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An integral inequality
05-06-2010, 08:46 AM
Post: #1
An integral inequality
Suppose that \(f > 0\) and satisfies the Lipschitz condition \(|f(x)-f(y)| \le L|x-y|\) on \([a,b]\)
Let \(a \le c \le d \le b\), \( \displaystyle{\int_c^d\frac{dx}{f(x)}} = \alpha\) and \( \displaystyle{\int_a^b\frac{dx}{f(x)}} = \beta\), prove that \[ \int_{a}^{b}f(x)dx\le\frac{e^{2L\beta}-1}{2L\alpha}\int_{c}^{d}f(x)dx \]
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05-10-2010, 12:52 AM
Post: #2
RE: An integral inequality
Let $D=d-c$ and $B=b-a$. Since $\int_c^d \frac{dx}{f(x)}=\alpha$, we must have $f\le \frac D\alpha$ at some point of $[c,d]$. Now, among all $L$-Lipschitz positive functions on $[a,b]$ whose minimum does not exceed $\frac D\alpha$, the function $f(x)=\frac D\alpha+L(x-a)$ gives the largest value to the integral $\int_a^b f(x)dx$ and the least value to the integral $\int_a^b \frac{dx}{f(x)}$. Hence $\int_a^b f(x)dx\le \frac{BD}{\alpha}+\frac{B^2L}2$ and $\beta\ge\frac1L\ln\left(1+LB\frac{\alpha }D\right)$. Also, by Cauchy-Schwartz, $\int_c^d f(x)\,dx\ge \frac{D^2}{\alpha}$. Thus, it suffices to prove that
\[ \frac{BD}\alpha+\frac{B^2L}2\le\frac{D^2}{\alpha} \frac{e^{2L\frac 1L\ln(1+LB\frac\alpha D)}-1}{2L\alpha}\,. \]
But it is an identity.

http://www.artofproblemsolving.com/Forum...84#p430984
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