Given the Following IVP: $$y'=f(x,y)=\frac 1 {2+{\cos(xy)}}$$ $$y(0)=\frac 1 2$$ I need to show that the IVP has a single solution $u(x)$ that is defined in $(-\infty,\infty)$, increasing, and that $u(x)<x$ for all $x>1$
I was able to show, using the fact that $f_y(x,y)=\frac {x \sin(xy)} {{2+\cos(xy)}^2}$ is bounded in $[a,b]\times(-\infty,\infty)$ for all $a$,$b$, that there exist a single solution, and to show that it's derivative is positive, so that it is increasing, but I couldn't get the inequality right.
I Tried using the Mean Value Theorem but the bound that I got wasn't tight enough ($u(x)<x+\frac 1 2$). I also tried to bound integral up to a general point $x>1$ but this didn't work either.
Any assistance would be appreciated.
Let $y(x)$ be the unique solution to the IVP, which is unique because as the OP said, $f(x,y)$ is locally Lipschitz in $y$ and continuous in $x$. Global existence follows because $y'(x)$ is bounded from above and below: it is impossible for the solution to blow up in finite time.
It suffices to show $y(1) \leq 1.$ By the fundamental theorem of calculus, $$y(1) - y(0) = \int_0^1 \frac{1}{2 + \cos(x\cdot y(x))}dx.$$ You deduced from the properties of cosine that $$\frac{1}{3} \leq y'(x) \leq 1$$ so in turn $$\frac{1}{3}x + \frac{1}{2} \leq y(x) \leq x+ \frac{1}{2} .$$ This implies for $x\in [0,1]$, $$0 < y(x) < \frac{3}{2} \leq \frac{\pi}{2}.$$ However, this is great news, because it implies $\cos(x\cdot y(x)) \geq 0$ for $x \in [0,1].$
From this point, you can conclude $$y(1) - y(0) \leq \frac{1}{2}$$ and therefore $y(1) \leq 1.$ With the bound $y'(x) \leq 1$ we can conclude the desired bound $y(x) \leq x.$