Assume that two random variables $X$ and $Y$ satisfy $X\ge 0$, $Y\ge0$ and $X+Y\ge 1$. Does it hold that $$\mathbb{E}\left[\frac{X}{X+Y}\right]\le \frac{\mathbb{E}[X]}{\mathbb{E}[X+Y]},$$ whereby $\mathbb{E}[\cdot]$ denotes the expected value?
Jensen's inequality may be helpful, but I don't know how to apply it here.
I want to add the additional condition that X increases if X+Y increases.
No. Both directions of the inequality are possible:
Case A: $X=1$ or $4$ with equal probability and independently $Y=1$ or $2$ with equal probability
Case B: $X=1$ or $2$ with equal probability and independently $Y=1$ or $4$ with equal probability
With further thought, finding examples having two directions for the inequality is trivial whenever you do not start with equality (you do not need independence); all you have to do is swap $X$ and $Y$, since both $E\big[\frac{X}{X+Y}\big]+E\big[\frac{Y}{Y+X}\big]=1$ and $\frac{E[X]}{E[X+Y]} +\frac{E[Y]}{E[Y+X]}=1$, so $$E\left[\frac{X}{X+Y}\right] < \frac{E[X]}{E[X+Y]} \iff E\left[\frac{Y}{Y+X}\right]> \frac{E[Y]}{E[Y+X]}.$$
Clearly $E\big[\frac{X}{X+Y}\big]$ and $\frac{E[X]}{E[X+Y]}$ are each between $0$ and $1$ so the difference between them is between $-1$ and $+1$. They can approach these bounds, for example with $P\big((X,Y)=(n^4,1)\big)=\frac1n$ and $P\big((X,Y)=(1,n^2)\big)=1-\frac1n$ which for large $n$ gives $E\big[\frac{X}{X+Y}\big] \approx \frac1n +\frac1{n^2}$ approaching $0$ and $\frac{E[X]}{E[X+Y]} \approx 1-\frac1n +\frac2{n^2}$ approaching $1$.