Does $\frac{1}{a}F(x)>\frac{1}{c}G(y)$ imply $\frac{1}{a}\int_0^bF(b)>\frac{1}{c}\int_0^dG(d)?$

Question

Suppose I have two functions $G,F: \mathbb{R}_+ \rightarrow \mathbb{R}_+$. I also know that $F'(x)=G'(x)$. Now suppose that the following inequality holds

$$\frac{1}{a}F(b)>\frac{1}{c}G(d)$$

For constants $a,c>0$, and $d>b>0$. Is it true that this implies that

$$\frac{1}{a}\int_0^bF(x)>\frac{1}{c}\int_0^dG(x)?$$

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Since $F'(x)=G'(x)$, the rates of decrease of functions $\frac{1}{a}F(b)$ and $\frac{1}{c}G(d)$ in their respective arguments are the same. Hence this implies that for all $x\leq b$ and all $y\leq d$, the following must hold

$$\frac{1}{a}F(x)>\frac{1}{c}G(y)$$

This implies that,

$$\frac{1}{a}\int_0^bF(x)>\frac{1}{c}\int_0^dG(x).$$

Is this correct? If so, how could you rephrase my argument to make it seem more convincing and rigorous? If it is wrong, why?

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Note that $a$ and $c$ are arbitrary positive constants.

Answer 1

No. Take $F\equiv 1, G\equiv 1$, $a=1,c=2,b=4,d=5$.