The problem is this: Prove by induction that $x_1^2+x_2^2+.....+x_n^2>0$ unless $x_1=...=x_n=0. $ I have tried:

If $x_n$> 0 then For $n=1$: $$x_1^2>0$$

We suppose that it is true for $n=k$:$$ x_1^2+x_2^2+...+x_k^2>0$$

We will prove for $n=k+1$:

As $x_{n}^2>0$ so $x_{k+1}^2 >0$

and $$ x_1^2+x_2^2+...+x_k^2>0 $$

therefore $$x_1^2+x_2^2+.....+x_{k+1}^2>0.$$ Q.E.D.

But I am not sure. Thank you.

  • 5
    $\begingroup$ Induction is really overkill here. All you need is that $a^2$ is greater than zero for any non-zero (real) $a$ and that a sum of positive numbers is positive. Q.E.D. $\endgroup$ – Brevan Ellefsen Dec 22 '16 at 4:56
  • $\begingroup$ Yes, what @BrevanEllefsen said is true. Also you do not need the fact that $x_n>0$ on your third line, because squaring a negative number is always positive $\endgroup$ – wesssg Dec 22 '16 at 4:58

For $n = 1, x_1^2 > 0$ unless $x_1 = 0$, thus the base case is cleared.Assume its true for $n = k$, then $x_1^2+x_2^2+\cdots x_{k+1}^2 \ge x_1^2+x_2^2+\cdots + x_k^2 > 0$, unless $x_{k+1} = 0 = x_1 = x_2 = .. = x_k$. Thus the statement is true for all $n \ge 1$.


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