An upper bound for a function I am trying to find an upper bound $b\ge f(x)~\forall x\ge0$ for the following function:
$$f(x)=\frac{x}{(w+ux^2)^2},$$
where $w,u>0$ are parameter values. I am interested in the positive domain $0\le x<\infty$.
As $x\to0$, $f(x)\to 0$, and as $x\to\infty$, $f(x)\to0$. Since $f(x)>0$ for all $x>0$ and since it is continuous, a maximum exists. 
If the function were simpler, I would solve for the arg max $x^*$ and substitute in the function to get the exact maximum (smallest upper bound) $f(x^*)$. However, setting $f'(x)=0$ yields a fourth degree polynomial, so substituting the exact $x^*$ that maximizes $f(x)$ is not practical (though technically possible).
How can I determine an upper bound, even if it is larger than the actual maximum? I am looking for an analytical technique, i.e., an approach that gives me a condition as a function of the parameter values $w,u$. Ideally, the bound would be close to the maximum. Thanks!
 A: Using Clayton's hint let's compute the derivative of $f$ and let's set it equal to $0$:
$$(w + ux^2)^2 - 4x^2u(w + ux^2) = 0.$$
Notice that I have already simplified the denominator. We can rewrite this as $$(w + ux^2)(w + ux^2 - 4ux^2) = 0.$$
We can further simplify since $w + ux^2 > 0$ by assumption. Then we are left with the following second order equation $$x^2 = \frac{w}{3u}$$ the positive root is the value we are looking for (as mentioned in the question, we are interested in $x \in [0,\infty)$).
$$x_{max} = \sqrt{\frac{w}{3u}}$$
$$f(x_{max}) = \frac{3\sqrt{3}}{16w\sqrt{uw}}$$
A: The function is of the form $x/p(x)$ where $p(x)=(v+ux^2)^2$ is a fourth degree polynomial and $v=e+t$. This function is maximum when $p(x)-xp'(x)=0$ This translates into
$$(v+ux^2)^2-4ux^2(v+ux^2)=0$$
We have $u\gt 0$ and $v\gt 0$ this means $(v+ux^2)\gt 0$ so our equation above becomes
$$v-3ux^2=0$$
This gives for the positive solution $x^*=\sqrt{{{e+t}\over 3u}}$ and the bound you're looking for is $f(x^*)$
A: Given that the logarithm function is monotonic, you could differentiate the logarithm of $f(x)$ (a natural base is assumed) w.r.t $x$ to obtain arg max $x^*$, 
$$\begin{align}\frac{d\log(f(x))}{dx}&=\frac{d}{dx}\left(\log(x)-2\log(w+ux^2)\right)\\&=\frac{1}{x}-\frac{4ux}{w+ux^2}\end{align}$$
Setting the derivative to $0$ and solving for $x$ will lead to a quadratic is $x$ which is easier to solve than the quartic were you to differentiate $f(x)$.   
