# Purely “algebraic” proof of Young’s Inequality

Young’s inequality states that if $a, b \geq 0$, $p, q > 0$, and $\frac{1}{p} + \frac{1}{q} = 1$, then (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”:

so

Example. If $$pp$$ and $$qq$$ are positive rationals such that $$1p+1q=1\frac1p + \frac1q = 1$$, then for positive $$xx$$ and $$yy$$ $$xpp+yqq≥xy.\frac{x^p}p + \frac{y^q}q \ge xy.$$
Since $$1p+1q=1\frac1p + \frac1q = 1$$, we can write $$p=m+nmp = \frac{m+n}m$$, $$q=m+nnq = \frac{m+n}n$$ where $$mm$$ and $$nn$$ are positive integers. Write $$x=a1/px = a^{1/p}$$, $$y=b1/qy = b^{1/q}$$. Then $$xpp+yqq=am+nm+bm+nn=ma+nbm+n.\frac{x^p}p + \frac{y^q}q = \frac a{\frac{m+n}m} + \frac b{\frac{m+n}n} = \frac{ma + nb}{m + n}.$$
However, by the AM–GM inequality, $$ma+nbm+n≥(am⋅bn)1m+n=a1pb1q=xy,\frac{ma + nb}{m + n} \ge (a^m \cdot b^n)^{\frac1{m+n}} = a^{\frac1p} b^{\frac1q} = xy,$$ and thus $$xpp+yqq≥xy.\frac{x^p}p + \frac{y^q}q \ge xy.$$