<s>[INST] 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?
 [/INST] 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 $a=k$
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