A proof of Young's inequality I need to prove that if $1 < p < \infty$ and $a, b \geqslant 0$ then $$ ab \leqslant \frac{a^{p}}{p} + \frac{b^{q}}{q}$$ 
where $\frac 1p+\frac 1q=1$.
I fix $b$ and maximize the function $f(a) = ab - \frac{a^{p}}{p}$, but the maximum I find is $b^{q}$ with $q = \frac{1}{p-1}$. I have no idea how to get $q$ in the denominator. Any suggestion?
 A: Let's prove that $f(x)=\frac {x^p}p+\frac{y^q}q-xy \geqslant 0$.
I'm going to calculate its minima for $x,y>0. $
$$f'(x)=x^{p-1}-y=0$$
So
$$x=y^{\frac 1{p-1}}=y^{q-1}$$
since $(p-1)(q-1)=1$ by assumption.
Hence
$$f(x) \geqslant \frac{y^{p(q-1)}}p+\frac{y^q}q-y^{(q-1)+1}=\frac{y^q}p-(1-\frac 1q)y^q=0$$
Q.E.D
A: This is standard theorem of means. If $(x + y) = 1$ then $a^{x}b^{y} \leq ax + by$. In this equation replace $a$ by $a^{p}$ and $b$ by $b^{q}$ and $x = 1/p$, $y = 1/q$. The condition $\dfrac{1}{p} + \dfrac{1}{q} = 1$ is missing in the question.
The proof of general theorem of means follows by the concavity of function $f(x) = \log x$. Clearly $f''(x) = -1/x^{2} < 0$ therefore the function $f(x) = \log x$ is concave (meaning the chord connecting two points on graph of $y = \log x$ will be below the part of the graph between those two points). Algebraically this means that if $ x, y \in [0, 1]$ with $x + y = 1$ then $f(ax + by) \geq xf(a) + xf(b)$. Thus we get $\log(ax + by) \geq x\log a + y\log b$ or $a^{x}b^{y} \leq ax + by$
