# Proving with the simplest method possible: $X^2+1 > X$ [closed]

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

The simpler the method the more welcome it is.

• Is $x$ a real number? – Axel Aug 18 '20 at 16:16
• Just complete the square. Hint: $\left(x-\frac 12\right)^2=x^2-x+\frac 14$. – lulu Aug 18 '20 at 16:16
• The derivative of $x^2-x+1$ has root at $\frac12$ which is positive. – Qi Zhu Aug 18 '20 at 17:24
• 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 ;) – Stinking Bishop Aug 18 '20 at 17:42

$$x^2+1>0 \Rightarrow x^2+1>\frac{x^2+1}{2}$$ $$\frac{x^2+1}{2}-x=\frac{(x-1)^2}{2}\ge 0$$ $$\hbox{Thus }x^2+1-x>\frac{x^2+1}{2}-x\ge 0$$ $$x^2+1>x,\hbox{ QED.}$$

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 $$X^2+1\ge1>X$$.

$$x^2-x+1=\frac{x^3+1}{x+1}$$ 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.

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

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

If $$x=0$$ , then

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

• $2|x|=|x|$ when $x=0$ – J. W. Tanner Aug 18 '20 at 17:55
• @J.W.Tanner Thanks for pointing out. – user-492177 Aug 18 '20 at 19:06

For $$x=1, 2>1$$. For $$x>1$$, $$x^2>x$$. For $$x<1,$$ $$x< 1$$ and $$x^2\geq 0$$.

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.

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

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

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 $$x^2+1=|x|^2+1=|x|\cdot (|x|+1/|x|)>|x|\ge x.$$