
#1
Aug808, 02:06 PM

P: 57

1. The problem statement, all variables and given/known data
My question is whether the following inequality can be proven. 2. Relevant equations [tex] \left\int_a^bg\left(x\right)dx\int_a^bh\left(x\right)dx\right\leq\int_a^b\leftg\left(x\right)h\left(x\right)\rightdx [/tex] 3. The attempt at a solution I tried to write down the inequality in the form of it's primitives, where [tex]G\left(x\right)[/tex] is the primitive of [tex]g\left(x\right)[/tex] and [tex]H\left(x\right)[/tex] is the primitive of [tex]h\left(x\right)[/tex]. The inequality then becomes: [tex] \leftG\left(b\right)G\left(a\right)H\left(b\right)+H\left(a\right)\right\leq\leftG\left(b\right)H\left(b\right)\right\leftG\left(a\right)H\left(a\right)\right [/tex] But what next, or are there other means of getting a proof? 



#2
Aug808, 02:59 PM

P: 1,106

Assuming [tex] a \leq b [/tex] and f is continuous on the interval [a,b], then
[tex] \left\int_a^bf\left(x\right)dx \right \leq \int_a^b\leftf(x)\rightdx[/tex] which follows from the fact that [tex] f(x) \leq \leftf(x)\right [/tex] and [tex] f(x) \leq \leftf(x)\right [/tex] and that If f,g are both continuous on the interval [a,b] and [tex] f(x) \leq g(x) [/tex] for all x in the interval. Then [tex] \int_a^b f(x)dx \leq \int_a^b g(x)dx [/tex] Rearranging and using the first inequality should give you the desired inequality. 



#3
Aug808, 03:08 PM

P: 57

Oh, I see it now, it is indeed not that difficult.
[tex] \left\int_a^bg\left(x\right)dx\int_a^bh\left(x\right)dx\right\leq\int_a^b\leftg\left(x\right)h\left(x\right)\rightdx [/tex] If we rearrange: [tex] \left\int_a^b\left(g\left(x\right)h\left(x\right)\right)dx\right\leq\int_a^b\leftg\left(x\right)h\left(x\right)\rightdx [/tex] Substituting [tex]f\left(x\right)=g\left(x\right)h\left(x\right)[/tex] and using the first formula of snipez90, we get the proof. Thanks! 


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