Young’s Inequality for Products

$$ab \leq \frac{a^p}{p} + \frac{b^q}{q}$$ where $\frac{1}{p} + \frac{1}{q} = 1$, with $p, q > 1$, and $a, b > 0$.

What the inequality is saying in plain English:

A product ab can be controlled by a sum of powers of a and b.

Why this matters
Products are often awkward to handle because they couple variables together while sums are usually easier because they keep the variables seperate. So this inequality lets you replace a difficult multiplicative term with something additive. When variables are separated, you can study each one independently. That matters because most mathematical tools work better on separated structure:

• integrals distribute over sums,
• derivatives distribute over sums,
• estimates can often be applied term-by-term,
• maxima/minima are easier to study when variables are uncoupled,
• error control is easier when each source of size appears in its own term.

Proof

See The tangent-line origin of Young’s inequality for how we got from concavity to the first inequality.

\[ \text{concavity of } x^{1/p} \;\to\; x^{1/p}\le \frac{x}{p}+\frac{1}{q} \;\to\; ab\le \frac{a^p}{p}+\frac{b^q}{q}. \]

Starting from the tangent-line inequality for the concave function \(x^{1/p}\),

$$ x^{1/p}\le \frac{x}{p}+\frac{1}{q}, \qquad \frac{1}{p}+\frac{1}{q}=1, $$

we derive Young’s inequality.

Assume first that

$$ a^p \le b^q. $$

Then

$$ \frac{a^p}{b^q}\le 1, $$

so the tangent-line inequality applies with

$$ x=\frac{a^p}{b^q}. $$

Substituting gives

$$ \left(\frac{a^p}{b^q}\right)^{1/p} \le \frac{1}{p}\frac{a^p}{b^q}+\frac{1}{q}. $$

Now simplify the left-hand side:

$$ \left(\frac{a^p}{b^q}\right)^{1/p} = \frac{(a^p)^{1/p}}{(b^q)^{1/p}} = \frac{a}{b^{q/p}}. $$

Using

$$ \frac{1}{p}+\frac{1}{q}=1 \quad\Longrightarrow\quad \frac{q}{p}=q-1, $$

we get

$$ b^{q/p}=b^{q-1}. $$

So the inequality becomes

$$ \frac{a}{b^{q-1}} \le \frac{a^p}{p\,b^q}+\frac{1}{q}. $$

Since \(\frac{a}{b^{q-1}}=ab^{1-q}\), this is

$$ ab^{1-q}\le \frac{a^p}{p\,b^q}+\frac{1}{q}. $$

Multiply both sides by \(b^q\):

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

Links to this note:
Hölder’s Inequality