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 $c\rightarrow3^+$.

For positive variables it's true and you wrote the proof.

By the way, you can not use the theorem 2.8 in the contest, otherwise, you need to write a proof of this theorem, which is not so easy. I mean the theorem from this book.


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