Cauchy- schwartz inequality states that: If u, v are vectors such that $u = (u_{1}, u_{2}, u_{3}, .........u_{n}) and v = (v_{1}, v_{2}, v_{3},..............v_{n})$ in $R^n$ in a real inner product space then

$$|u.v| \geq ||u|| * ||v||..........................(1)$$

Proof: We know that if u and v are any two vectors in $R^2$ and $R^3$ then

$$|u.v| \geq ||u|| * ||v|| cos\theta$$

But $cos\theta \leq 1 \\ \therefore |u*v| \leq ||u|| * ||v||$

Let $u = (x,y,z)$ & $v = (3,4,12)$ where x,y,z are positive

$$\therefore ||u|| = \sqrt{x^2 + y^2 + z^2} \\ ||v|| = \sqrt{3^2 + 4^2 + 12^2} = \sqrt{169} = 13 \\ \therefore ||u|| * ||v|| = 13 * \sqrt{x^2 + y^2 + z^2}$$

Now, $|u * v| = |u_{1}v_{1} + u_{2}v_{2} + u_{3}v_{3} + u_{4}v_{4} | = |3x + 4y + 12z| = 3x + 4y + 12z $

From (1)

$$(3x + 4y + 12z) \leq 13 \sqrt{x^2 + y^2 + z^2}$$

i.e. $(x^2 + y^2 + z^2)^{\frac{1}{2}} \leq \frac{1}{13} (3x + 4y + 12z), x,y,z$ are positive.