# Use an indirect proof to show that if $x^3+x-1 \gt 10$ then $x > 1$.

Use an indirect proof to show that if $x^3+x-1 \gt 10$ then $x > 1$.

Let if possible $x\le 1,$ then how to proceed?

EDITED:

Whole question boils down to how to show that if $x\lt 1$ then $x^3\lt 1$?(without using functions, derivatives )

• Of course, the argument should work for all $x\leq 1$ @George, so "subbing in" doesn't quite work. Commented Feb 16, 2016 at 21:55
• If $x \le 1$ then clearly $x^3 \le 1$ and so $x^3 + x -1$ would be less or equal to $1$. Commented Feb 16, 2016 at 21:55
• @FrancescoChini, then we need to show that $x\le 1 \Rightarrow x^3\le 1$
– User
Commented Feb 16, 2016 at 21:57
• True. but $x \to x^3$ is a non decreasing function and $1^3 = 1$. Commented Feb 16, 2016 at 21:58
• @Francesco Chini True, but how to convince someone who do not know about functions.
– User
Commented Feb 16, 2016 at 22:00

If $x\le 1$ then either $x\le 0$ or $0 < x \le 1$. If $x\le 0$ then obviously $x^3\le 0 \le 1$. If $0 < x \le 1$ then multiply the inequality three times with itself (you can do that since both $x$ and $1$ are positive numbers) and you get that $x^3\le 1$.

Thus in any case: $x\le 1$ implies $x^3\le 1$. Summation by parts gives $$x^3+x-1\le 1$$ Thus $$x^3+x-1 \gt 10 \gt 1 \Rightarrow x\gt 1$$

Note $f'(x) = 3x^2 + 1 > 0$ hence the function is strictly increasing. We have $f(1) = 1$. Suppose for some $x<1$ , $f(x) > 10$. This contradicts the function being strictly increasing.

Edit: OP asks for a solution without calculus. We have $1^3 + 1 - 1 = 1 < 10$. Suppose $x \in (0,1)$. Then $x^3\in (0,1)$ clearly. This implies $0<x^3 + x < 2$, or $0<x^3 + x - 1 <1<10$.

For $x<0$, $x^3<0$ and $x<0$ so $x^3 + x - 1$ is at most $-1<10$.

$x \leq 1 \Longrightarrow x^3 + x - 1 < 10$. The assertion follows by contraposition.

We just need to show that if $x \le 1$, then $x^3 \le 1$ and then we are done. We can't use any notion about calculus, as required in the comments. If $x$ is negative, then we would get immediately $x^3 <0 \le 1$. So let's assume $x \ge0$ We know that $$x \le 1.$$ Let's multiply each side by $x$. $$x^2 \le x \le 1.$$ Again let's multiply every term by $x$. $$x^3 \le x^2 \le x \le 1.$$

Suppose $x < 1$, then $x - 1 < 0 \implies 0 > (x -1)\left[(x + \frac{1}{2})^{2} + \frac{3}{4}\right] = x^{3} - 1$.

• I want an indirect proof.
– User
Commented Feb 16, 2016 at 22:27
• I have offered that now in the edited version. Commented Feb 16, 2016 at 22:36