Quasiconcavity and homogeneity How to prove that if $f$ is strictly quasi-concave and homogeneous of degree 1, then $f$ is concave? It was left as an exercise by Silberberg &amp; Suen (2001), p.140.

I simply could not elaborate any sketches to leave here as a starting point.
Note that strictness of quasi-concavity is not required, unless you want to obtain strict concavity as a result.

Take any $x,y\in\mathbb R^n$. Observe that homogeneity of degree 1 (HD1) implies that 
\begin{equation}
f(x/f(x))=f(x)/f(x)=1=f(y/f(y)).
\end{equation}
For any $\alpha\in(0,1)$, let 
\begin{equation}
\theta=\frac{\alpha f(x)}{\alpha f(x)+(1-\alpha)f(y)}.\tag{1}
\end{equation}
Note that $\theta$ also lives in the (open) unit interval.
Thus, by quasi-concavity, we have for every $\theta\in(0,1)$,
\begin{equation}
f\left(\theta\frac{x}{f(x)}+(1-\theta)\frac{y}{f(y)}\right)\ge\min\left\{\frac{x}{f(x)},\frac{y}{f(y)}\right\}=1. 
\end{equation}
Expanding the LHS using $(1)$, we get 
\begin{equation}
f\left(\frac{\alpha x+(1-\alpha)y}{\alpha f(x)+(1-\alpha)f(y)}\right)\ge1.
\end{equation}
Invoking HD1 again, we have
\begin{equation}
\frac{f(\alpha x+(1-\alpha)y)}{\alpha f(x)+(1-\alpha)f(y)}\ge1
\quad\Leftrightarrow\quad 
f(\alpha x+(1-\alpha)y)\ge \alpha f(x)+(1-\alpha)f(y),
\end{equation}
which means $f$ is concave.