Finding the best constant in an inequality. I want to find the smallest constant C such that for all x and y we have$$\frac{\cosh(x)e^y + e^{-y}}{e^{Cx^2}+1} \leq e^{Cy^2}$$
Playing around with a graphing utility online, it looks like $C \approx .6$
Is there a good analytic way to approach these types of questions?
I don't know if it helps, but when $x = 0$ we get $\cosh(y) \leq e^{Cy^2}$ which is true when $C \geq .5$
 A: First, note the following: We have to optimize the globally minimal difference between two multivariable functions, subject to a constraint. Such problems often require a sufficient amount of brute force computation to render an analytical approach impractical.
Generalizing the problem a bit, we have a global inequality between two functions $f:\mathbb{R}^{2} \rightarrow \mathbb{R}^{2}$ and $g:\mathbb{R}^{2} \rightarrow \mathbb{R}^{2}$: 
$$f_{C}(x,y) \leq g_{C}(x,y),\ \forall (x,y) \in \mathbb{R}^{2},$$
where the subscript denotes the particular constant $C$ used. We can simplify by defining a function $h_{C}(x,y) = g_{C}(x,y)-f_{C}(x,y)$, and require that 
$$h_{C}(x,y) \geq 0, \ \forall (x,y) \in \mathbb{R}^{2}.$$
Let's not forget about the end goal of this model: We want the minimal constant $C_{min}$ such that the above is satisfied. Define a function $H:\mathbb{R} \rightarrow \mathbb{R}$ which maps $C$ to the global min of $h_{C}$:
$$H(C) = \min_{(x,y)\in \mathbb{R}^{2}}\{h_{C}(x,y)\}.$$
Then the entire problem can be formulated as 
$$C_{min} = \min_{C \in \mathbb{R}}\{C\},$$
subject to 
$$H(C_{min}) \geq 0.$$
Why is this approach often impractical? First of all, finding $H(C)$ and then minimizing $H(C)$ are tasks which contain several messy steps. Let's just start with the first: Deriving $H(C)$.
To find $H(C)$, we have to find the global minimum $(x_{min},y_{min})$ for $h_{C}$ in terms of $C$, where $h_{C}(x,y)$ for your problem is 
$$h_{C}(x,y) = e^{Cy^2}-\frac{\cosh(x)e^y + e^{-y}}{e^{Cx^2}+1}.$$
On the first calculation (solving $\frac{\partial h_{C}}{\partial x} = 0, \frac{\partial h_{C}}{\partial y}=0$) even Wolfram Alpha runs out of allowed time, which is usually a good indicator that the  problem's brute force requires computational methods. 

