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I am trying to learn how to prove inequalities by only using basic ordering axioms. I have the following inequality $$\frac1{1-x} \le 1+2x \ \ \ \forall x\in\mathbb{R},0\le x \le \frac12$$

I came to this point: $$x \ge 2x^2$$ But here ie where I get stuck. If I only use axioms, how can I prove that statment? I know that for $x=0 $ and $x= \frac12$ the inequality holds. But for the values in between, how do I formulate a formal prove?

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  • $\begingroup$ Consider 0<x<1/2. Ie:1/2-x>0. By closure of postiives 1/2x-x^2>0,x-2x^2>0. $\endgroup$ – Jacob Wakem Nov 6 '16 at 21:48
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$x \le \dfrac{1}{2} \implies x - \dfrac{1}{2} \le 0 \implies \dfrac{2x-1}{2} \le 0\implies 2x -1 \le 0$. Also, $x \ge 0 \implies x(2x-1) \le 0\implies 2x^2 - x\le 0\implies x-2x^2 \ge 0\implies (2x-x) - 2x^2 \ge 0\implies 1 + 2x - x - 2x^2 \ge 1\implies (1+2x) - x(1+2x) \ge 0 \implies (1-x)(1+2x) \ge 1\implies \dfrac{1}{1-x} \le 1+2x$ since $1 - x > 0$.

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Since you already know it holds for $x=0$, we can assume $x\neq 0$. Then we can divide by $x$ and get:
$$ 1\geq 2x $$ so: $$ x\leq \frac{1}{2} $$ Which we assumed in the beginning.

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