Inequality for concave functions This shouldn't be too hard, but I'm stuck.  Suppose $f$ is a concave function on the interval $[a,b]$, meaning
$$\lambda f(x) + (1-\lambda) f(y) \leq f(\lambda x + (1-\lambda) y)$$
for every $x,y \in [a,b]$ and every $\lambda \in [0,1]$.  I want to prove that for any $p, q, r \in [a,b]$ with $q \geq r \geq 0$ we have:
$$f(p + q) + f(p - q) \leq f(p + r) + f(p - r)$$
This inequality comes up in a paper that I'm reading on random walks, and in that context the function $f$ is piecewise linear.  So I'm not willing to assume that $f$ is differentiable, but piecewise smooth is fine if it helps (I don't think it will).
 A: Let's calculate first the specific $\lambda$ that "averages" the endpoints $p\pm q$ to give $p\pm r$, i.e. solve for $\lambda$,
\begin{align}
\lambda (p+q) + (1-\lambda)(p-q) = \pm r \\
\implies \lambda_{\pm} = \frac {\pm r+q}{2q} \tag{1}
\end{align}
Now, using the fact that $q\geq r\geq 0\implies p \pm r\in[p- q,p+q]$,
\begin{align}
f(p\pm r) &= f( \lambda_\pm(p+q) + (1-\lambda_\pm)(p-q) )\\
&\geq \lambda_\pm f(p+q) + (1-\lambda_\pm)f(p-q) \\
\end{align}
Summing these and substituting the values in $(1)$,
\begin{align}
f(p+r)+f(p-r) &\geq (\lambda_++\lambda_-)f(p+q) + (2-\lambda_+-\lambda_-)f(p-q) \\
&= f(p+q) + f(p-q)
\end{align}
Aside
You shouldn't be assuming $p,q,r\in[a,b]$, but rather that $p\pm q\in[a,b]$
A: Since p+q>p+r>p-r>p-q, by concavity, there exists a $\lambda_1, \lambda_2$ such that
$$\lambda_1 f(p+q)+(1-\lambda_1)f(p-q)<f(p+r)$$ and 
$$\lambda_2 f(p+q)+(1-\lambda_2)f(p-q)<f(p-r)$$
and so 
\begin{equation}(\lambda_2+\lambda_1)f(p+q)+(2-\lambda_1-\lambda_2)f(p-q)<f(p+r)+f(p-r).\end{equation}
Lets now solve for $\lambda_1, \lambda_2$. 
We know that
$\lambda_1(p+q)+(1-\lambda_1)(p-q)=p+r$
and so
$(2\lambda_1-1)q=r.$
Further, $\lambda_2(p+q)+(1-\lambda_2)(p-q)=p-r$
and so
$(1-2\lambda_2)q=r.$
Thus we can conclude $\lambda_1+\lambda_2=1$. Inserting this into equation 3, we get
\begin{equation}f(p+q)+f(p-q)<f(p+r)+f(p-r).\end{equation} 
A: Oops, this is pretty easy:
Define $\ell(t) = t(p-q) + (1-t)(p+q)$.  It is easy to check that:
$$\ell \left(\frac{q+r}{2q}\right) = p-r \hspace{1cm} \text{and} \hspace{1cm} \ell \left(\frac{q-r}{2q}\right) = p+r$$
Note that $(q+r)/(2q)$ and $(q-r)/(2q)$ are both in $[0,1]$, and their sum is $1$.  It follows that:
$$f(p-r) = f \left(\frac{q+r}{2q}(p-q) + \frac{q-r}{2q}(p+q) \right) \geq \frac{q+r}{2q}f(p-q) + \frac{q-r}{2q}f(p+q)$$
and similarly:
$$f(p+r) \geq \frac{q-r}{2q}f(p-q) + \frac{q+r}{2q}f(p+q)$$
Adding these two inequalities together gives the result.
