Minimum value of $\frac{4}{x}+2x+10+\frac{3+x}{4x^2+1}$

Question

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.

Answer 1

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

Answer 2

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.