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This is a simple inequality problem that I have found in a book (as part of a solution of a larger problem), however I have failed to prove it.

The only proof provided is that it is true since $(x-2)^2\ge0$ (I hope this could help you).

Can anyone please provide me with a proof or even the start of it.

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  • $\begingroup$ Hint: $\sqrt {x-1} >0$ at least for $x>1$ so you can cross multiply safely. $\endgroup$
    – lulu
    Commented Aug 28, 2016 at 23:37
  • $\begingroup$ Do you mean $x\ge2$ ? But how does this prove it ? $\endgroup$ Commented Aug 28, 2016 at 23:37
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    $\begingroup$ Necessarily $x > 1$. Now, multiply by $\sqrt{x-1}$ and take the squares. $\endgroup$
    – Crostul
    Commented Aug 28, 2016 at 23:38
  • $\begingroup$ Try working backwards. Keep in mind $x>1>0$ or else the expression is not defined over the real numbers. $\endgroup$ Commented Aug 28, 2016 at 23:38
  • $\begingroup$ I meant $x>1$, else the left hand is undefined. I didn't say it proved it, it was a hint not a complete solution. Useful first step. $\endgroup$
    – lulu
    Commented Aug 28, 2016 at 23:39

5 Answers 5

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Note that for every $ x > 1 $, we have \begin{align} \frac{x}{\sqrt{x - 1}} & = \frac{x - 1}{\sqrt{x - 1}} + \frac{1}{\sqrt{x - 1}} \\ & = \sqrt{x - 1} + \frac{1}{\sqrt{x - 1}} \\ & \geq 2 \sqrt{\sqrt{x - 1} \cdot \frac{1}{\sqrt{x - 1}}} \qquad (\text{By the AM-GM Inequality.}) \\ & = 2. \end{align}

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  • $\begingroup$ How is $\frac{x - 1}{\sqrt{x - 1}} + \frac{1}{\sqrt{x - 1}} = \sqrt{x - 1} + \frac{1}{\sqrt{x - 1}}$ $\endgroup$ Commented Aug 28, 2016 at 23:54
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    $\begingroup$ Observe that $ \dfrac{x - 1}{\sqrt{x - 1}} = \dfrac{\left( \sqrt{x - 1} \right)^{2}}{\sqrt{x - 1}} = \sqrt{x - 1} $ for every $ x > 1 $. $\endgroup$ Commented Aug 29, 2016 at 0:00
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    $\begingroup$ @BasemFouda Note that $\frac{a}{\sqrt a}$ = $\frac{a^{2/2}}{a^{1/2}} = a^{2/2 - 1/2} = \sqrt a$. $\endgroup$
    – MathMajor
    Commented Aug 29, 2016 at 0:00
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An elementary approach: squares are positive so $(x-2)^2\ge 0$. Dividing by $4x^2$ you obtain $$\frac{x^2-4x+4}{4x^2}\ge 0$$ Separating the fractions, this becomes $$\frac{1}{4}\ge\frac{1}{x}-\frac{1}{x^2}$$ Since both quantities are positive if $x>1$, we can take the square root and $$\frac{1}{2}\ge\sqrt{\frac{1}{x}-\frac{1}{x^2}}=\frac{\sqrt{x-1}}{x}$$ Taking reciprocals yields the result: $$\frac{x}{\sqrt{x-1}}\ge 2$$

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  • $\begingroup$ I like your solution. Frankly speaking, it is more elementary than mine as it is cast entirely in the language of ordered fields. $\endgroup$ Commented Aug 29, 2016 at 0:07
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A solution which uses the hint, keep in mind, everything is positive and we have $x>1$:

We start with \begin{align} \frac{x}{\sqrt{x-1}} \ge 2&\iff x\ge2{\sqrt{x-1}}\\ &\iff x^2\ge4(x-1)\\ &\iff x^2-4x+4\ge0 \end{align} and now we see that $$ x^2-4x+4=(x-2)^2\ge0 \text{ (hint)} $$ and therefore we have $$ \frac{x}{\sqrt{x-1}} \ge 2\iff(x-2)^2\ge0 $$ which is true and therefore proves the above statement.

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Hint: differentiate $\frac{x}{\sqrt{x-1}}$, find the minimum (which is $2$ at $x=2$).

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Note that $$\frac{x}{\sqrt{x-1}} = \frac{t^2+1}{t} = t+\frac{1}{t} \ge 2$$ where $t = \sqrt{x-1} \ge 0$.

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