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Basic Facts:

Let $a,b,c\in\Bbb R$, then

  1. If $a<b$ and $c>0$ then $ac< bc$
  2. $a^2\ge0\land a^2=0$ iff $a=0$
  3. If $a\ge0$, there exists a unique number $\sqrt a$ whose square is $a$
  4. If $a<b$ and $b<c$ then $a<c$

My new attempt:

New attempt

Could someone tell me if my approach in proving the inequality is correct? If not could you please tell me where I went wrong?

Thanks in advance.

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  • $\begingroup$ It's a circular argument, you start from the inequality you want to prove, that's wrong. A correct way is much simpler, too: $\endgroup$ – Professor Vector May 10 '17 at 6:38
  • $\begingroup$ Your very first line is false. if $\frac {\sqrt x}{x+ 1} > 1/2$ then $\frac 12*\frac {\sqrt x}{x+ 1} > \frac 12* \frac 12 = \frac 14$ And we have no way of knowing at this point if $\frac {\sqrt x}{x+ 1} > 1/2$ or $\frac {\sqrt x}{x+ 1} \le 1/2$ $\endgroup$ – fleablood May 10 '17 at 6:54
  • $\begingroup$ @OP Please do not vandalize your question. $\endgroup$ – dxiv May 10 '17 at 21:29
  • $\begingroup$ To be honest, with just those four facts I don't think it can be proven. In your text was any information given defining ">"? I suspect there are more we can be using. Example: Have we been told that for all non-zero x that either x < 0 or x > 0. (That'd be a "fact") or that $a < b \iff b-a > 0$? $\endgroup$ – fleablood May 10 '17 at 22:36
  • $\begingroup$ $ We\ proved\ that\ if\ 0<\ a<b\ then\ a^2<b^2\ and\ \sqrt{a}<\sqrt{b} $ $\endgroup$ – user444945 May 10 '17 at 22:39
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Using (2) applied to $a=\sqrt{x}-1$ gives, using that, by (3), $\sqrt{x}^2=x>0$,

$$0\leq (\sqrt{x}-1)^2=x-2\sqrt x+1$$ which immediately yields the desired, by means of (1).

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  • $\begingroup$ Hi, I tried to prove again using your idea. Could you please have a look? $\endgroup$ – user444945 May 10 '17 at 18:24
  • $\begingroup$ How would that lead to the result. We can't claim $0 \le x - 2\sqrt{x} + 1 \implies 2\sqrt{x}\le x + 1$ with only the facts given. $\endgroup$ – fleablood May 10 '17 at 22:41
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Using your argument, we can prove:

$$\frac{\sqrt{x}}{x+1}\le\frac14$$

As we have:

$$\frac14\frac{\sqrt{x}}{x+1}\le\frac1{16}$$

and

$$\frac{x}{(x+1)^2}\le\frac14\frac{\sqrt{x}}{x+1}$$

so

$$\frac{x}{(x+1)^2}\le\frac1{16}$$

and

$$\frac{\sqrt{x}}{x+1}\le\frac14$$

But this isn't true - take $x=1$ for example.

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Your logic is circular. You start with the conclusion and then arrive at that conclusion.

Here is a valid proof:
$0 \leq (x-1)^2$ by fact 2
$\implies 0 \leq \frac{1}{4}(x-1)^2$
$\implies 0 \leq \frac{x^2}{4}-\frac{x}{2}+\frac{1}{4}$
$\implies x \leq \frac{x^2}{4}+\frac{x}{2}+\frac{1}{4}$
$\implies x \leq \frac{1}{4}(x+1)^2$
$\implies \frac{x}{(x+1)^2} \leq \frac{1}{4}$
$\implies \frac{\sqrt x}{x+1} \leq \frac{1}{2}$

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