# Inequalities - Tangent line trick

I will first state the "trick":

we fix $$a=\frac{a_1+a_2+...+a_n}{n}, \$$If $$f$$ is not convex we can sometimes prove:$$f(x)\ge f(a)+f'(a)(x-a)$$ If this manages to hold for all x, then summing up the inequality will give us the desired conclusion.

Then we have the following example: We want to show, given that $$a+b+c=3$$: $$\sum_{cyc}(\frac{18}{(3-c)(4-c)}-c^2) \ge 6$$Using the tangent line trick we get the following:$$\frac{18}{(3-c)(4-c)}-c^2 \ge \frac{(c+3)}{2}\Leftarrow \Rightarrow$$ $$c(c-1)^2(2c-9) \le 0$$then the proof claims "and the conclusion follows by summing", now, two things are really confusing to me, first of I don't understand how the "tangent line trick" is applyed here i.e. where does the $$\frac{c+3}{2}$$ comes from...

Apart from that our last inequality, i.e. $$c(c-1)^2(2c-9) \le 0$$ holds only for $$0 \le c \le \frac{9}{2}$$, so I'm confused since we don't have such bounds on c (nor on a and b), am I not getting this right or is the proof incomplete?

You can find everything I wrote here (page 5): Olympiad Inequalities by Evan Chen

The inequality, which you want to prove is wrong. Try $$c\rightarrow3^+$$.