I have lost the reference of this beautiful gem. Can anyone help me?
Let $I$ any interval, $f:I\to\mathbb R$ be a function and $0<\theta<1$.
Definition. —
Let us say that $f$ is $\theta$-convex if
\begin{align}
f\big(\theta x+(1-\theta)y\big)
\le \theta f(x) + (1-\theta)f(y),\quad \forall x,y\in I.
\label{conv-reduite}
\end{align}
Lemma. — Let $f:I\to\mathbb R$ be a function and $0<\theta<1$. Then $$ \mbox{$f$ is $\theta$-convex} \ \Longrightarrow \ \hbox{$f$ is $\frac12$-convex}. $$
Proof. — Let $m=\tfrac12(x+y)$. One ''notices'' that $ ((\theta^2+(1-\theta)^2)m=m- \theta(1-\theta)(x+y) $ so that $$ m=\theta\big[\theta m+(1-\theta)y\big] +(1-\theta)\big[\theta x+\ (1-\theta)m)\big]. $$ Applying $\theta$-convexity three times, one get easily \begin{align*} f(m) & \le \theta f\big(\theta m+(1-\theta)y\big) +(1-\theta)f\big(\theta x+ (1-\theta)m)\big) \\ & \le \theta^2 f(m)+\theta(1-\theta)f(y) +\theta(1-\theta) f( x)+ (1-\theta)^2f(m) \end{align*} So we have $ [1-\theta^2-(1-\theta)^2]f(m)\le \theta(1-\theta)[f(x)+f(y)], $ that is to say \begin{align*} 2\theta(1-\theta)f(m)\le\theta(1-\theta)\big[f(x)+f(y)\big]. \end{align*}