Purely "algebraic" proof of Young's Inequality
This proof is from "Mathematical Toolchest" published by the Australian Mathematics Trust (image).
Example. If $p$ and $q$ are positive rationals such that $\frac1p + \frac1q = 1$, then for positive $x$ and $y$ $$\frac{x^p}p + \frac{y^q}q \ge xy.$$
Since $\frac1p + \frac1q = 1$, we can write $p = \frac{m+n}m$, $q = \frac{m+n}n$ where $m$ and $n$ are positive integers. Write $x = a^{1/p}$, $y = b^{1/q}$. Then $$\frac{x^p}p + \frac{y^q}q = \frac a{\frac{m+n}m} + \frac b{\frac{m+n}n} = \frac{ma + nb}{m + n}.$$
However, by the AM–GM inequality, $$\frac{ma + nb}{m + n} \ge (a^m \cdot b^n)^{\frac1{m+n}} = a^{\frac1p} b^{\frac1q} = xy,$$ and thus $$\frac{x^p}p + \frac{y^q}q \ge xy.$$
Yes, at least for rational $p, q$. There is a general statement here, which can be summarized as follows:
Every polynomial inequality is a consequence of the trivial inequality $x^2 \ge 0$.
In more detail, to prove Young's inequality for general $p, q$, it suffices by a continuity argument to prove it for $p, q$ rational. By making an appropriate substitution of the form $a = x^n, b = y^m$ where $n, m$ are even integers, Young's inequality for a fixed choice of rational $p, q$ becomes equivalent to the statement that a certain polynomial with real coefficients always takes on non-negative real values when fed real inputs.
By Artin's solution to Hilbert's 17th problem, a polynomial with this property is a sum of squares of rational functions. An expression of this polynomial as a sum of squares of rational functions constitutes a proof of the corresponding case of Young's inequality from repeated application of the trivial inequality.
I don't see how you could avoid analysis for irrational $p, q$, since you can't even define the relevant functions without analysis.
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 $p\gt0$.