TheoremStatement: AI-adaptedProof: AI-generatedSession-authored (Fable 5 assisted)precheck 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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The two-point convexity inequality for the exponential function

Statement

For all $x,y\in\mathbb R$ and $0\le t\le1$ , $\exp((1-t)x+ty)\le(1-t)\exp x+t\exp y.$ 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: $(f+g)'(c) = f'(c) + g'(c)$ , $(\alpha f)'(c) = \alpha f'(c)$ , $(fg)'(c) = f'(c)g(c) + f(c)g'(c)$ , and $(f/g)'(c) = \bigl(f'(c)g(c) - f(c)g'(c)\bigr)/g(c)^{2}$ when $g(c) \ne 0$ ).

[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 $g(s)=1-s+s\exp d-\exp(sd)$ . Then $g(0)=g(1)=0$ and $g''(s)=-d^2\exp(sd)\le0$ .

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 $g(s)/s\ge-g(s)/(1-s)$ , hence $g(s)\ge0$ .

step 1.1L1L2

3.1

Multiplying by $\exp x>0$ converts $g(t)\ge0$ into the displayed inequality.

step 2.1L3

4.1

When $0<t<1$ and $d\ne0$ , $g''<0$ , 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

The weighted arithmetic-geometric mean inequality for real weights Theorem

Dependency tree · next 3 levels

Direct dependencies and their dependencies through the next three levels: 81 results over 21 levels. An arrow runs from a result to what uses it, and this result sits at the bottom with a heavier outline. Click 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)