Show that $y$ is convex in $x$ provided that $y=h(I)$ and $x=g(I)$ 
Suppose that $\phi:[0,\infty)\to[0,1]$ is strictly increasing, infinitely differentiable such that $I\mapsto(1-\phi(I))I$ is injective. Define
  $$
y=\phi(I)I,\quad x=(1-\phi(I))I.
$$
  I would like to show that $y$ is a convex function of $x$.

This is my second attempt at this question, which is a simplified version of something in an article I am reading. (This question is self-contained so you don't need to check the article.) The authors provided an intuitive argument:
$$
\frac{y}{x}=\frac{\phi(I)}{1-\phi(I)}
$$
is increasing in $I$ so when $I\uparrow$, $y$ becomes proportionally larger than $x$. The argument included a $y$ in terms of $x$ plot in the $xy$-plane which also made sense. But I would like to learn a rigorous way to handle this; perhaps by showing $y=f(x)$ for some $f''\geq 0$. Can someone help please?
Edit: (a response to Michael Grant's comment below)


*

*$\phi(z)=\frac{z}{z+1}$  implies that $x=\frac{I}{1+I}\in[0,1)$. Moreover, $y=\frac{I^2}{I+1}=\frac{x^2}{1-x}$ is convex on $x\in[0,1)$.

*$\phi(z)=1-\exp(-z)$ does not satisfy $I\mapsto(1-\phi(I))I$ being injective. Moreover, if we restrict $I$ to be in $[0,1]$ so that $I\mapsto(1-\phi(I))I$ is injective, then $y=-x-\text{ProductLog}[-x]$ is convex for $x\in[0,\frac{1}{e}]$. Here ProductLog denotes the single-valued branch $W_0$ of the Lambert W function.
 A: But this is not true. For $y(x)$ to be convex, you need $\frac{dy}{dx}$ to increase, not $\frac{y}{x}$. And
$$
\frac{dy}{dx}= \frac{dy}{dI}:\frac{dx}{dI} = \frac{\frac{dy}{dI}}{1-\frac{dy}{dI}}
$$
(since $x = I-y$). So $\frac{dy}{dx}$ iff $\frac{dy}{dI}$ increases, or, in other words, $y(x)$ is convex iff $y(I)$ is convex.
Consider the following example (it is differentiable only once, but you can smooth it out):
$$
\phi(I) = \begin{cases}\phi_1(I)=\frac{I}{4} &\text{for  } 0\leq I\leq 1;\\ \phi_2(I)=\frac{3}{4} -\frac{I}{8} - \frac{3}{8I} & \text{for  }1\leq I \leq \frac{3}{2}; \\ \phi_3(I)=1 - \frac{63}{32I} + \frac{45}{32I^2} & \text{for  }I\geq \frac 32. \end{cases}
$$
The idea is as follows: $\frac{dy}{dI}$ increases from $0$ to $1/2$ on $[0,1]$, while $\phi$ increases from $0$ to $1/4$; then $\frac{dy}{dI}$ decreases from $1/2$ to $3/8$ on $[1,3/2]$, while $\phi$ still increases from $1/4$ to $5/16$; and on $[3/2,+\infty)$ both $\frac{dy}{dI}$ and $\phi$ both increase to $1$. (I really hope I have not miscalculated anywhere...) Since $\frac{dy}{dI}<1$ everywhere, we have $\frac{dx}{dI} = 1 - \frac{dy}{dI} >0$, so $x(I)$ is injective. But since $\frac{dy}{dI}$ decreases on $[1,3/2]$, $y(x)$ is concave on the corresponding interval, which is $[3/4, 33/32]$.
Update. Why $\phi(I)$ is increasing in this example:
$\phi'_1(I) = \frac 14>0$;
$\phi'_2(I) = -\frac 18 + \frac{3}{8I^2}$, which is positive for $0<I<\sqrt{3}$, and the interval $[1,3/2]$ lies within these boundaries;
$\phi'_3(I) = \frac{63}{32I^2} - \frac{45}{16I^3} = \frac{1}{I^2}(\frac{111}{32} - \frac{45}{16I})$, which is positive for $I>\frac{45}{16}:\frac{63}{32} = \frac{10}{7}<\frac 32$.
Why $\phi$ is continuous: we only need to check this on the borders of intervals:
$\phi_1(1) = \frac 14$, and $\phi_2(1) = \frac 34 - \frac 18 - \frac 38 = \frac 14$.
$\phi_2(\frac 32) = \frac 34 - \frac 3{16} - \frac 14 = \frac 5{16}$, and $\phi_3(\frac 32) = 1- \frac{21}{16} + \frac{5}{8} = \frac 5{16}$.
Why $\phi$ is differentiable:
$\phi'_1(1) = \frac 14$, and $\phi'_2(1) = -\frac 18 +\frac 38 = \frac 14$.
$\phi'_2(\frac 32) = -\frac 18 + \frac{1}{6} = \frac{1}{24}$, and $\phi'_3(\frac 32) = \frac 78 - \frac 56 = \frac{1}{24}$.
