# A Triangle inequality question

1. Jun 21, 2016

### LagrangeEuler

In the derivation of triangle inequality $$|(x,y)| \leq ||x|| ||y||$$ one use some $z=x-ty$ where $t$ is real number. And then from $(z,z) \geq 0$ one gets quadratic inequality
$$||x||^2+||y||^2t^2-2tRe(x,y) \geq 0$$
And from here they said that discriminant of quadratic equation
$$D=4(Re(x,y))^2-4 ||y||^2|x||^2 \leq 0$$
Could you explain me why $<$ sign in discriminant relation? When discriminant is less then zero solutions are complex conjugate numbers. But I still do not understand the discussed inequality. What about for example in case
$$||x||^2+||y||^2t^2-2tRe(x,y) \leq 0$$?

2. Jun 21, 2016

### blue_leaf77

$$||x||^2+||y||^2t^2-2tRe(x,y) \geq 0$$
must always be zero or positive for all real $t$. This means the function $f(t)=||x||^2+||y||^2t^2-2tRe(x,y)$ must always be above the $t$ axis, which further implies that it can have at most one solution (which occurs when the minimum of this function touches the $t$ axis): one solution or no solution at all. In terms of the discrimant, the discrimant of $f(t)$ is either zero or negative.