Solve the inequality $\sqrt x+\sqrt{x+1}>\sqrt 3$.

I want to make sure my method is correct:

The condition is that $x\geq 0$

$x+2\sqrt x\times \sqrt{x+1} +x+1>3$






$x\in (\frac{1}{3},\infty)$

I know my final solution is fine, but is everything written properly? Should I put $\iff$ at the beginning of each row?

  • 1
    $\begingroup$ A $\iff$ would not be correct as you are squaring the equation, and $y=x^2$ is a many-one function. $\endgroup$ – GoodDeeds Nov 1 '16 at 18:19
  • $\begingroup$ In step three how do you know that 2-2x >= 0 or that 2-2x < 0 but |2-2x| < 2\sqrt{x(x+1)? $\endgroup$ – fleablood Nov 1 '16 at 18:48
  • $\begingroup$ It might just be me, but I think worry about $\iff$. Sometimes the conclusions will only go one way. You will have to worry about addding extraneous information, especially when you square. The results of which will certainly not be an if and only if. $\endgroup$ – fleablood Nov 1 '16 at 18:55

Over its domain (the set of non-negative real numbers) the function $f(x)=\sqrt{x}+\sqrt{x+1}$ is an increasing function, since it is the sum of two non-negative increasing functions. It follows that $f(x)>\sqrt{3}$ holds as soon as $x>x_0$, where $x_0$ is the only positive number such that $$ \sqrt{x_0}+\sqrt{x_0+1} = \sqrt{3}.\tag{1} $$ $x_0=\frac{1}{3}$ is clearly a solution of $(1)$, hence the given inequality holds for $\color{red}{\large x>\frac{1}{3}}$.

  • $\begingroup$ I guess you meant $ \sqrt{3} $ and not $ 3 $ . $\endgroup$ – Sylvain Julien Nov 1 '16 at 19:46
  • $\begingroup$ @SylvainJulien: sure, thanks. Now fixed. $\endgroup$ – Jack D'Aurizio Nov 1 '16 at 19:50

We have

$x\geq 0$ .


$\sqrt{x}+\sqrt{x+1}>\sqrt{3} \implies$

$\sqrt{x+1}>\sqrt{3}-\sqrt{x} \implies $

$x+1>3+x-2\sqrt{3x} \implies$

$\sqrt{3x}>1 \implies x>\frac{1}{3}$

In the other direction,

$x>\frac{1}{3} \implies$

$\sqrt{x+1}+\sqrt{x}>\sqrt{\frac{4}{3}}+\sqrt{\frac{1}{3}} =\sqrt{3}$


  • $\begingroup$ But what if $\sqrt{3} - \sqrt{x}$ is negative? Then we can't square both sides of the inequality, let alone use the $\iff$ sign. $\endgroup$ – Stefan4024 Nov 1 '16 at 18:27
  • $\begingroup$ If it is negative, the inequality is satisfied since $\sqrt{x+1}\geq 0$. $\endgroup$ – hamam_Abdallah Nov 1 '16 at 18:29
  • $\begingroup$ Yeah, but you have to explicitly mentioned that. Otherwise you're just squaring, which might not be true after all. $\endgroup$ – Stefan4024 Nov 1 '16 at 18:31
  • $\begingroup$ What @Stefan4024 is getting at is that 1 > -2, but squaring both sides doesn't preserve the inequality. $\endgroup$ – Turambar Nov 1 '16 at 18:34

Your answer is fine and correct, and you can write it more concise with key steps. In the end, you can simply write $x > 1/3$ without re-write it as $x \in (1/3, \infty)$.


You've made a mistake by assuming that $2 - 2x \ge 0$, when squaring in the second row. To fix this consider the two cases when $x < 1$ and $x \ge 1$.

This will enable you to add the $\iff$ signs, which you're required, as otherwise you have proven that $\sqrt{x} + \sqrt{x+1} > \sqrt{3} \implies x > \frac 13$ instead of $x > \frac 13 \implies \sqrt{x} + \sqrt{x+1} > \sqrt{3}$

Assume that $x > 1$. Then we have that $2-2x < 0 < 2\sqrt{x(x+1)}$, so the inequality is true for any $x \ge 1$.

On the other side when $x < 1$ you can continue in your way and you will get that $x \in \left(\frac 13,1\right)$. Now combining the answers you will get that the solution set is $\left(\frac 13,\infty\right)$

  • $\begingroup$ if $x\geq 1$ then I have $4x(x+1) < 4-8x+4x^{2}$. I change the inequality sign? $\endgroup$ – lmc Nov 1 '16 at 18:32
  • $\begingroup$ @Now_now_Draco_play_nicely You can quickly discard the case $x\ge 1$, as the RHS is negative, but the LHS is positive. It's a trivial thing, but you have to be careful about it. $\endgroup$ – Stefan4024 Nov 1 '16 at 18:37
  • $\begingroup$ I'm really confused now. Can you write out how you would solve this problem step by step? $\endgroup$ – lmc Nov 1 '16 at 18:41
  • 1
    $\begingroup$ @Now_now_Draco_play_nicely You can check now $\endgroup$ – Stefan4024 Nov 1 '16 at 18:44
  • $\begingroup$ Let me ask you just one more thing if I had for example $2\sqrt{x(x+1)}<2-2x$ then I wouldn't need to check for $x\geq 1$ since on the RHS I would have a non-negative expression? $\endgroup$ – lmc Nov 1 '16 at 19:15

Note that $\sqrt{\frac13}+\sqrt{\frac13+1}=\sqrt{3}$ . The inequality follows trivially.

  • $\begingroup$ Also, the same argument is essentially given by Jack D'Aurizio, but in an answer of much higher quality. $\endgroup$ – wythagoras Nov 1 '16 at 20:54
  • $\begingroup$ @wythagoras it doesn't. I am comfortable with your edit. $\endgroup$ – Jacob Wakem Nov 1 '16 at 20:54
  • $\begingroup$ @wythagoras I was not aware of his solution. I am not comfortable with an intersubjective standard of quality. $\endgroup$ – Jacob Wakem Nov 1 '16 at 20:56

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