※ 引述《deryann (星辰)》之銘言:
: 設 f(0)=0 f(1)=1
: 試證明:
: 1
: 積分 [f'(x)]^2 dx >=1
: 0
Let g(x)=f(x)-x
g(0)=0, g(1)=0
g'(x)=f'(x)-1
f'^2=(g'+1)^2=g'^2+2g'+1
∫[0,1]f'^2 dx =∫[0,1]g'^2 dx +2∫[0,1]g' dx+∫[0,1] dx
=∫[0,1]g'^2 dx + 1
Done.
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