find the $\text{Min}(f)=?$

Question

let $f(x)=\sqrt{x}+\dfrac{1}{\sqrt{x}}+\sqrt{k-x}+\dfrac{1}{\sqrt{k-x}} \ \ : k \in \mathbb{R^+}$

then find the $\text{Min}(f)=?$

Answer 1

Assuming $f$ is defined for $\{x \mid x \in \mathbb{R}, 0<x<k\}$ and that you want a global minimum over all $k$.

Note that $f(x)>0$ for all $x$ in the domain of $f$. Thus, by AM-GM, $\dfrac{\sqrt{x}+\frac{1}{\sqrt{x}}+\sqrt{k-x}+\frac{1}{\sqrt{k-x}}}{4} \geq \sqrt[4]{1} = 1$, so $f(x)=\sqrt{x}+\frac{1}{\sqrt{x}}+\sqrt{k-x}+\frac{1}{\sqrt{k-x}}\geq 4$. Therefore, the minimum value of $f$ is $4$, with equality at $\sqrt{x}=\frac{1}{\sqrt{x}} = \sqrt{k-x}$, so $x=1$, $k=2$.

For specific values of $k$, there is no clean way I can see to do it, but from Wolfram Alpha, we get some interesting results. For $0 < k \leq 6$, we have the only minimum occurring at $x=\frac{k}{2}$, giving $f(x)=\sqrt{2k}+\sqrt{\frac{2}{k}}$. However, for $k > 6$, we have the minima occuring at $x=\frac{k}{2} \pm \frac{1}{2} \sqrt{k^2-4k-2\sqrt{4k+1}-2}$. The corresponding values for $f$ is too convoluted for me to calculate. :P