{10}>b$ then we consider the secant line through $\left(0,f(0)\right)$ and $\left(1,f(1)\right)$ .
$\bullet$ If $\frac{9}{13}>c$ and $b\ge\frac{9}{10}$ then we use convexity of $f$ on $\left[\frac{9}{10},3\right)$ and Jensen.

A very similar approach works also for Evan's one from #3, that one is a bit weaker.Click to reveal hidden text

The following stronger one is still true but even uglier calculation wise, at least with the method outlined above.

For $a,b,c\ge 0$ such that $a+b+c=3$ and $ab+bc+ca>0$ prove
$\sum\frac{20}{(4-a)(3-a)}+3(ab+bc+ca)\ge 19.$
Any idea for a nicer/more efficient solution?

This post has been edited 2 times. Last edited by Victoria_Discalceata1, May 20, 2022, 3:53 AM

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Taco12

#15 h Dec 11, 2022, 5:30 PM

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Note that $9=(a+b+c)^2=2(ab+bc+ac)+a^2+b^2+c^2$ , so subtracting $9$ from the RHS and $2(ab+bc+ac)+a^2+b^2+c^2$ from the LHS reduces the inequality to $\frac{18}{(3-a)(4-a)}-a^2+\frac{18}{(3-b)(4-b)}-b^2+\frac{18}{(3-c)(4-c)}-c^2 \geq 6.$ Let $f(x)=\frac{18}{(3-x)(4-x)}-x^2$ .

Claim: $f(x) \geq \frac{x+3}{2}$ .
Proof. The inequality reduces as $36 \geq (2x^2+x+3)(3-x)(4-x) \rightarrow x(x-1)^2(2x-9) \leq 0,$ which is clearly true. $\blacksquare$

Summing $f(a), f(b), f(c)$ with our claim gives $f(a) + f(b) + f(c) \geq \frac{a+b+c+9}{2} = 6. \blacksquare$

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#16 h Mar 19, 2023, 5:28 PM

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It suffices to show that $\sum_{\mathrm{cyc}} \frac{18}{(3-a)(4-a)} - a^2 \geq 15-(a+b+c)^2 = 6.$ But using tangent line trick, observe that $\frac{18}{(3-a)(4-a)} - a^2 \geq \frac a2 + \frac 32 \iff \frac{-a(a-1)^2(2a-9)}{(a-4)(a-3)} \geq 0,$ which is true. Sum cyclically to get the result.

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eibc

#17 h Apr 23, 2023, 2:29 AM

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Since $9 = (a + b + c)^2 = a^2 + b^2 + c^2 + 2(ab + bc + ca)$ , the inequality we wish to prove can be rewritten as
$\sum_{\text{cyc}} \frac{18}{(3 - a)(4 - a)} + 3 - a^2 \ge 15.$ The tangent line trick then gives us the inequality
$\frac{18}{(3 - x)(4 - x)} + 3 - x^2 \ge \frac{x + 9}{2} \iff x(x - 1)^2(2x - 9) \le 0,$ which clearly holds for $0 < x < 3$ . Therefore, we have
$\sum_{\text{cyc}} \frac{18}{(3 - a)(4 - a)} + 3 - a^2 \ge \frac{a + b + c + 27}{2} = 15,$ as desired.

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#18 h May 30, 2023, 11:29 PM

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posting for my own storage

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#19 h Dec 8, 2023, 10:45 AM

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Set $f(x):= \frac{18}{(3-x)(4-x)}-x^2$ . Now since $2(ab+bc+ca)=9-(a^2+b^2+c^2)$ , the problem is equivalent to :
$\sum_{cyc} f(a) \geq 6$
This is true by tangent trick lemma since:

$f(x)\geq f'(1)(x-1)+f(1)$ $\iff \frac{18}{(3-x)(4-x)}-x^2\geq \frac{x+3}{2}$ $\iff x(2x-9)(x-1)^2\le 0$
Which is true for all $0<x<3$

$\mathbb{Q.E.D.}$

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#20 h Feb 27, 2024, 2:13 AM

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We subtract $(a+b+c)^2$ from the LHS and $9$ from the RHS. We have $\sum_{\text{cyc}} \frac{18}{(3-a)(4-a)} -a^2 \geq 6$ We claim that the identity $\frac{18}{(3-a)(4-a)} -a^2 \geq \frac{a+3}{2}$ is true for all $a$ in $[0, 3]$ . Multiplying out we get $-2x^4+13x^3-20x^2+9x = x(x-1)^2(-2x+9) \geq 0$ Which is obviously true over $[0, 3]$ . Thus, we have the result. $\blacksquare$ It's so funny how random tangent line is.

This post has been edited 1 time. Last edited by Math4Life7, Feb 27, 2024, 2:27 AM

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#21 h Jan 30, 2025, 3:47 AM

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Note that $2(ab+bc+ca) = 9-(a^2+b^2+c^2),$ thus it suffices to show that $\sum \left( \frac{18}{(3-a)(4-a)} - a^2 \right) \geq 6.$ Let $f(x) = \frac{18}{(3-x)(4-x)} - x^2.$ Observe that $f(x) \geq \frac12 x + \frac32$ for all $0 \leq x \leq 4.5.$ Thus summing this for $x=a, b, c$ yields the desired result. QED

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#22 h Apr 1, 2025, 1:16 AM

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Good problem, TLT k1lls though

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