How to prove that $x^2 +1 \geq 2x$? 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.
 A: 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).
A: 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!
A: $x^2 +1 \geq 2x\iff x^2 -2x+1 \geq0$
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 $x^2 +1 \geq 2x$
A: $$x^2 + 1 = x^2 - 2x + 1 + 2x = \underbrace{(x-1)^2}_{\geq 0} + 2x \geq 2x$$
A: 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.$\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\qquad\,\,\,\,\,\,\Box$
