# Why is solution to inequality $\sqrt{1 - x} - \sqrt{x} > \frac{1}{\sqrt{3}}$ equal to interval $[0, \frac{3 - \sqrt{5}}{6})$?

Given inequality $\sqrt{1 - x} - \sqrt{x} > \frac{1}{\sqrt{3}}$ we can easily determine, that it's domain is $D = [0, 1]$. Because each term is real, we can take square of the inequality, which yields: $$\frac{1}{3} > \sqrt{x}\sqrt{1 -x}.$$ Squaring it again we get inequality: $$9x^2 - 9x + 1 > 0,$$ solution to which is a domain $[0, \frac{3 - \sqrt{5}}{6}) \cup (\frac{3 + \sqrt{5}}{6}, 1]$. But, solution to original inequality is just $[0, \frac{3 - \sqrt{5}}{6})$. What am I omitting/where I'm doing mistakes?

• See that when you make the quadratic, you assumed $x^2=a\implies x=\sqrt{a}$, when actually $x^2=a\implies x=\pm\sqrt a$. The $\pm$ sneaks up on you, and this is called an extraneous solution. – Simply Beautiful Art Aug 26 '16 at 20:48
• The mistake is in the "which yields" step, because it doesn't yield that – user290300 Aug 26 '16 at 20:51
• @SimpleArt how can I take care of extraneous solutions in case of the inequalities? We haven't studied them in HS, so I'm asking... – Accelerate to the Infinity Aug 26 '16 at 20:52
• @AcceleratetotheInfinity To take care of extraneous solutions, simply check your answer. Like see if it works for $x=0.9$ – Simply Beautiful Art Aug 26 '16 at 20:56
• You are missing a step right before squaring both sides. Since the LHS must be greater than $\frac 1 {\sqrt 3}$, this means $\sqrt{1-x} > \sqrt x$ and therefore $x < \frac 1 2$. With this condition, both sides are positive, so the inequality is EQUIVALENT to the one obtained by squaring both sides. Without this step, you have the forward implication, but not the implication back to the original inequality. Now you can solve but remember $x < \frac 1 2$. – mathguy Aug 26 '16 at 21:08

The number $\sqrt{1-x}-\sqrt{x}$ can be negative and, in this case, the inequality doesn't hold. Squaring both sides can be done only if both sides are nonnegative. Thus your inequality becomes $$\begin{cases} 0\le x\le 1 \\[8px] \sqrt{1-x}-\sqrt{x}\ge0 \\[4px] 1-x-2\sqrt{x(1-x)}+x>\dfrac{1}{3} \end{cases}$$ The second inequality becomes $1-x\ge x$, so $x\le1/2$. The third inequality is $$\sqrt{x(1-x)}<\frac{1}{3}$$ so $9x^2-9x+1>0$ that has the solutions $$x<\frac{3-\sqrt{5}}{6}\qquad\text{or}\qquad x>\frac{3+\sqrt{5}}{6}$$ that should be combined with $0\le x\le 1/2$. Thus the solution set of the original equation is $$\left[0,\frac{3-\sqrt{5}}{6}\right)$$ because $(3+\sqrt{5})/6>1/2$.

A different strategy is to write the inequality as $$\sqrt{3}\sqrt{1-x}>1+\sqrt{3}\sqrt{x}$$ With the constraint $0\le x\le1$, this becomes $$3-3x>1+2\sqrt{3}\sqrt{x}+3x$$ or $$3x+\sqrt{3}\sqrt{x}-1<0$$ Setting $t=\sqrt{x}$ (with $0\le t\le1$), this is quadratic: $3t^2+\sqrt{3}t-1<0$. So we have the solution set \begin{cases} 0\le t \le 1 \\[8px] \dfrac{-\sqrt{3}-\sqrt{15}}{6}<t<\dfrac{-\sqrt{3}+\sqrt{15}}{6} \end{cases} that becomes $$0\le t<\dfrac{-\sqrt{3}+\sqrt{15}}{6}$$ and therefore $$0\le x<\left(\dfrac{-\sqrt{3}+\sqrt{15}}{6}\right)^2=\frac{3-\sqrt{5}}{6}$$

Observe first that if must be $\;0\le x\le 1\;$ , so the given interval is within this constraint. Second, square:

$$\sqrt{1-x}-\sqrt x\ge\frac1{\sqrt3}\implies1-x-2\sqrt{x(1-x)}+x\ge\frac13\implies$$

$$2\sqrt{x(1-x)}\le\frac23\implies x(1-x)\le\frac19\implies 9x^2-9x+1\ge0\implies$$

The last quadratic's roots are given by

$$x_{1,2}=\frac{9\pm\sqrt{45}}{18}=\frac{3\pm\sqrt5}6\implies$$

$$9x^2-9x+1\ge0\iff \left(x-\frac{3-\sqrt5}6\right)\left(x-\frac{3+\sqrt5}6\right)\ge0\iff$$

$$x\le\frac{3-\sqrt5}6\;\;\text{or}\;\;x\ge\frac{3+\sqrt5}6\;\;\;\color{red}{(***)}$$

and we get what we want from the leftmost inequality.

