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 $$ab \leq \frac{a^2}{2} + \frac{b^2}{2}.$$
Best Answer
This proof is from "Mathematical Toolchest" published by the Australian Mathematics Trust (image).