Showing that a function has a certain absolute minimum. Suppose we have the function $$f(x) = \frac{x}{p} + \frac{b}{q} - x^{\frac{1}{p}}b^{\frac{1}{q}}$$ where $x,b \geq 0 \land p,q > 1 \land \frac{1}{p}+\frac{1}{q} = 1$
I am trying to show that $b$ is the absolute minimum of $f$. 
I proceeded as follows:
$$\frac{df(x)}{dx} = \frac{1}{p} - \frac{x^{\frac{1}{p}-1}}{p} b^{\frac{1}{q}} = \frac{x - x^{\frac{1}{p}} b^{\frac{1}{q}}}{px}$$
Now I will look for critical points by searching for the zeros of this function.
$$\frac{x - x^{\frac{1}{p}} b^{\frac{1}{q}}}{px} = 0 \iff x - x^{\frac{1}{p}} b^{\frac{1}{q}} = 0 \iff x = x^{\frac{1}{p}} b^{\frac{1}{q}}$$. 
Now I can see that $b$ is a critical point. 
How ever when I continue my calculations to check whether there are any other critical points
$$x = x^{\frac{1}{p}} b^{\frac{1}{q}} \implies x^p = b^{\frac{p}{q}}x \implies x^{p-1} = b^{\frac{p}{q}} \implies x = b^{\frac{p}{(p-1)q}}$$
But this could not be equal to $b$, where did I go wrong?
 A: Note that $$\dfrac{p}{q(p-1)}=1$$ since $\dfrac1p + \dfrac1q = 1$. This is because, we have $\dfrac1q = 1 - \dfrac1p = \dfrac{p-1}p \implies \dfrac{p}{q(p-1)} = 1$.
A: This is an instance of the Weighted Arithmetic Mean-Geometric Mean Inequality. It uses the concavity of $\log(x)$ and can be generalized to
$$
\prod_{i=1}^nx_i^{\alpha_i}\le\sum_{i=1}^n\alpha_ix_i\tag{1}
$$
where $\alpha_i\ge0$ and $\sum\limits_{i=1}^n\alpha_i=1$.
$(1)$ follows from Jensen's Inequality because $\log(x)$ is concave. Just exponentiate the following instance of Jensen
$$
\sum_{i=1}^n\alpha_i\log(x_i)\le\log\left(\sum_{i=1}^n\alpha_ix_i\right)\tag{2}
$$
A: Let  $\dfrac{1}{p}=a$ and $\dfrac{1}{q}=k$
$ax+bk-x^a \cdot b^k=f(x)  $
$\dfrac{x + \dots x_{ath}+ b+ \dots b_{kth}}{a+k} \ge (x^ab^k)^{1/(a+k)}$ (By AM-GM inequality)
$x+ \dots x (a$ times)$=ax$ and $b+ \dots b (k$ times)$=bk$
$a+k=1 \implies ax+bk \ge x^a\cdot b^k \implies ax+bk-x^ab^k \ge 0$
Now you have $f(x) \ge 0$, the minimum is achieved when $x=b$ and $a=k$
