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?

You could use the identity:
$$(ap+br+cs)^2+(aq+bs-cr)^2$$
$$=(a^2p^2+b^2r^2+c^2s^2+2abpr+2acps+2bcrs)+(a^2q^2+b^2s^2+c^2r^2+2abqs-2acqr-2bcrs)$$
$$=(a^2p^2+b^2r^2+c^2s^2+2abpr+2acps)+(a^2q^2+b^2s^2+c^2r^2+2abqs-2acqr)$$
$$=a^2(p^2+q^2)+(b^2+c^2)(r^2+s^2)+2ab(pr+qs)+2ac(ps-qr)$$
If $a^2=b^2+c^2$, dividing by $a^2$ leaves the four squares intact.
