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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∈Q with p>1 and q>1 (Order on the rationals) be conjugate exponents, that is

1p+1q=1,equivalentlyq=p p−1 .

Then for all a,b∈R with a≥0 and b≥0,

ab  ≤  app+bqq,

with the rational powers of Rational powers ar of a positive base (its supplementary clause gives 0p=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 1q=1−1p=p−1p 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≥0.

[L1]

Weighted AM-GM with rational weights (Weighted AM-GM inequality with rational weights): for x0,x1>0 and rationals w0,w1≥0 with w0+w1=1, x0 w0x1 w1≤w0x0+w1x1.

[L2]

Rational power laws (Laws of rational exponents, Rational powers ar of a positive base): for u>0 and rationals r,s, ur>0 and (ur)s=urs, and u1=u; and 0r=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⋅(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→R is an order-preserving field homomorphism, so ι(1/p)=ι(p)−1>0 (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
1.1

The pair w0:=1/p, w1:=1/q is a legitimate system of rational weights: both are rational and positive because p,q>1>0, and w0+w1=1 by hypothesis.

givenL3L4
1.2

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

givenL2L3L4
1.3

For a>0 and b>0, which is the only case in which this step is used, the left-hand factors simplify: (ap)1/p=ap⋅(1/p)=a1=a and, in the same way, (bq)1/q=b.

L2L3
2.1

Assume now a>0 and b>0, and put x0:=ap>0 and x1:=bq>0; applying weighted AM-GM with the weights of step 1.1 gives (ap)1/p(bq)1/q≤1pap+1qbq.

step 1.1L1L2
3.1

Substituting, ab≤app+bqq for all a,b>0, and together with the degenerate cases this proves the inequality for all a,b≥0.

step 2.1step 1.3step 1.2∎

Depends on

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