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.