v_Enhance wrote: Let $a, b, c$ be positive reals such that $a+b+c=3$. Prove that \[18\sum_{\text{cyc}}\frac{1}{(3-c)(4-c)}+2(ab+bc+ca)\ge 15 \] $18\sum_{\text{cyc}}\frac{1}{(3-c)(4-c)}+2(ab+bc+ca)-15=\sum_{cyc}\left(\frac{1}{(3-a)(4-a)}+3-a^2-5\right)=$ $=\sum_{cyc}\frac{(a-1)(-a^3+6a^2-8a+6)}{a^2-7a+12}=\sum_{cyc}\left(\frac{(a-1)(-a^3+6a^2-8a+6)}{a^2-7a+12}-\frac{a-1}{2}\right)=$ $=\sum_{cyc}\frac{a(9-2a)(a-1)^2}{2(3-a)(4-a)}\geq0$.

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 $.

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: \[\sum_{cyc}\frac{90}{(3-c)(4-c)}+13(ab+bc+ca)\ge 84\]

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

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 \[ \sum 18 \frac{1}{(3-c)(4-c)}+c(3-c) \ge \frac{7}{2}((a+b+c)-3)+3 \cdot 5 = 15 \]

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.

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. $\square$

thanks its seems simple

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

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

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.

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.

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?

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 $$f(a) + f(b) + f(c) \geq \frac{a+b+c+9}{2} = 6. \blacksquare$$

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.

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.

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 $0<x<3$ $$\mathbb{Q.E.D.}$$

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.