Elementary inequalities exercise - how to 'spot' the right sum of squares? I have been working through CJ Bradley's Introduction to Inequalities with a high school student and have been at loss to see how one could stumble upon the solution given for Q3 in Exercise 2c.
Question:
If $ad-bc=1$ prove that $a^2+b^2+c^2+d^2+ac+bd \geq \sqrt{3}$.
Solution:
Make the inequality homogenous by multiplying both sides by $ad-bc=1$. [That seems sensible.] Take everything onto one side so we now want to show
$$a^2+b^2+c^2+d^2+ac+bd -\sqrt{3}(ad-bc) \geq 0 \tag{1}.$$
[That's also a reasonable thing to do. The trouble is coming next...]
Now play around until you notice the left hand side can be written as
$$\frac{1}{4}(2a+c-\sqrt{3}d)^2 + \frac{1}{4}(2b+d+\sqrt{3}c)^2 \tag{2}.$$
[What??]
I played around for a fair while and didn't get to this. I have a suspicion that this question was created by reverse-engineering. What thought processes get you from (1) to (2), without knowing (2) beforehand? How can you get to the solution without pulling a rabbit out of a hat?

There is a geometric interpretation. Let $v=(a,b),\ w=(c,d)$ denote two vectors in $\mathbb{R}^2.$ with lengths  $r$ and $s$ respectively, and the angle $\phi$ between them. Then $$ac+bd=rs\, \cos\phi, \quad ad-bc= rs\, \sin \phi.$$
The left hand of the inequality takes the form \begin{multline*} r^2+s^2 +rs\,\cos\phi \ge 2rs +rs\,\cos\phi \\ =rs(2 +\cos\phi)    \ge rs(2-|\cos\phi|).\end{multline*}
By the Cauchy-Schwarz inequality we get $$\sqrt{3}|\sin\phi| +|\cos\phi|\le 2.$$
Thus $$rs(2 -|\cos\phi|)\ge \sqrt{3}\,rs\,|\sin\phi|.$$ The last expression is equal to the right hand side of the desired inequality.
