Minimum value of $\frac{4}{x}+2x+10+\frac{3+x}{4x^2+1}$ If the minimum value of $\frac{4}{x}+2x+10+\frac{3+x}{4x^2+1}$ when $x>0$ is $\frac{p}{q}$ where $p,q\in N$ then find the least value of $(p+q).$
I can find the minimum value of $\frac{4}{x}+2x$ using $AM- GM$ inequality but that too comes irrational and when i tried to find the minimum value of $\frac{3+x}{4x^2+1}$ using first derivative test,i am facing difficulty.Please help me in solving this problem.

The derivative of $f(x) = \frac{4}{x}+2x+10+\frac{3+x}{4x^2+1}$ is
$$
\begin{align}
f'(x) &= \frac{-4}{x^2}+2+\frac{(4x^2+1)(1) - (3+x)(8x)}{(4x^2+1)^2}\\
&= \frac{-4}{x^2}+2+\frac{-4x^2-24x+1}{(4x^2+1)^2}\\
\end{align}
$$
So when $f'(x) = 0$, we have
$$
\begin{align}
-4(4x^2+1)^2 + 2x^2(4x^2+1)^2 + (-4x^2-24x+1)x^2 &= 0\\
32x^6-52x^4-24x^3-29x^2-4&=0
\end{align}
$$
There is a solution at $x \approx 1.577$, which can be found numerically.
