# Is there a means of analytically proving the following identity?

Okay, so before I begin, my background is more in the world of applied, rather than pure, mathematics, so this question is motivated by a physics problem I'm looking at just now. Mathematically, it boils down to looking for a minimum of a real valued function of a single, positive, real-valued variable, $u$. The part of differentiating the function and finding the condition for a stationary point is straightforward. I can, somewhat heuristically, convince myself that the function must be a minimum, but this is speaking from a physical standpoint. The condition for a minimum rests on the truth (or otherwise) of the following inequality:

$\cosh^2(u) \geq \frac{u^2}{8}.$

Now, I can plot this on a graph, and it clearly holds up for the range of values over which the plot is carried out (and would appear to be true in general). I can then say that it's true for all sensible values of the physical parameter $u$, which is simply a combination of a bunch of physical constants and a variable. But obviously I cannot draw an infinite graph, and would rather like a concrete proof to show that this is true for all positive, real-valued $u$. Is there a method that is recommended for dealing with such a problem?

I don't expect a full solution, as I realise it's quite elementary-looking, and once pointed in the right direction I could no doubt take care of it myself. I'm just curious to know what the best analytic method would (in your opinion) be to deal with it so the proof looks a wee bit neater and more rigorous anyway.

• Thanks to all three of you for your answers, that deals with it pretty comprehensively, I think :) – Ben Snow Nov 19 '13 at 14:25

Expanding in a Taylor series, $cosh^2(x)=1+x^2+$ (higher order even terms) $\geq 1+x^2 \geq \frac{x^{2}}{8}$.

Notice that your inequality is equivalent to $\cosh^2(u) - \frac{u^2}{8} \geq 0$.

So any points which satisfy $\cosh^2(u) - \frac{u^2}{8} = 0$ will be bounds of the intervals for which the inequality is true.

Because there is no $u \in \Bbb R$ such that $\cosh^2(u) - \frac{u^2}{8} = 0$, the inequality is either always true or always false for all $u \in \Bbb R$.

Since a test case satisfies the inequality, the inequality is then true $\forall u \in \Bbb R$.

• Very neat, and straightforward to boot... it shows that I have no real pure maths knowledge, I guess... – Ben Snow Nov 19 '13 at 14:28

A way to do this is to define a function $f: f(u)=\cosh^2(u)-u^2/8$ and to show that $\forall u, f(u)\geq 0$.

And how to do this? Take the derivative of $f(u)$, which is $2 \cosh(u) \cdot \sinh(u)- u/4$, find the minimum value of $f$ according to the value which makes the derivative $0$. As it's greater or equal than $0$, you're done.

• I'm actually annoyed that I didn't think to do this myself now... thanks for that. – Ben Snow Nov 19 '13 at 14:30
• 1.: The derivative is $2 cosh(u)sinh(u)-u/4$ (you have $\cdot$ instead of $-$). 2.: This argument is only correct if you also show that the limits of your function as $x$ goes to $\pm\infty$ are positive or $\infty$. As an example where the local minimum is positive and the function is not always $>0$ take $1-x^2$. – MichalisN Nov 19 '13 at 14:35