LemmaStatement: AI-generatedProof: AI-generated

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AI-generated: a genuinely novel statement formulated by AI, with no source for the claim itself.

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A rational entropy test above the critical degree ratio two

Statement

For $t > 0$ write $I (t) = t \log t - t + 1$ . Then $I (9 / 4) \leq 5 / 8 < 2 / 3$ . Define $Λ (t) = t / \sqrt{t - 1}$ for $t \geq 2$ , using the positive square root. This function is nondecreasing and $Λ (t) \geq 9 / (2 \sqrt{5}) = 2 + g$ for $t \geq 9 / 4$ , where $g = 9 / (2 \sqrt{5}) - 2 > 0$ .

Statement references

Entropy rates, inverse branches and shifted root contracts

Order, continuity, range, and the product, quotient, and reciprocal laws for the natural logarithm

The natural logarithm has derivative 1/x and equals the integral from 1 to x of 1/t

On an interval I, for f continuous on I and differentiable at every interior point: f' ≥ 0 throughout gives f nondecreasing, f' > 0 gives f increasing, f' ≤ 0 and f' < 0 give the two decreasing forms; conversely a nondecreasing f has f' ≥ 0 and a nonincreasing f has f' ≤ 0 wherever it is differentiable, and no strict converse is claimed

Nonnegative real square roots exist uniquely

For a natural n ≥ 1 the function x ↦ xⁿ is differentiable everywhere with derivative ι(n) x^ n-1; for n = 0 it is the constant 1, with derivative 0; for a natural n ≥ 1 the function x ↦ x⁻ⁿ is differentiable at every x ≠ 0 with derivative -ι(n) x⁻ⁿ⁻¹; consequently every polynomial function is differentiable at every real, with the derivative computed term by term

Sums, scalar multiples, products and quotients: (f+g)'(c) = f'(c) + g'(c), (α f)'(c) = α f'(c), (fg)'(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) ≠ 0

A function differentiable at c is continuous at c

Facts & Assumptions

[F1]

The entropy function is continuous and strictly increasing on $(1, \infty)$ . Entropy rates, inverse branches and shifted root contracts

[F2]

The natural logarithm obeys the product law, has $\log 1 = 0$ and has derivative $1 / t$ for $t > 0$ ; a continuous function on an interval with nonnegative derivative at every interior point is nondecreasing. Order, continuity, range, and the product, quotient, and reciprocal laws for the natural logarithm The natural logarithm has derivative 1/x and equals the integral from 1 to x of 1/t On an interval I, for f continuous on I and differentiable at every interior point: f' ≥ 0 throughout gives f nondecreasing, f' > 0 gives f increasing, f' ≤ 0 and f' < 0 give the two decreasing forms; conversely a nondecreasing f has f' ≥ 0 and a nonincreasing f has f' ≤ 0 wherever it is differentiable, and no strict converse is claimed

[F3]

The power rule gives derivatives $1$ and $- t^{- 2}$ for $t$ and $t^{- 1}$ at $t > 0$ . Sums and scalar multiples of differentiable functions are differentiable, with the corresponding derivative formulas. Differentiability implies continuity. For a natural n ≥ 1 the function x ↦ xⁿ is differentiable everywhere with derivative ι(n) x^ n-1; for n = 0 it is the constant 1, with derivative 0; for a natural n ≥ 1 the function x ↦ x⁻ⁿ is differentiable at every x ≠ 0 with derivative -ι(n) x⁻ⁿ⁻¹; consequently every polynomial function is differentiable at every real, with the derivative computed term by term Sums, scalar multiples, products and quotients: (f+g)'(c) = f'(c) + g'(c), (α f)'(c) = α f'(c), (fg)'(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) ≠ 0 A function differentiable at c is continuous at c

[F4]

Every nonnegative real number has a unique nonnegative square root. If the number is positive its root is positive, since a zero root would have square zero. Nonnegative real square roots exist uniquely

Proof

Given: The data and hypotheses in the Statement.

Goal: Establish the stated conclusions.

1.1F1F2F3algebra

Set $h (t) = (t - t^{- 1}) / 2 - \log t$ on $(0, \infty)$ . By F2 and F3, all three terms are differentiable there, since $t > 0$ makes the reciprocal and logarithm admissible. Thus $h$ is differentiable, hence continuous, on $(0, \infty)$ , and $h' (t) = (1 + t^{- 2}) / 2 - t^{- 1} = (t - 1)^{2} / (2 t^{2}) \geq 0$ . Its restriction to the interval $[1, \infty)$ is continuous, including at $1$ , and differentiable at every interior point, so F2 makes it nondecreasing there. Since $h (1) = 0$ , evaluating at $3 / 2$ gives $\log (3 / 2) \leq 5 / 12$ . By the product law, $\log (9 / 4) = 2 \log (3 / 2) \leq 5 / 6$ . Hence $I (9 / 4) \leq (9 / 4) (5 / 6) - 9 / 4 + 1 = 5 / 8 < 2 / 3$ .

2.1F4algebra∎

For $t \geq s \geq 2$ , F4 gives positive roots of $t - 1$ and $s - 1$ , so $Λ (t)$ and $Λ (s)$ are well-defined and positive. Their squares satisfy $Λ (t)^{2} - Λ (s)^{2} = (t - s) (1 - ((t - 1) (s - 1))^{- 1}) \geq 0$ , because $(t - 1) (s - 1) \geq 1$ . Factoring the difference of squares and dividing by $Λ (t) + Λ (s) > 0$ gives $Λ (t) \geq Λ (s)$ . At $s = 9 / 4$ , uniqueness in F4 gives $\sqrt{5 / 4} = \sqrt{5} / 2$ , since the positive right-hand side has square $5 / 4$ . Thus $Λ (9 / 4) = 9 / (2 \sqrt{5})$ ; its square is $81 / 20 > 4$ , and positivity gives $Λ (9 / 4) > 2$ . The bound for every $t \geq 9 / 4$ and the fixed positive gap follow.

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