Take the number $n$ of stages of a binomial variable $X \sim B (n, p)$ to be fixed, and allow $p$ to vary. $p$ is not the random variable of this binomial experiment - $X$ is - but "allow $p$ to vary" means "consider a binomial experiment $X \sim B (n, p)$ for different values of $p$". Then the question is... show that $\sigma^2 < np $ and $ \sigma^2 < nq$
I got that $$ \begin{split} \sigma^2 &= npq \\ &= np(1-p)\\ &= n(p-p^2)\\ &= -p^2 n + pn \end{split} $$ where I use the binomial distribution formula for variance (SD squared). And I was hoping some graphical method of solving the above problem would emerge but all I can think of drawing the line np and then establishing with calculus somehow that the curve always has lower gradient than line n=p as p approaches 1. I s there any other more obvious way?
and this looked like a graph 
which is symmetric about $p = 1/2$
We want to show that given $X \thicksim \operatorname{Bin}(n,p)$ for fixed $n$, that $\sigma^2 < np$ and $\sigma^2 < nq$ when $q = 1 - p$. Presuably here, $\sigma^2 = \operatorname{Var}(X) = npq$. This is less than both $np$ and $nq$, since $p, q < 1$. Multiplying any number by a number less than one would reduce it, so it trivially follows.
A graphical method of solving the problem would show that a graph of $p(1-p)$ is below $p$ and $(1-p)$ for any value of $0 < p < 1$. For example:
Clearly shows that the equalities follow (simply divide by $n$ to obtain them). But this is an overly complex way of showing it, and it's much easier to do the first. I'm actually not convinced anyone would accept this.
Calculus wise, we seek to prove that for $f(x) = x(1-x) = x - x^2$, $g(x) = x$, and $h(x) = 1-x$, that $f(x) < g(x)$ and $f(x) < h(x)$, $x \in (0,1)$. $f'(x) = 1 - x$, while $g'(x) = 1$. Then, it is clear that since $f(0) = g(0) = 0$, $f(x) < g(x)$ for all $x \in (0,1)$. Next, since $h(0) = 1$, we want to find the intersection of the two functions, by: $$f(x) = h(x) \\ x - x^2 = 1 - x \\ 0 = x^2 - 2x + 1 \\ 0 = (x-1)(x-1)$$ And the intersection is at $1$, so our inequality holds. These are silly ways of solving it though, since it follows trivially from the first proof.