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Let f be a positive continuous function on [0,1].

Set $A=\int_0^1 fdx$

Prove that

$\sqrt{1+A^2}\le\int_0^1\sqrt{1+f^2}dx\le1+A$

I proved R.H.S inequality. So my question is L.H.S inequality.

I guess ... The cauchy-Schwarz‘s inequality is necessary

Please help me

user128766
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1 Answers1

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Let $F(x)=\int_0^xf.$ The graph of $F$ starts at $(0,0)$ and ends at $(1,A).$ The straight line distance between these points, which is $\sqrt {1+A^2},$ cannote be greater than the arc length of the graph of $F,$ which is $\int_0^1\sqrt {1+(F'(x)^2}\,dx = \int_0^1\sqrt {1+(f(x)^2}\,dx.$

zhw.
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