Inverse Inequality!

We know that the tangent function is convex for all $\tan \theta \geq 0$ .

Therefore, its inverse function is concave, i.e. $\arctan (x)$ is concave for all positive $x$ .

By Jensen's Inequality,

$\large \arctan (\frac{A+B+C}{4}) \geq \frac{1}{2} \arctan (\frac{A}{2}) + \frac{\arctan (B)}{4} + \frac{\arctan (C)}{4}$

Multiplying both sides by $4$ gives us

$4 \arctan (\frac{A+B+C}{4}) \geq 2 \arctan (\frac{A}{2}) + \arctan (B) + arctan (C)$

Substituting $A+B+C = 4$ gives us

$4 \arctan (1) \geq 2 \arctan (\frac{A}{2}) + \arctan (B) + arctan (C)$

$4 \frac{\pi}{4} = \pi \geq 2 \arctan (\frac{A}{2}) + \arctan (B) + arctan (C)$

So, the maximum value of $P$ is $\pi$ . Therefore, our answer is $\lfloor 10000 \pi \rfloor = \boxed {31415}$ .

What I did was mostly due to intuition. You can write $P=arctan \frac{A}{2}+arctan \frac{A}{2}+arctan B+arctan C$ You can see A/2, B & C have equal weight. And, $A/2+A/2+B+C=4$ Here, we can distribute 4 to these 4 terms which leads to $A=2$ and $B=C=1$ . These values corrrspond to the max value of P. [tan is an increasing function in the domain 0 to π/2.] So, $max P=4 \times arctan 1= π$