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Let some $k>1$ - since for $k=1$ we have nothing to prove ($x=x$). Starting from the fact that $a<b∧ c<d\Rightarrow a+c<b+d$ we arrive to the fact that for any $x>0$ we have $x+x>0$. Similarly, we can have $x+x+x>0$ and, so on, applying this "trick" $k-1$ times we arrive to: $$\underbrace{x+x+\ldots+x}_{k-1\text{ times}}>0.$$ Now, ...