Added on request: If we take

$$f(x)=\sqrt{1-x}-\sqrt x\implies f'(x)=-\frac1{2\sqrt{1-x}}-\frac1{2\sqrt x}=-\frac12\left(\frac1{\sqrt{1-x}}+\frac1{\sqrt x}\right)<0\implies$$

$\;f\;$ is monotone descending, and since

$$f\left(\frac{3-\sqrt5}6\right)=\sqrt{1-\frac{3-\sqrt5}6}-\sqrt\frac{3-\sqrt5}6=\frac1{\sqrt6}\left(\sqrt{3+\sqrt5}-\sqrt{3-\sqrt5}\right)\le\frac1{\sqrt3}\;(*)$$

since

$$(*)\iff \left(\sqrt{3+\sqrt5}-\sqrt{3-\sqrt5}\right)^2\le\left(\sqrt\frac63\right)^2\iff6-4\le2\;\;\color{green}\checkmark$$

so in $\;\color{red}{(***)}\;$ only the left inequality applies.

• I think you want to explain why $x\in(\frac{3+\sqrt{5}}{6},1)$ is not included in the solution. – user84413 Aug 26 '16 at 20:59
• You cannot reduce the solution to the leftmost inequality ; the rightmost inequality provides solutions : interval $(3+\sqrt{5})/6,1]$ that the OP also finds... – Jean Marie Aug 26 '16 at 21:00
• @JeanMarie You're right, thanks. Added explanation why the rightmost interval is not a solution. – DonAntonio Aug 26 '16 at 21:18
• @user84413 Correct, explanation added. Thanks. – DonAntonio Aug 26 '16 at 21:18

Here is a solution mostly based on a graphical understanding (nevertheless written with rigorous arguments).

We acknowledge first that the domain of definition of the inequation is $(0,1)$.

Let us write this inequation under the form:

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

As it is an inequality between positive numbers, it is equivalent to the "squared" inequality:

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

Re-arranging this inequality in the following way:

$$\tag{1}\dfrac{1}{3}-x>\dfrac{1}{\sqrt{3}}\sqrt{x}$$

(1) can be expressed under the form: $\tag{2}f(x)>g(x)$ where

• $f(x):=\dfrac{1}{3}-x$, decreasing on $[0,1]$ with, in particular, $f(0)=1/3$ and $f(1/3)=0.$

• $g(x):=\dfrac{1}{\sqrt{3}}\sqrt{x}$, increasing on $[0,1]$ with, in particular, $g(0)=0$ to $g(1/3)=\dfrac{1}{3}>0.$

Thus, their associated curves possess a unique intersection point $A$ whose abscissa $x_0$ is the solution to $f(x)=g(x)$ (see figure below). A little computation (already done by the OP) shows that $x_0$ is one of the roots $x_0=\dfrac{3\pm\sqrt{5}}{6}$ of

$$\tag{3} 9x^2-9x+1=0$$

This root has to be $<1/3$ ; thus it is $x_0=\dfrac{3-\sqrt{5}}{6}$.

Thus the looked for region corresponds to the values of $x$ that are in $[0,x_0)$ (See curves below).

Remark: an interesting fact is that the "fake" second solution of the quadratic equation (3) can be visualized (and then understood) as the intersection $B$ of the straight line with "mirror" (dotted) curve with equation $y=-\sqrt{x/3}$, which appears into the play when $\sqrt{x}$ is squared...

Well, there is this but I'm not sure how to generalize it.

$\sqrt{1-x} - \sqrt{x} > 0$ so $1-x > x$ so $x < \frac 12$ and $x \in [0,1/2)$.

When you square but sides of $\sqrt{1-x} - \sqrt{x} > 1/\sqrt{3}$ to get $\frac{1}{3} > \sqrt{x}\sqrt{1 -x}$ you are extraneously adding the possibility that $\sqrt{x} - \sqrt{1-x} > 1/\sqrt{3}$ (i.e that maybe $x \in (1/2, 1]$) which we know can not be true.

When we square a second time to get $9x^2 - 9x + 1 > 0$ we are extraneously adding the possiblities that $x < 0$ or $x > 1$. (You caught those).

The solution to $9x^2 - 9x + 1 > 0$ is $(\infty, \frac{3-\sqrt{5}}{6})\cup (\frac{3+\sqrt{5}}{6}, \infty)$.

You caught it should be $[(\infty, \frac{3-\sqrt{5}}{6})\cup (\frac{3+\sqrt{5}}{6}, \infty)]\cap [0,1]$ but you didn't catch it should be

$[(\infty, \frac{3-\sqrt{5}}{6})\cup (\frac{3+\sqrt{5}}{6}, \infty)] \cap [0,1/2)$.

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I guess the way to generalize this to note every time you square an inequality make a note to combine with the known inequality of sign.

i.e.

$\sqrt{1-x} - \sqrt{x} > 1/\sqrt{3}$

Square both sides and conclude

$\sqrt{x}\sqrt{1-x} < 1/3$ AND $\sqrt{1-x} - \sqrt{x} \ge 0$.

So $\sqrt{1-x} \ge \sqrt{x}$.

Square those both side to get $1-x \ge x$ AND $x \ge 0$ and $x \le 1$

So $\sqrt{x}\sqrt{1-x} < 1/3$ And $x \in (-\infty, 1/2]) \cap [0,\infty) \cap (-\infty, 1] = [0,1/2]$.

Square both sides to get

$9x^2 - 9x + 1 > 0$ AND $\sqrt{x}\sqrt{1-x} \ge 0$ and $x \in [0,1/2]$

So $9x^2 - 9x + 1 > 0$ and $x \in [0,\infty) \cap (-\infty,1] \cap [0,1/2] = [0,1/2]$.

So $x \in (\infty, \frac{3-\sqrt{5}}{6})\cup (\frac{3+\sqrt{5}}{6}, \infty)\cap [0,1/2]= [0,\frac{3-\sqrt{5}}{6})$