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I am trying to prove that $x^2 +1 \geq 2x$ without using circular logic (meaning first assuming that this inequality is true and then moving to the $2x$ to the left side and factoring it). Thanks.

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    $\begingroup$ $(x-1)^2 > 0$ .. $\endgroup$ – Donald Splutterwit Sep 10 '17 at 21:51
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    $\begingroup$ @DonaldSplutterwit for |x|>1 $\endgroup$ – user451844 Sep 10 '17 at 21:53
  • $\begingroup$ It's not true for $x=1$. You can show $x^2+1\geq 2x$, however. $\endgroup$ – Thomas Andrews Sep 10 '17 at 21:55
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    $\begingroup$ What you should actually be doing is showing that $x^2 + 1 \geq 2x$ is equivalent to $(x-1)^2 \geq 0$, which is true. This is not circular. $\endgroup$ – mechanodroid Sep 10 '17 at 21:55
  • $\begingroup$ okay sorry not equal to 1 it should be. $\endgroup$ – user451844 Sep 10 '17 at 21:58
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$$x^2 + 1 = x^2 - 2x + 1 + 2x = \underbrace{(x-1)^2}_{\geq 0} + 2x \geq 2x$$

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    $\begingroup$ beautiful...........!!!! $\endgroup$ – Stu Sep 10 '17 at 23:05
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Work backwards . . .

\begin{align*} &(x-1)^2 \ge 0&&\text{[since squares are nonnegative]}\\[4pt] \implies\;&x^2-2x+1 \ge 0\\[4pt] \implies\;&x^2 + 1 \ge 2x\\[4pt] \end{align*} Alternatively, work forwards, but use $\iff$ . . . \begin{align*} &x^2+1 \ge 2x \qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\;];\;\;\; \\[4pt] \iff\;&x^2-2x + 1 \ge 0\\[4pt] \iff\;&(x-1)^2 \ge 0\\[4pt] &\text{which is true since squares are nonnegative.}\\[4pt] \end{align*} This allows the argument to be reversed without actually reversing it.

But to use the $\iff$ version, you need to be careful to make sure that each line is equivalent to the previous one (i.e., each implies the other).

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If $x \le 0$, it is true.

If $x=1$, there is equality.

If $x>1$, you put $x=1+\epsilon $ with $\epsilon>0$.

then $$x^2+1=(1+\epsilon)^2+1$$ $$=1+\epsilon^2+2\epsilon+1$$ $$=2 (1+\epsilon)+\epsilon^2>2x $$

If $0 <x <1$ put $y=\frac {1}{x}>1 $. then $$y^2+1=\frac {1}{x^2}+1$$ $$=\frac {x^2+1}{x^2}>2\frac {1}{x} $$ $$\implies 1+x^2>2x $$ Done!

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$x^2 +1 \geq 2x\iff x^2 -2x+1 \geq0$

let $f(x):= x^2 -2x+1 \implies f'(x)= 2x-2 $

$f'(x)=0 \implies x=1$

$f$ is decreasing over $(-\infty,1]$ and increasing over $[1,+\infty)$ and $f(1)=0$ so $f(x)\ge 0$

So we have proved $f(x)=x^2 -2x+1\ge 0$ so $x^2 +1 \geq 2x$

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Proof: $$\begin{align} x^2 + 1 &\geqslant 2x \\ \iff x^2 + 1 - 2x &\geqslant 0 \\ \iff x^2 + 1^2 - 2\cdot x\cdot 1 &\geqslant 0 \\ \iff (x - 1)^2 &\geqslant 0,\end{align}$$ as desired.$\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\,\,\,\,\,\,\Box$

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  • $\begingroup$ Pardon me if I'm wrong, but $P\Rightarrow Q$ doesn't necessarily imply $Q\Rightarrow P$. For example, $(x=\pi)\Rightarrow(\sin(x)=0)$ but the converse isn't always true. For instance, $\sin(x)=0$ could mean $x=0$ or $x=\pi$, etc. So I think there needs to be another step to show that $x-1\geq0\Rightarrow x^2+1\geq 2x$. $\endgroup$ – Jam Sep 14 '17 at 15:01
  • $\begingroup$ Also, $(x-1)^2\geq0\Rightarrow (x\geq 1)$ isn't necessarily true (take $x=0$ for example). You've got to consider both sides of the parabola. $\endgroup$ – Jam Sep 14 '17 at 15:05
  • $\begingroup$ Woah sorry about that. I am going to re-edit. $\endgroup$ – George N. Missailidis Sep 22 '17 at 22:04
  • $\begingroup$ @GeorgeN.Missailidis No I will $\endgroup$ – user477343 Sep 22 '17 at 22:07

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