Happy New Year 2016

Let
$y = \sin\theta$
$x = \cos\theta \sin \alpha$
$z = \cos \theta \cos \alpha$
$\therefore axy + byz = \sin\theta \cos \theta \times \left( a\sin\alpha + b\cos\alpha \right)$
$\therefore axy + byz = \dfrac{\sqrt{a^2 + b^2} \sin2\theta}{2} \times \sin(\alpha + \beta)$ where $\beta = \arccos\left(\dfrac{a}{\sqrt{a^2+b^2}}\right)$
This value is equal to $\dfrac{\sqrt{a^2 + b^2}}{2}$ when $\alpha + \beta = \dfrac{\pi}{2}$
& $2\theta = \dfrac{\pi}{2} \rightarrow \theta = \dfrac{\pi}{4}$
$\therefore y = \sin\dfrac{\pi}{4} = \dfrac{1}{\sqrt{2}}$
Comparing, A = 1, B = 1, C = 2.
$A + B + C = 1+1+2 = 4$

By the arithmetic-geometric inequality of absolute happiness,

$\frac{x^2+\frac{a^2y^2}{a^2+b^2}}{2} + \frac{\frac{b^2y^2}{a^2+b^2}+z^2}{2} \geq \sqrt{\frac{a^2x^2y^2}{a^2+b^2}} + \sqrt{\frac{b^2z^2y^2}{a^2+b^2}}$

Which results in

$\frac{\sqrt{a^2+b^2}}{2} \geq axy+byz$

wherein equality is achieved if and only if $\frac{a^2y^2}{a^2+b^2} = x^2$ and $\frac{b^2y^2}{a^2+b^2} = z^2$

Now, with more happiness, we substitute these into the equation

$x^2+y^2+z^2 = 1$

$\frac{a^2y^2}{a^2+b^2} +y^2 +\frac{b^2y^2}{a^2+b^2}=1$

$2y^2=1$ , $y=\frac{1}{\sqrt{2}}$

Now, after repeatedly doubting the answer which you have derived because it doesn't seem to fit the answer format (why does the answer format have to be like that???) we get A=1, B=1, C=2 ( $y=\frac{1}{1\sqrt{2}}$ )

Then, $A+B+C=4$

using the c-s inequity: $y^2(a^2+b^2)(x^2+z^2)≥(axy+bzy)^2$ $(axy+byz)≤\sqrt{y^2(a^2+b^2)(1-y^2)}=\sqrt{(a^2+b^2)(y^2-y^4)}=\sqrt{(a^2+b^2)(-(y^2-.5)+(.5)^2)}$ we know from a square≥0, or $-(y^2-.5)^2≤0$ or $-(y^2-.5)^2+.25≤.25$ . equity occurs at $y^2=.5\Longrightarrow y=\dfrac{1}{1\sqrt{2}}$ . we put the max in $axy+byz≤\sqrt{(a^2+b^2)(y^2-y^4)}≤\sqrt{(a^2+b^2).25}=\dfrac{\sqrt{a^2+b^2}}{2}$ so y must be that value or else the equity cant be achieved.