TheoremStatement: AI-adaptedProof: AI-adaptedprecheck passjudge pass (z-ai/glm-5.2)verified 2026-07-26 (claude-opus-5)

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Young's inequality for products (rational conjugate exponents)

Statement

Let $p, q \in Q$ with $p > 1$ and $q > 1$ (Order on the rationals) be conjugate exponents, that is

$\frac{1}{p} + \frac{1}{q} = 1, equivalently q = \frac{p}{p - 1} .$

Then for all $a, b \in R$ with $a \geq 0$ and $b \geq 0$ ,

$a b \leq \frac{a^{p}}{p} + \frac{b^{q}}{q},$

with the rational powers of Rational powers $a^{r}$ of a positive base (its supplementary clause gives $0^{p} = 0$ , since $p > 0$ ) and with the rationals $p, q$ acting on $R$ through the canonical embedding (The unique embedding of ℚ into an ordered field).

The conjugate exponent is rational because $p$ is. From $\frac{1}{q} = 1 - \frac{1}{p} = \frac{p - 1}{p}$ one gets $q = p / (p - 1)$ , a quotient of rationals with nonzero denominator (Arithmetic on the rationals), hence a rational. This is the observation that keeps Hölder and Minkowski inside the rational world on this page.

Facts & Assumptions

Given: Rationals $p, q > 1$ with $1 / p + 1 / q = 1$ , and reals $a, b \geq 0$ .

[L1]

Weighted AM-GM with rational weights (Weighted AM-GM inequality with rational weights): for $x_{0}, x_{1} > 0$ and rationals $w_{0}, w_{1} \geq 0$ with $w_{0} + w_{1} = 1$ , $x_{0}^{w_{0}} x_{1}^{w_{1}} \leq w_{0} x_{0} + w_{1} x_{1}$ .

[L2]

Rational power laws (Laws of rational exponents, Rational powers $a^{r}$ of a positive base): for $u > 0$ and rationals $r, s$ , $u^{r} > 0$ and $(u^{r})^{s} = u^{r s}$ , and $u^{1} = u$ ; and $0^{r} = 0$ for rational $r > 0$ .

[L3]

Rational arithmetic (The rationals as equivalence classes of pairs of integers, Arithmetic on the rationals, Order on the rationals): $q = p / (p - 1)$ is a quotient of rationals with nonzero denominator, hence rational, and $p \cdot (1 / p) = 1$ . Moreover $Q$ is itself a totally ordered field (The rationals form a totally ordered field), which is what licenses the order arithmetic used on $p$ and $q$ ; being an ordered field it has $1 > 0$ (The multiplicative identity is positive, which is where that fact is proved: The rationals form a totally ordered field states totality, compatibility with addition and closure of the positives under multiplication, and not this), so $p > 1 > 0$ gives $p > 0$ by transitivity and hence $1 / p > 0$ , and likewise $1 / q > 0$ (Inverses of positives are positive, and reciprocation reverses order, claim 1, applied in $Q$ ).

[L4]

The embedding $ι : Q \to R$ is an order-preserving field homomorphism, so the field image of the rational $1 / p$ is $p^{- 1} > 0$ , with p in that inverse denoting its field image (The unique embedding of ℚ into an ordered field, Inverses of positives are positive, and reciprocation reverses order, Sign rules for products and monotonicity of multiplication).

Proof

technique · direct

1.1givenL3L4

The pair $w_{0} : = 1 / p$ , $w_{1} : = 1 / q$ is a legitimate system of rational weights: both are rational and positive because $p, q > 1 > 0$ , and $w_{0} + w_{1} = 1$ by hypothesis.

1.2givenL2L3L4

Degenerate cases: if $a = 0$ then the left-hand side is $0$ while the right-hand side is $0^{p} / p + b^{q} / q = b^{q} / q \geq 0$ , since $0^{p} = 0$ for $p > 0$ and $b^{q} \geq 0$ ; the case $b = 0$ is symmetric, so the inequality holds whenever $a = 0$ or $b = 0$ .

1.3L2L3

For $a > 0$ and $b > 0$ , which is the only case in which this step is used, the left-hand factors simplify: $(a^{p})^{1 / p} = a^{p \cdot (1 / p)} = a^{1} = a$ and, in the same way, $(b^{q})^{1 / q} = b$ .

2.1step 1.1L1L2

Assume now $a > 0$ and $b > 0$ , and put $x_{0} : = a^{p} > 0$ and $x_{1} : = b^{q} > 0$ ; applying weighted AM-GM with the weights of step 1.1 gives $(a^{p})^{1 / p} (b^{q})^{1 / q} \leq \frac{1}{p} a^{p} + \frac{1}{q} b^{q}$ .

3.1step 2.1step 1.3step 1.2∎

Substituting, $a b \leq \frac{a^{p}}{p} + \frac{b^{q}}{q}$ for all $a, b > 0$ , and together with the degenerate cases this proves the inequality for all $a, b \geq 0$ .

Young's inequality for products (rational conjugate exponents)

Statement

Facts & Assumptions

Proof

Depends on

Dependency tree · 2 levels

Sources