TheoremStatement: 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.

These labels describe origin, not correctness: citations and verification chips remain separate evidence.

The two-point convexity inequality for the exponential function

Statement

For all $x, y \in R$ and $0 \leq t \leq 1$ , $\exp ((1 - t) x + t y) \leq (1 - t) \exp x + t \exp y .$ If $0 < t < 1$ , equality holds exactly when $x = y$ .

Facts & Assumptions

Given: Reals $x, y$ and $t \in [0, 1]$ .

[L1]

$\exp^{'} = \exp$ ; the chain rule differentiates compositions, and the algebra of derivatives differentiates finite linear combinations (The exponential function is smooth and $(\exp)^{'} = \exp$ , The chain rule, in one line from Carathéodory: if $g$ is differentiable at $c$ and $f$ is differentiable at $g (c)$ , then $f \circ g$ is differentiable at $c$ with $(f \circ g)^{'} (c) = f^{'} (g (c)) g^{'} (c)$ , Sums, scalar multiples, products and quotients: $(f + g)^{'} (c) = f^{'} (c) + g^{'} (c)$ , $(α f)^{'} (c) = α f^{'} (c)$ , $(f g)^{'} (c) = f^{'} (c) g (c) + f (c) g^{'} (c)$ , and $(f / g)^{'} (c) = (f^{'} (c) g (c) - f (c) g^{'} (c)) / g (c)^{2}$ when $g (c) \neq 0$ ).

[L2]

The mean value theorem applies on a closed interval to a function differentiable in its interior (The mean value theorem, as the case $g (x) = x$ of Cauchy's: for $f$ continuous on $[a, b]$ with $a < b$ and differentiable on $(a, b)$ there is $c \in (a, b)$ with $f (b) - f (a) = f^{'} (c) (b - a)$ ).

[L3]

The exponential is positive and strictly increasing (The exponential function is strictly increasing).

Proof

technique · direct

1.1

Put $d = y - x$ and $g (s) = 1 - s + s \exp d - \exp (s d)$ . Then $g (0) = g (1) = 0$ and $g^{''} (s) = - d^{2} \exp (s d) \leq 0$ .

L1L3

2.1

If $0 < s < 1$ , apply the mean value theorem to $g$ on $[0, s]$ and $[s, 1]$ ; since $g^{'}$ is nonincreasing by the mean value theorem applied to $g^{'}$ , their slopes give $g (s) / s \geq - g (s) / (1 - s)$ , hence $g (s) \geq 0$ .

step 1.1L1L2

3.1

Multiplying by $\exp x > 0$ converts $g (t) \geq 0$ into the displayed inequality.

step 2.1L3

4.1

When $0 < t < 1$ and $d \neq 0$ , $g^{''} < 0$ , so the slope comparison is strict and $g (t) > 0$ ; when $d = 0$ , equality is immediate.

step 1.1step 2.1L3∎

Depends on

Used by

Two closed convex sets can have no strong separator Counterexample
Conditional Hoeffding bound for bounded martingale differences Lemma
The real Gamma function is strictly log-convex Theorem
The weighted arithmetic-geometric mean inequality for real weights Theorem

Dependency tree · two levels

26 results within two dependency steps of this one, each drawn at its shortest distance from it. An arrow runs from a result to what uses it, so the chart reads left to right and ends at this result, which carries a heavier outline. Every node is a link to that result. Click elsewhere on the chart to enlarge it.

Sources

J. Lebl, Basic Analysis, Logarithm and Exponential (standard reference, not scraped)
MIT OpenCourseWare 18.100B Real Analysis, Spring 2025 full lecture notes (standard reference, not scraped)