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I need a method to prove $X^2+1 >X$ using simple algebra.

The simpler the method the more welcome it is.

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    $\begingroup$ Is $x$ a real number? $\endgroup$ – Axel Aug 18 '20 at 16:16
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    $\begingroup$ Just complete the square. Hint: $\left(x-\frac 12\right)^2=x^2-x+\frac 14$. $\endgroup$ – lulu Aug 18 '20 at 16:16
  • $\begingroup$ The derivative of $x^2-x+1$ has root at $\frac12$ which is positive. $\endgroup$ – Qi Zhu Aug 18 '20 at 17:24
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    $\begingroup$ I would love to see a question where we would be asked to prove $X^2+1\gt X$ in the most roundabout way possible. Hopefully using some deep results from group theory applied to some obscure group. I may ask such a question myself, later ;) $\endgroup$ – Stinking Bishop Aug 18 '20 at 17:42
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$$x^2+1>0 \Rightarrow x^2+1>\frac{x^2+1}{2}$$ $$\frac{x^2+1}{2}-x=\frac{(x-1)^2}{2}\ge 0$$ $$\hbox{Thus }x^2+1-x>\frac{x^2+1}{2}-x\ge 0$$ $$x^2+1>x,\hbox{ QED.}$$

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For $X\ge1$ it's true because in that case $X^2\ge X$ so $X^2+1>X^2\ge X$.

For $X\lt1$ it's true because in that case $X^2+1\ge1>X$.

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$x^2-x+1=\frac{x^3+1}{x+1}$ when $x\neq -1$. Notice that $x^3+1$ and $x+1$ are both positive on $(-1, \infty)$ and negative on $(-\infty,-1)$. So, the fraction is always positive.

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If $x\neq 0$ ,by $AM-GM $ Inequality

$x^2+1\ge 2|x| \gt |x|\ge x $

If $x=0$ , then

$1=x^2+1\gt x=0$

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  • $\begingroup$ $2|x|=|x|$ when $x=0$ $\endgroup$ – J. W. Tanner Aug 18 '20 at 17:55
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    $\begingroup$ @J.W.Tanner Thanks for pointing out. $\endgroup$ – user-492177 Aug 18 '20 at 19:06
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For $x=1, 2>1$. For $x>1$, $x^2>x$. For $x<1,$ $x< 1$ and $x^2\geq 0$.

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Notice that $x^2 \geq 0$. Break this into three cases.

Case 1: $x <0$. Then $x < 0 < 1 \leq x^2 + 1$.

Case 2: $x\in[0,1]$. then $x \leq 1 \leq 1 + x^2$ because again $x^2\geq 0$.

Case 3: $x > 1$. Since $1 < x $ and since $x$ is positive, then multiplying by $x$ preserves the inequality. So $1<x \implies x < x^2 < x^2 + 1$.

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For $x\le 0$ the inequality is trivial since LHS is always positive.

For $x>0$ we have

$$x^2+1>x \iff x^2-x+1 >0$$

which is true indeed

$$x^2-x+1 >x^2-2x+1=(x-1)^2\ge 0$$

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If $x=0$ then $x^2+1=1>0=x.$

If $x\ne 0$ then $|x|>0\,$ so $(|x|\ge 1 \lor 1/|x|\ge 1)\,$

so $\max (|x|,1/|x|)\ge 1$ and $\min (|x|,1/|x|)>0$ $$\text {so}\quad |x|+1/|x|= \max (|x|,1/|x|)+\min (|x|,1/|x|)>1$$ so $x^2+1=|x|^2+1=|x|\cdot (|x|+1/|x|)>|x|\ge x.$

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