Using Jensen's inequality to prove another inequality? Suppose $u(\cdot)$ and $v(\cdot)$ are two differentiable, strictly increasing, and strictly concave real functions. Specifically, $v(\cdot)$ is "more concave" than $u(\cdot)$ in the sense that there exists an increasing and strictly concave function $\phi(\cdot)$ such that $v(x)=\phi(u(x))$ at all $x$. It is also equivalent to
\begin{equation}
\frac{v''(x)}{v'(x)}&lt;\frac{u''(x)}{u'(x)} \textrm{ for any }x\,.
\end{equation}

Let $p_i\in(0,1), \sum_{i\in I}p_i=1$ be probabilities and $|I|&gt;2$. Let $x_i$ and $y_i$ be strictly positive for all $i\in I$. Assume
\begin{equation}
\sum_{i\in I}p_ix_i&lt;\sum_{i\in I}p_iy_i,
\end{equation}
and
\begin{equation}
\sum_{i\in I}p_iu(x_i)=\sum_{i\in I}p_iu(y_i).
\end{equation}

Conjecture:
\begin{equation}
\sum_{i\in I}p_iv(x_i)&gt;\sum_{i\in I}p_iv(y_i).
\end{equation}
I believe this is right (after trying many numerical examples) and I think a clever use of Jensen's inequality (or its variants) will do this. But I'm stuck on doing it formally. Any hints/thoughts on providing a formal proof?

Remark: this is related to my other post: Proving an inequality of the expectation of concave functions?

Update: after some more attempts, I believe some techniques in convex analysis would be helpful. Geometrically, the middle equation represents a hyperplane in the $R^{|I|}$ space, and the desired result (very roughly) says that a concave transformation of that hyperplane should be separated from a convex transformation of it. 

To be clear, I wasn't saying the conjecture should be generally true. Any thoughts on finding any sufficient conditions to make it work would be very helpful.
The reverse inequality holds. A  differentiable  function of one variable is concave on an interval $J$ if and only if the function lies below all of its tangent, i.e., 
$$f(x) \leq f(y) + f'(y)(x-y)$$
for all $x,y\in J$. In particular, if $f′(c) =0$, then $c$ is a global maximum  of $f(x)$.

Applying this for $v$, we have
$$p_i(v(x_i) - v(y_i)) \le p_i v'(y_i)(x_i-y_i)$$ for all $p_i\in(0,1)$ and all $i\in I$.

Summing up over $i$
\begin{align}
\sum_{i\in I}p_i(v(x_i) - v(y_i)) &amp;\le \sum_{i\in I} p_i v'(y_i)(x_i-y_i) \\
&amp;\le  \sum_{i\in I} p_i |v'(y_i)|(x_i-y_i) \qquad \text{since} \,\,  w\le |w|, \forall w \in \mathbb{R} \\
&amp;\le \min_i |v'(y_i)|\sum_{i\in I} p_i  (x_i-y_i) &lt;0.
\end{align}
The second inequality holds since $v$ is increasing so that  $|v'|=v'$ iff $v'&gt;0$.
The reverse inequality holds. A  differentiable  function of one variable is concave on an interval $J$ if and only if the function lies below all of its tangent, i.e., 
$$f(x) \leq f(y) + f'(y)(x-y)$$
for all $x,y\in J$. In particular, if $f′(c) =0$, then $c$ is a global maximum  of $f(x)$.

Applying this for $v$, we have
$$p_i(v(x_i) - v(y_i)) \le p_i v'(y_i)(x_i-y_i)$$ for all $p_i\in(0,1)$ and all $i\in I$.

Summing up over $i$
\begin{align}
\sum_{i\in I}p_i(v(x_i) - v(y_i)) &amp;\le \sum_{i\in I} p_i v'(y_i)(x_i-y_i) \\
&amp;\le  \sum_{i\in I} p_i |v'(y_i)|(x_i-y_i) \qquad \text{since} \,\,  w\le |w|, \forall w \in \mathbb{R} \\
&amp;\le \min_i |v'(y_i)|\sum_{i\in I} p_i  (x_i-y_i) &lt;0.
\end{align}
The second inequality holds since $v$ is increasing so that  $|v'|=v'$ iff $v'&gt;0$.