Intuition behind $(\{a, b, p, q\} \subset \mathbb{R}^{+} \;\wedge\;\; 1/p +1/q = 1) \Rightarrow a^p/p + b^q/q \geq ab$

Question

If $p$ and $q$ are positive real numbers with¹ $$ \frac{1}{p} + \frac{1}{q} = 1,$$ then, for any non-negative real numbers $a, b$,

$$ \frac{a^p}{p} + \frac{b^q}{q} \geq ab$$

My textbook offers a totally unenlightening (albeit fairly clear) proof of this fact.²

What's the intuition behind it?

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¹ _{Is there a name for pairs of positive numbers $p, q$ in that satisfy $\frac{1}{p} + \frac{1}{q} = 1\;$ ?}

² _{Said textbook does not give any name for this theorem, or for the inequality. I'd love to know what they are.}

Answer 1

The inequality is called Young's inequality, and $(p,q)$ is called a Hölder pair.

Answer 2

Think about the log function. It's concave so $$\log(\lambda x + (1 - \lambda)y)) \gt \log(x) + (1 - \lambda)\log(y).$$ Now take the right values for $x$, $y$ and $\lambda$ and it's a go.

Answer 3

If you are familiar with the AM-GM inequality, this is a consequence of the weighted version:

If $\alpha_1,..,\alpha_n, a_1,..,a_n$ are positive then

$$\frac{\alpha_1a_1+..+\alpha_na_n}{\alpha_1+..+a_n}\geq \sqrt[\alpha_1+..+\alpha_n]{a_1^{\alpha_1}a_2^{\alpha_2}...a_n^{\alpha_n}}$$

Now, if you set $n=2, \alpha_1=\frac{1}{p}, \alpha_2=\frac{1}{q}, a_1=a^p, a_2 =b^q$ you get your equality.

Comment 1: The weighted AM-Gm is an immediate consequence of the concavity of log function, already mentioned.

Comment 2: For rational weights, the weighted version is just the standard AM-GM with some numbers equal. One can prove it in the rational case from the standard one, and then in general by simple "any real number is a limit of rationals" argument.