Prove all cost functions are concave in input prices and demand for inputs is downward
Let $x(w, q)$ denote the solution to the cost minimization problem :

\begin{eqnarray*} \min_{x} &amp; \ w\cdot x \\ \text{s.t.} &amp; \ \ f(x) \geq q \end{eqnarray*}
where $f$ is the production function.
Since $x(w, q)$ minimizes cost at $(w, q)$, following holds for all $w$ and for all $q$ :
\begin{eqnarray*} w\cdot x(w, q) \leq w\cdot x(w', q) \ \ \  \forall w' \end{eqnarray*}

Also we know that cost function is cost of cost minimizing input choice :
\begin{eqnarray*} C(w, q) = w\cdot x(w, q) \end{eqnarray*}

First, we'll show that $C$ is concave in $w$. Consider any arbitrary $w'$, $w''$ and an arbitrary $\lambda \in [0, 1]$.
\begin{eqnarray*} C(\lambda w' + (1-\lambda) w'', q) &amp; = &amp; (\lambda w' + (1-\lambda) w'')\cdot x(\lambda w' + (1-\lambda) w'', q) \\  &amp; = &amp; \lambda w' \cdot x(\lambda w' + (1-\lambda) w'', q) + (1-\lambda) w''\cdot x(\lambda w' + (1-\lambda) w'', q) \\  &amp; \geq &amp; \lambda w' \cdot x(w', q) + (1-\lambda) w''\cdot x(w'', q)  \\  &amp; = &amp; \lambda C(w' , q) + (1-\lambda) C(w'', q) \end{eqnarray*}
Therefore, $C$ is concave in $w$.

To show that input demand is downward sloping, consider any arbitrary $w'$, and $w''$,
\begin{eqnarray*} w'\cdot x(w', q) &amp; \leq &amp; w'\cdot x(w'', q) \\ w''\cdot x(w'', q) &amp; \leq &amp; w''\cdot x(w', q) \end{eqnarray*}
Adding them we get
\begin{eqnarray*} (w'-w'')\cdot (x(w', q) - x(w'', q)) &amp; \leq &amp; 0 \end{eqnarray*}
We now consider a special case where $w'$ and $w''$ differ only in the $j$th input's price. Consequently,
\begin{eqnarray*} (w'-w'')\cdot (x(w', q) - x(w'', q)) = (w_j'-w_j'') (x_j(w', q) - x_j(w'', q)) &amp; \leq &amp; 0 \end{eqnarray*}
Therefore, demand for $j$th input is inversely related to its price.