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.
 A: While the answers that have been proposed already are great, I think that rather than trying to find the positive roots of the following polynomial :
\begin{align}
32x^6-52x^4-24x^3-29x^2-4&=0,
\end{align}
your problem could be solved more easily by using the rational root theorem.

Rational root theorem :
  Let $P(x)=\sum\limits_{k=0}^n a_kx^k$ where every $a_k$ is an integer and $a_n\neq0$. If $\alpha$ is a rational root of $P(x)$, then $\alpha=\frac{p}{q}$ where :
  
  
*
  
*$p$ divides $a_0$,
  
*$q$ divides $a_n$,
  
*$\gcd(p,q)=\pm1$.
  

In your case, you are looking for a positive rational root hence we can narrow the possible values to $4$, $2$, $1$, $\frac{1}{2}$, $\frac{1}{4}$, $\frac{1}{8}$, $\frac{1}{16}$, $\frac{1}{32}$.
As none of these values is a root of our polynomial, you can conclude that the minimum of your function is not reached at a positive rational.
A: 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.
A: HINT: There is another approach using calculus
let $$y=\frac{4}{x}+2x+10+\frac{3+x}{4x^2+1}$$
$$\frac{dy}{dx}=-\frac{4}{x^2}+2+\frac{(4x^2+1)(1)-(3+x)(8x)}{(4x^2+1)^2}$$
Now, for maxima or minima, put $$\frac{dy}{dx}=0$$
$$\frac{(4x^2+1)(1)-(3+x)(8x)}{(4x^2+1)^2}-\frac{4}{x^2}+2=0$$
Since, $x>0$ then find the positive roots of the equation for minima & proceed 
