# Is this inequality true and provable?

by 3029298
Tags: inequality, integral, mathematics, proof
 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 $$\left|\int_a^bg\left(x\right)dx-\int_a^bh\left(x\right)dx\right|\leq\int_a^b\left|g\left(x\right)-h\left(x\right)\right|dx$$ 3. The attempt at a solution I tried to write down the inequality in the form of it's primitives, where $$G\left(x\right)$$ is the primitive of $$g\left(x\right)$$ and $$H\left(x\right)$$ is the primitive of $$h\left(x\right)$$. The inequality then becomes: $$\left|G\left(b\right)-G\left(a\right)-H\left(b\right)+H\left(a\right)\right|\leq\left|G\left(b\right)-H\left(b\right)\right|-\left|G\left(a\right)-H\left(a\right)\right|$$ But what next, or are there other means of getting a proof?
 P: 1,104 Assuming $$a \leq b$$ and f is continuous on the interval [a,b], then $$\left|\int_a^bf\left(x\right)dx \right| \leq \int_a^b\left|f(x)\right|dx$$ which follows from the fact that $$f(x) \leq \left|f(x)\right|$$ and $$-f(x) \leq \left|f(x)\right|$$ and that If f,g are both continuous on the interval [a,b] and $$f(x) \leq g(x)$$ for all x in the interval. Then $$\int_a^b f(x)dx \leq \int_a^b g(x)dx$$ Rearranging and using the first inequality should give you the desired inequality.
 P: 57 Oh, I see it now, it is indeed not that difficult. $$\left|\int_a^bg\left(x\right)dx-\int_a^bh\left(x\right)dx\right|\leq\int_a^b\left|g\left(x\right)-h\left(x\right)\right|dx$$ If we rearrange: $$\left|\int_a^b\left(g\left(x\right)-h\left(x\right)\right)dx\right|\leq\int_a^b\left|g\left(x\right)-h\left(x\right)\right|dx$$ Substituting $$f\left(x\right)=g\left(x\right)-h\left(x\right)$$ and using the first formula of snipez90, we get the proof. Thanks!