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Need help of proving an inequality of intergrals

Integration

New member
Jan 16, 2013
2
Let [TEX]f(x) \in C[a,b] [/TEX] and let [TEX]f(x)>0 [/TEX] on [TEX][a,b][/TEX]. Prove that
[TEX]\exp \Big(\frac{1}{b-a}\int_a^b \ln f(x) dx \Big)\leq \frac{1}{b-a}\int_a^b f(x) dx[/TEX]

I have learnt Gronwall's Inequality and Jensen's Inequality(and inequality deduced from it like Cauchy Schwarz Inequality) but i couldn't use them to fit the condition.
Would you help me please?Thank you.
 
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GJA

Well-known member
MHB Math Scholar
Jan 16, 2013
271
Hi Integration.

Note that $e^{x}$ is a convex function. Now we write
\[ e^{\frac{1}{b-a}\int_{a}^{b}\ln(f(x))dx}=e^{\int_{a}^{b}\frac{1}{b-a}\ln(f(x))dx}\]

The next step is to apply Jensen's inequality (with $e^{x}$ as the convex function) to the right side of the above equation. Doing this will get us what we're after.

Does this clear things up? Let me know if anything is unclear. Good luck!

Edit: This argument is only valid in the case where $f(x)\geq 1$ because we need $\ln(f(x))$ to be nonnegative to apply Jensen's inequality in the manner that is outlined above (Jensen's inequality - Wikipedia, the free encyclopedia)
 
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Integration

New member
Jan 16, 2013
2
Thank you so much for inspiration.
Maybe using the convex function [TEX]- \ln (x) [/TEX] would solve the problem?

Another question: How can i deduce the integral from of Jensen's inequality from the finite one? Cause I have only learnt the latter one
 
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