<s>[INST] 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.
 [/INST] 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 
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