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

Question

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 )

Answer 1

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 $$x^3+x-1 \gt 10 \gt 1 \Rightarrow x\gt 1$$

Answer 2

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.

Answer 3

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 $x$. $$ x^3 \le x^2 \le x \le 1. $$

Answer 4

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