LemmaStatement: AI-adaptedProof: AI-generatedprecheck passaudited 2026-08-02

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The first chip identifies the source of the statement or construction; the second identifies the source of its local proof or verification.

Literature-sourced: the exact statement appears in a cited source; only wording and notation differ.
AI-adapted: a semantically identical restatement of literature-sourced material, modulo indexing, notation, and boundary cases adopted by the library.
AI-generated: a genuinely novel statement formulated by AI, with no source for the claim itself.

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For a convex function and $x < y < z$ , the three secant slopes satisfy $s (x, y) \leq s (x, z) \leq s (y, z)$

Statement

Let $f : I \to R$ be convex on an interval and, for distinct $a, b \in I$ , write $s (a, b) : = (f (b) - f (a)) / (b - a)$ . If $x < y < z$ lie in $I$ , then

$s (x, y) \leq s (x, z) \leq s (y, z) .$

Facts & Assumptions

Given: A convex $f : I \to R$ and $x < y < z$ in $I$ .

[L1]

A function is convex when the convex-combination inequality holds for every weight in $[0, 1]$ (Convex, strictly convex, concave, strictly concave, and midpoint-convex real functions on an interval).

Proof

technique · direct

1.1

Put $λ = (z - y) / (z - x) \in (0, 1)$ ; then $y = λ x + (1 - λ) z$ , so convexity gives $f (y) \leq λ f (x) + (1 - λ) f (z)$ .

2.1

Multiplying this inequality by the positive number $z - x$ and rearranging gives $(z - y) (f (y) - f (x)) \leq (y - x) (f (z) - f (y))$ .

step 1.1algebra

3.1

Dividing step 2.1 successively by the positive products $(y - x) (z - x)$ and $(z - y) (z - x)$ gives $s (x, y) \leq s (x, z) \leq s (y, z)$ .

step 1.1step 2.1∎

Depends on

Convex, strictly convex, concave, strictly concave, and midpoint-convex real functions on an interval

Used by

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Sources

R. Gardner, Convex Functions, Notes 6.6 (standard reference, not scraped)

For a convex function and x<y<zx<y<z, the three secant slopes satisfy s(x,y)≤s(x,z)≤s(y,z)s(x,y)≤s(x,z)≤s(y,z)