is this inner product positive-definite? $$\left \langle u, v \right \rangle = pu_{1}v_{1}+qu_{1}v_{2}+qu_{2}v_{1}+pu_{2}v_{2}\\\text{ for }\\ \text{p >0} \text{ and } p^{2}\geq q^{2}$$
The solution breaks down $$\left \langle u, u \right \rangle = pu_{1}u_{1}+qu_{1}u_{2}+qu_{2}u_{1}+pu_{2}u_{2} $$ into
$$p(u_{1}+\frac{q}{p}u_{2})^{2}+(p-\frac{q^{2}}{p})u_{2}^{2}$$ 
In this instance, Mathematica will not usually give me the desired answer to solve the question. I can't imagine having to 'guess' under a time constraint during examinations. Is there a procedural manner to obtain this solution?
This is a trivial question. The tedious part is the algebra manipulation involved which is a fairly 'unfair' question should it pops out during any examination. But just in the case it does, are there any tricks around getting to the solution?

$p(u_1+\frac{q}{p} u_2)^2$ is nonnegative because it is the product of $p$ (which is strictly positive) with a square.
$(p-\frac{q^2}{p}) u_2^2$ is also nonnegative, being $\frac{1}{p} (p^2 - q^2) u_2^2$, and we're given that $p>0$ and $p^2 \geq q^2$.
Hence the sum is nonnegative.
When is it zero? Consider the second term, which must be zero (since the sum of nonnegative quantities being zero means that the summands are all zero). Either $p^2 = q^2$, or $u_2 = 0$. In the latter case, now consider the first term; obtain that $u_1 = 0$. In the former case: have $\frac{q}{p} = \pm 1$, so from the first term we need $u_1 \pm u_2 = 0$. That may not always be positive, so if $p^2 = q^2$ then the inner product is not positive definite.
