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April 20th, 2015, 01:10 PM   #1
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Tangent line trick

If $a,b,c \in \mathbb{R}$ such that $a+b+c=3$, prove that
$18\sum_{cyc}\frac{1}{(3-c)(4-c)}+2(ab+ac+bc) \geq 15$

So a solution I saw to this problem said:
First replace $2(ab+ac+bc)$ with$9-a^2-b^2-c^2$ to get
$\sum_{cyc}\frac{18}{(3-c)(4-c)}-c^2 \geq 6$

After that it said applying the "tangent line trick" you can obtain
$\frac{18}{(3-c)(4-c)}-c^2 \geq \frac{c}{2}+\frac{3}{2}$ which yields $c(c-1)^2(2c-9) \leq 0$, and then summing will yield the desired inequality

However, I'm not sure how to apply that here, as what function did the solution to apply the tangent line trick to (it didn't say, all it said was what resulted from it)? Is it $f(x)=\frac{18}{(3-x)(4-x)}-x^2$? How would that yield $\frac{18}{(3-c)(4-c)}-c^2 \geq \frac{c}{2}+\frac{3}{2}$? Because if you take the tangent line at $x=0$, you do get the intercept is $\frac{3}{2}$, but the slope would be $f'(0)=\frac{7}{8}$, not $\frac{1}{2}$.

Quick explanation of the "tangent line trick"
Apparently, the tangent line trick can be applyed to inequalities containing convex functions (or if you have bounds on what the variables can be the function just needs to be convex over that interval), stating that a line drawn tangent to the function at a certain point will always be "lower" than the graph (every point on the graph will be above or equal to - at the tangent point- the corresponding point on the line), and the opposite holds for concave functions.

Last edited by USAMO Reaper; April 20th, 2015 at 01:54 PM.
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