证明sinπx≤π/2乘以x(1-x),其中x∈【0,1】
g(x) = xπ^2( 1 - x ) - sinxπ = uπ( 1 - u/π ) - sinu = uπ - u^2 - sinu;{ u = xπ,u∈[ 0,π ] };
g'(u) = π - 2u - cosu = 0,u = π/2;
g''(u) = -2 + sinu < 0,故 g(u) 是凸函数,u = π/2 是 g(u) 最大点;
g(π/2) = π/2 * π - (π/2)^2 - sin(π/2) = π^2/2 - π^2/4 - 1 = π^2/4 - 1 > 0;
g(0) = 0 - 0 - 0 = 0;g(π) = π^2 - π^2 - sinπ = 0;
故 xπ^2( 1 - x ) - sinxπ ≥ 0,xπ^2( 1 - x ) ≥ sinxπ ;证毕 。
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