Proof of $ca^2 < cb^2$ in an ordered field

Question

Let $(K,+,\cdot, P)$ be a totally ordered field.

How can one prove that for $a,b,c \in K$ it holds that if $a < b$ and $c > 0$ then $ca^2 < cb^2$?

I know that if $a < b \wedge c > 0$ it implies that $ac < bc$ because from $a < b$ it follows that $a+c \leq b+c$ and if $a+c=b+c $ it would follow that $a=b$ which contradicts $a < b$.

But how can I apply the same analogy to $ca^2 < cb^2$?

Answer 1

It is false. In $\Bbb Q$ we have $1>0$, $-2<-1$ and $1\cdot(-2)^2>1\cdot(-1)^2$.