8

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:$$ \frac{18}{(3-c)(4-c)}-c^2 \ge \frac{(c+3)}{2}\iff$$ $$ c(c-1)^2(2c-9) \le 0$$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

Daniele Tampieri
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Spasoje Durovic
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2 Answers2

2

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.

Michael Rozenberg
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1

After

$$\sum_{cyc}(\frac{18}{(3−c)(4−c)}−c^2)≥6$$

I believe it should be something like

$$a = \frac{a + b + c}3 = 1;f(c) = \frac{18}{(3−c)(4−c)}−c^2$$ $$g(c) =\frac{df}{dc} = 18(\frac{7-2c}{(3-c)^2(4-c)^2})$$ $$f(a) = 2; g(a) = \frac{1}{2}; (c-a) = (c-1)$$ So, we first prove that the function is convex and use the Tangent Trick Formula from the book:

1) I found it frustrating to find the double derivative of a function manually, so I just used graphical calculator

2) Now we know that it's convex, we see that the sign of the equation is $\ge$ $$\frac{18}{(3−c)(4−c)}−x^2 \ge \frac{1}{2}(c-1) + 2$$

Summing everything up gives us:

$$\sum_{cyc}(\frac{18}{(3−c)(4−c)}−c^2)≥\frac{1}{2}\sum_{cyc}(c-1)+\sum_{cyc}2 = \frac{1}{2}(a+b+c-3)+2*3 = 6$$

I apologise for the way I tried to explain it, it's my very first time exploring Olympic math and commenting on this platform.

Nurasik12
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