Question: If $x^2 + y^2 + z^2 = 1$ , then maximum value of $x + 2y + 3z$ is

I have solved this question using Cauchy-Schwarz inequality as follows

$$(x+2y+3z)^2 \leq (1^2+2^2+3^2)(x^2+y^2+z^2)$$

$$\implies (x+2y+3z)^2 \leq 14$$

My question is that is there an alternate to this solution.

Answer: 
Maximum value happens when the plane $x+2y+3z=c$ touches the sphere $x^2+y^2+z^2=1$. At the point of contact, the normal $n=(1,2,3)$ is proportional to the point itself $p=(x,y,z)$. Which means $p=n/|n|$ and $c=p\cdot n = n^2/|n| = |n|$
You can consider it as an optimization problem with a constraint, and use the method of Lagrange multipliers; see:
<https://en.wikipedia.org/wiki/Lagrange_multiplier>

So $g(x,y,z)=x^2+y^2+z^2-1$, and the Lagrange function is $$L(x,y,z,\lambda) = x+2y+3z-\lambda(x^2+y^2+z^2-1)$$

The derivatives are

$\frac{d}{dx} L(x,y,z,\lambda) = 1-2\lambda x=0$,

$\frac{d}{dy} L(x,y,z,\lambda) = 2-2\lambda y=0$,

$\frac{d}{dz} L(x,y,z,\lambda) = 3-2\lambda z=0$, and of course

$\frac{d}{d\lambda} L(x,y,z,\lambda) = x^2+y^2+z^2-1=0$ (providing no new information).

Clearly $\lambda\neq 0$, otherwise the first three equations are contradictory. Thus these three equations can be rearranged to obtain $x=\frac{1}{2\lambda}, y=\frac{2}{2\lambda}, z=\frac{3}{2\lambda}$, and putting these into the fourth equation results in $\frac{14}{4\lambda^2}-1 = 0$, that is, $\lambda = \pm\sqrt{\frac{7}{2}}$. Thus the only stationary points are $(x,y,z)=(\frac{1}{\sqrt{14}}, \frac{2}{\sqrt{14}}, \frac{3}{\sqrt{14}})$ and $(x,y,z)=(-\frac{1}{\sqrt{14}}, -\frac{2}{\sqrt{14}}, -\frac{3}{\sqrt{14}})$. It is quite clear that the original function must have a maximum and a minimum, so these must be the loci of them. The actual maximum value is $\sqrt{14}$ (and the minimum value is $-\sqrt{14}$).