05/01/21 09:27:43
>981
0<x<√2 ⇒ -cos(x)>-1, -sin(x)>-x, cos(x)>1-(1/2)x^2.
x/cos(x) < x/{1-(1/2)x^2}.
∫_[0,a] x/cos(x) dx < ∫_[0,a] x/{1-(1/2)x^2} dx = [-Ln{1-(1/2)x^2}](x=0→a) = Ln{2/(2-a^2)}.
>982-984
f(θ)は上に凸だから、f(θ) ≧ {(π/2-θ)f(0) + θf(π/2)}/(π/2) = 1.
>985
(a^3+b^3+c^3)/3 ≧ [(a+b+c)/3]・[(a^2+b^2+c^2)/3] … (2)
ぢゃない？