Valid proof of Young's inequality? Part of an exercise to prove Holder's inequality in Rudin involves proving Young's Inequality... That is, given $\frac{1}{p}+\frac{1}{q} = 1$, prove 
$$ab \leqslant \frac{a^p}{p} + \frac{b^q}{q}.$$   
Here's my attempt at a proof:    
Let $$f(x) = \frac{x^p}{p} + \frac{b^q}{q} -bx$$
  then, $$f'(x) = x^{p-1} -b$$ so that $f$ attains a minimum at $$x=b^{\frac{1}{p-1}}.$$ Since $\frac{p}{p-1} = q$, this is equivalent to saying that $f$ attains its minimum when $x^p = b^q$. Now, we need show that $$f(b^{\frac{1}{p-1}}) = 0.$$ We compute
$$
\begin{align}
f(b^{\frac{1}{p-1}}) &= \frac{(b^{\frac{1}{p-1}})^p}{p} + \frac{b^q}{q} - bb^{\frac{1}{p-1}} \\
&= \frac{b^q}{p} + \frac{b^q}{q} - b^{\frac{1}{p-1} +1} \\
&= \frac{b^q(p+q)}{pq} - b^{\frac{1}{p-1} +1} \\
&= b^q - b^{\frac{1}{p-1} +1} \\
&= b^q - b^q \\
& = 0
\end{align}$$
where $b^{\frac{1}{p-1} +1} = b^q$ since $$\frac{1}{p-1} +1 = \frac{1}{p-1} +\frac{p-1}{p-1} = \frac{p}{p-1} = q.$$ Thus, $f(x) = 0$ only when $x^p = b^q$. This is the global minimum of $f$ since $f^{''} \geq 0$ and analysis of concavity. Therefore, if $x > b^{\frac{1}{p-1}}$, $f(x) > 0$. That is, if $x^p > b^q$, the inequality holds. A similar analysis for $g(y) = \frac{a^p}{p} + \frac{y^q}{q} -ay$ shows that $g(y) > 0$ if $y^q > a^p$. Combining these two statements yields that, if $a^p \neq b^q$, the inequality holds, so we're done.  
Is this a valid proof?
If so, if anyone could provide any alternative proofs, I'd be more than interested to see them. 

As mentioned above, the standard proof uses concavity. You said you were interested in other proofs so here is how I learned it: 
We will first assume $a,b \ne 0$ as this then becomes trivial. An equivalent statement to the one we want to prove is the following:
$$\frac{a}{b^{q-1}} \le \frac{1}{p} \cdot \frac{a^p}{b^q} + \frac{1}{q}$$
(divide Young's Inequality by $b^q$).
Let $t = \frac{a^p}{b^q}$. Then $t^{\frac{1}{p}} = \frac{a}{b^{q-1}}$, since $\frac{q}{p} = q - 1$ , we want to show that 
$$t^{\frac{1}{p}} - \frac{1}{p} t \le \frac{1}{q}$$
Let $f(t) = t^{\frac{1}{p}} - \frac{1}{p} t$. Then $f(1) = 1 - \frac{1}{p} = \frac{1}{q}$. If we can show that this is the maximum of $f(t)$ when $t > 0$, we are done. 
$$f'(t) = \frac{1}{p} t^{\frac{1}{p} - 1} - \frac{1}{p} = \frac{1}{p} \left(t^{-\frac{1}{q}} - 1\right)$$
When $0 < t < 1$, we see that $f'(t) > 0$ and similarly when $t > 1$, we see that $f'(t) < 0$. Thus, $\frac{1}{q}$ is a local max of $f(t)$ and so we have shown that 
$$t^{\frac{1}{p}} - \frac{1}{p} t \le \frac{1}{q}$$
Now substituting back in $t = \frac{a^p}{b^q}$ we get Young's Inequality to come out with some algebraic manipulations. 
