Minimize a particular function in one variable

Question

For given $a,b$, what is the minimum value of the following expression?

$$ \frac{a}{x^2+b}+x,\qquad x>=0, a>0,b>0 $$

1. Differentiating the above gives a messy polynomial.
2. I tried plugging this into wolframalpha - It gives back the above polynomial.

I am looking for a reasonable scaling of the lower bound for the expression for $a,b \rightarrow \infty$

Answer 1

This is a possible approach.

$$ \frac{a}{x^2+b} + x \\ = \frac{a}{x^2+b} + x +\sqrt{b} - \sqrt{b} \\ > \frac{a}{x^2+b} + \sqrt{x^2+b} - \sqrt{b}\\ >a^{1/3} - \sqrt{b} $$

Can we improve the lower-bound?

Answer 2

Differentiating, we get $$\frac{-2 x (2 a x^2-(b+x^4)^2)}{(b+x^4)^2}$$ Setting the derivative to be zero, we get the following $$-2 x (2 a x^2-(b+x^4)^2) = 0$$ Assuming $x$ can't be zero (make sure to check to see if $x=0$ is a minimum when finished) we get that $$2 a x^2 = (b+x^4)^2$$ Can you solve from here? You can use the quartic formula to get the answer exactly (let $u=x^2$ first)