TITLE: Purely "algebraic" proof of Young's Inequality
QUESTION [49 upvotes]: Young's inequality states that if $a, b \geq 0$, $p, q > 0$, and $\frac{1}{p} + \frac{1}{q} = 1$, then $$ab\leq \frac{a^p}{p} + \frac{b^q}{q}$$ (with equality only when $a^p = b^q$).  Back when I was in my first course in real analysis, I was assigned this as homework, but I couldn't figure it out.  I kept trying to manipulate the expressions algebraically, and I couldn't get anywhere.  But every proof that I've seen since uses calculus in some way to prove this.  For example, a common proof is based on this proof without words and integration.  The proof on Wikipedia uses the fact that $\log$ is concave, which I believe requires the analytic definition of the logarithm to prove (correct me if I'm wrong).
Can this be proven using just algebraic manipulations?  I know that that is a somewhat vague question, because "algebraic" is not well-defined, but I'm not sure how to make it more rigorous. But for example, the proof when $p = q = 2$ is something I would consider to be "purely algebraic":
$$0 \leq (a - b)^2 = a^2 + b^2 - 2ab,$$ so $$ab \leq \frac{a^2}{2} + \frac{b^2}{2}.$$

REPLY [20 votes]: With $\dfrac1p+\dfrac1q=1$, $u=x^p$, $v=y^q$, and $1+pt=u/v$, the following are equivalent:
$$
\begin{align}
xy&\le\frac{x^p}{p}+\frac{y^q}{q}\\
u^{1/p}v^{1/q}&\le\frac{u}{p}+\frac{v}{q}\\
(u/v)^{1/p}&\le\frac{u/v}{p}+\frac{1}{q}\\
(1+pt)^{1/p}&\le\frac{1+pt}{p}+\frac{1}{q}\\
1+pt&\le(1+t)^p\tag{1}
\end{align}
$$
Where $(1)$ is the rational version of the Bernoulli inequality, proven below.

Bernoulli's Inequality for Rational Exponents
Using the integral version of the Bernoulli Inequality, proven at the end of this answer, we get that for $x\gt-n$,
$$
\begin{align}
\frac{\left(1+\frac{x}{n+1}\right)^{n+1}}{\left(1+\frac{x}{n}\right)^n}
&=\left(\frac{(n+x+1)n}{(n+1)(n+x)}\right)^{n+1}\frac{n+x}{n}\\
&=\left(1-\frac{x}{(n+1)(n+x)}\right)^{n+1}\frac1{1-\frac{x}{n+x}}\\
&\ge\left(1-\frac{x}{n+x}\right)\frac1{1-\frac{x}{n+x}}\\[8pt]
&=1\\[8pt]
\left(1+\frac{x}{n+1}\right)^{n+1}
&\ge\left(1+\frac{x}{n}\right)^n\tag{2}
\end{align}
$$
where the inequality is strict if $x\ne0$ and $n\ge1$.
Applying induction with $(2)$, we get that for $x\gt-m$ and integers $n\ge m$,
$$
\left(1+\frac{x}{n}\right)^n\ge\left(1+\frac{x}{m}\right)^m\tag{3}
$$
Letting $t=\frac{x}{n}\gt-\frac{m}{n}$ and taking $m^\text{th}$ roots gives
$$
\left(1+t\right)^{n/m}\ge1+\frac{n}{m}t\tag{4}
$$
Note that $(4)$ is trivially true for $-1\le t\le-\frac{m}{n}$. Thus, for all $t\ge-1$ and rational $p\ge1$,
$$
\left(1+t\right)^p\ge1+pt\tag{5}
$$
where the inequality is strict if $t\ne0$ and $p\gt1$.

Negative Exponents
As shown in $(2)$, both $\left(1+\frac xn\right)^n$ and $\left(1-\frac xn\right)^n$ are increasing in $n$, as long as $|x|\le n$. Thus, for $m=\max(k,n)$,
$$
\begin{align}
\left(1+\frac xk\right)^k\left(1-\frac xn\right)^n
&\le\left(1+\frac xm\right)^m\left(1-\frac xm\right)^m\\
&=\left(1-\frac{x^2}{m^2}\right)^m\\[3pt]
&\le1\tag6
\end{align}
$$
Thus, substituting $x\mapsto kx$, and taking $n^\text{th}$ roots, we get
$$
(1+x)^{k/n}\left(1-\tfrac knx\right)\le1\tag7
$$
Finally, setting $p=\tfrac kn$ yields
$$
1-px\le(1+x)^{-p}\tag8
$$
where the inequality is strict when $x\ne0$ and $p\gt0$.