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 $c=p\cdot n = n^2/|n| = |n|$
$$x^2 + y^2 +z^2 =1$$

$x + 2y + 3z$ can be represented as dot product of two vectors

$$x + 2y + 3z = ( xi + yj + zk) \cdot ( i + 2j + 3k )$$

Using $u \cdot v = |u | | v| \cos \theta $

$$( xi + yj + zk) . ( i + 2j + 3k ) = |\sqrt{x² + y² + z²} |. |\sqrt{1² + 2² + 3²}| | \cos \theta $$

$$ x + 2y + 3z = 1 \cdot \sqrt{14} \cos \theta$$

Maximum value of $\cos \theta = 1$

Hence maximum value of $x + 2y + 3z = \sqrt{14}$