需要證明 ∫ (0 to 1/2) f '(x)^2 dx ≥ 12 * (∫ (0 to 1/2) f(x) dx)^2。
f(x)=∫ (0 to x) f '(y) dy, 因此
∫ (0 to 1/2) f(x) dx = ∫ (0 to 1/2) ∫ (0 to x) f '(y) dy dx = ∫ (0 to 1/2) ∫ (y to 1) f '(y) dx dy = ∫ (0 to 1/2) (1-y) f '(y) dy。
根據Cauchy-Schwarz不等式，
(∫ (0 to 1/2) f(x) dx)^2 ≤ ∫ (0 to 1/2) (1-x)^2 dx *∫ (0 to 1/2) f '(x)^2 dx = (1/12)∫ (0 to 1/2) f '(x)^2 dx
f(x)=∫ (0 to x) f '(y) dy, 因此
∫ (0 to 1/2) f(x) dx = ∫ (0 to 1/2) ∫ (0 to x) f '(y) dy dx = ∫ (0 to 1/2) ∫ (y to 1) f '(y) dx dy = ∫ (0 to 1/2) (1-y) f '(y) dy。
根據Cauchy-Schwarz不等式，
(∫ (0 to 1/2) f(x) dx)^2 ≤ ∫ (0 to 1/2) (1-x)^2 dx *∫ (0 to 1/2) f '(x)^2 dx = (1/12)∫ (0 to 1/2) f '(x)^2 dx
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