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?


$a^2+b^2+c^2+d^2+ac+bd -\sqrt{3}(ad-bc) \geq 0 \tag{1}$

Completing the squares is made easier by introducing a factor of $\,2\,$, so
let $\,\lambda = \frac{1}{2}\,$ and $\,\mu = \frac{\sqrt{3}}{2}\,$, then the LHS of $(1)$ can be written as:
$$
a^2+b^2+c^2+d^2+ 2\lambda(ac+bd) - 2\mu(ad-bc)
$$
$$\begin{align}
= \;\;\;& a^2 + 2a(\lambda c - \mu d) \color{red}{+(\lambda c - \mu d)^2-(\lambda c - \mu d)^2}
\\ + \;&b^2+2b(\mu c + \lambda d) \color{blue}{+(\mu c + \lambda d)^2 -(\mu c + \lambda d)^2}
\\ + \;& c^2 + d^2
\end{align}
$$
$$
\require{cancel}
\begin{align}
= \;\;\;& (a+\lambda c - \mu d)^2 + (b + \mu c + \lambda d)^2
\\ - \;& \color{red}{\lambda^2c^2+ \cancel{2 \mu\lambda cd} - \mu^2 d^2} - \color{blue}{\mu^2 c^2 - \cancel{2 \mu\lambda cd} - \lambda^2d^2} + c^2 + d^2 
\end{align}
$$
$$
= (a+\lambda c - \mu d)^2 + (b + \mu c + \lambda d)^2 + (1 - \lambda^2 - \mu^2)(c^2+d^2)
$$
The expression reduces to the sum of two squares when $\,\lambda^2+\mu^2=1$, as in OP's question.
