Lemma 1. For all positive integers $n$, and rational numbers $x_1$,..., $x_{2^n}$ in the interval $(0,1)$, the inequality \[f(\frac{x_1+...+x_{2^n}}{2^n})\le \frac{f(x_1)+...+f(x_{2^n})}{2^n}\]holds. Proof. The proof is by induction on $n$. Base Case. For $n=1$, note that \[f(\frac{x_1+x_2}{2})\le \frac{f(x_1)+f(x_2)}{2}\]by the given condition. Inductive Step. Suppose the claim holds for $n=k$. Then, for $n=k+1$,\[f(\frac{x_1+...+x_{2^{k+1}}}{2^{k+1}})\le f(\frac{\frac{x_1+...+x_{2^k}}{2^k}+\frac{x_{2^k+1}+...+x_{2^{k+1}}}{2^k}}{2})\le \frac{f(\frac{x_1+...+x_{2^k}}{2^k})+f(\frac{x_{2^k+1}+...+x_{2^{k+1}}}{2^k})}{2} \le \frac{f(x_1)+...+f(x_{2^{k+1}})}{2^{k+1}}.\]Thus, by induction, the desired inequality holds for all positive integers $n$. $\blacksquare$ Lemma 2. If \[f(\frac{x_1+...+x_{n+1}}{n+1})\le \frac{f(x_1)+...+f(x_{n+1})}{n+1},\]holds for all rational numbers $x_1$,..., $x_{n+1}$ in the interval $(0,1)$, then \[f(\frac{x_1+...+x_n}{n})\le \frac{f(x_1)+...+f(x_n)}{n}.\] Proof. Let \[x_{n+1}=\frac{x_1+...+x_n}{n}\]and note that $x_{n+1}\in (0,1)$ is a rational number. Also, note that \[\frac{x_1+...+x_{n+1}}{n+1}=\frac{x_1+...+x_n}{n}.\]Thus, \[f(\frac{x_1+...+x_n}{n})=f(\frac{x_1+...+x_{n+1}}{n+1})\le \frac{f(x_1)+...+f(x_{n+1})}{n+1}=\frac{f(x_1)+...+f(x_n)+f(\frac{x_1+...+x_n}{n})}{n+1}.\]Rearranging gives the desired inequality. $\blacksquare$ Lemma 1 and Lemma 2 imply that for every positive integer $k$ and rational numbers $a_1$,..., $a_k$ that are in the interval $(0,1)$, \[f(\frac{a_1+...+a_k}{k})\le \frac{f(a_1)+...+f(a_k)}{k}.\]Now, suppose $\lambda=\frac{p}{q}\in (0,1)$ for positive integers $p$ and $q$. Since $p<q$, $q-p>0$. Let $x_1$ and $x_2$ be rational numbers in the interval $(0,1)$. Let $k=q$, $a_1=...=a_p=x_1$, and $a_{p+1}=...=a_{q}=x_2$. Then, \[f(\lambda x_1+(1-\lambda) x_2 )=f(\frac{px_1+(q-p)x_2}{q}) \le \frac{pf(x_1)+(q-p)f(x_2)}{q}=\lambda f(x_i)+(1-\lambda) f(x_2).\]