IMO 1962 Problem 2
We seek all real numbers $x$ satisfying
Proposed by: -
Verified: yes
Verdicts: PASS + PASS
Solve time: 25m30s
Problem
Find all real numbers $x$ satisfying $$\sqrt{3-x} - \sqrt{x+1} > \frac{1}{2}.$$
Problem Understanding
We seek all real numbers $x$ satisfying
$$\sqrt{3-x}-\sqrt{x+1}>\frac12.$$
Because square roots are involved, the first step is to determine the domain. After that, the natural strategy is to study the related equation
$$\sqrt{3-x}-\sqrt{x+1}=\frac12,$$
since the inequality is expected to change sign at the equality point. Solving this equation requires squaring, so every candidate obtained after squaring must be checked in the original equation to eliminate extraneous roots.
Finally, once the equality point is identified, monotonicity of the function
$$f(x)=\sqrt{3-x}-\sqrt{x+1}$$
determines exactly on which side of the threshold the inequality holds.
Key Observations
The expressions under the square roots must be nonnegative:
$$3-x\ge0,\qquad x+1\ge0.$$
Hence
$$-1\le x\le3.$$
Define
$$f(x)=\sqrt{3-x}-\sqrt{x+1}.$$
On the interval $(-1,3)$,
$$f'(x) = -\frac1{2\sqrt{3-x}} -\frac1{2\sqrt{x+1}} <0.$$
Therefore $f$ is strictly decreasing on $[-1,3]$. Consequently, the inequality $f(x)>\tfrac12$ can hold only on an interval of the form $[-1,a)$, where $a$ is the unique solution of
$$f(x)=\frac12.$$
Thus the main task is to solve the equality equation correctly and verify which roots are genuine.
Solution
The domain is
$$-1\le x\le3.$$
Now solve
$$\sqrt{3-x}-\sqrt{x+1}=\frac12.$$
Rearrange:
$$\sqrt{3-x}=\sqrt{x+1}+\frac12.$$
Squaring both sides gives
$$3-x = \left(\sqrt{x+1}+\frac12\right)^2.$$
Expanding carefully,
$$\left(\sqrt{x+1}+\frac12\right)^2 = (x+1)+\sqrt{x+1}+\frac14.$$
Hence
$$3-x = x+1+\sqrt{x+1}+\frac14.$$
Move all nonradical terms to the left:
$$3-x-x-1-\frac14 = \sqrt{x+1}.$$
Thus
$$\frac74-2x=\sqrt{x+1}.$$
Since the square root is nonnegative, we must have
$$\frac74-2x\ge0.$$
Now square again:
$$x+1=\left(\frac74-2x\right)^2.$$
Expanding,
$$x+1 = \frac{49}{16}-7x+4x^2.$$
Therefore
$$0 = 4x^2-8x+\frac{33}{16}.$$
Multiplying by $16$,
$$64x^2-128x+33=0.$$
Using the quadratic formula,
$$x = \frac{128\pm\sqrt{(-128)^2-4\cdot64\cdot33}}{128}.$$
Compute the discriminant:
$$(-128)^2-4\cdot64\cdot33 = 16384-8448 = 7936 = 256\cdot31.$$
Hence
$$\sqrt{7936}=16\sqrt{31},$$
so
$$x = \frac{128\pm16\sqrt{31}}{128} = 1\pm\frac{\sqrt{31}}8.$$
We must now check which roots satisfy the original equation.
First consider
$$x=1+\frac{\sqrt{31}}8.$$
Then
$$\frac74-2x = \frac74-2\left(1+\frac{\sqrt{31}}8\right) = \frac74-2-\frac{\sqrt{31}}4 = -\frac14-\frac{\sqrt{31}}4 <0.$$
But
$$\sqrt{x+1}\ge0,$$
so the equation
$$\frac74-2x=\sqrt{x+1}$$
cannot hold. Therefore
$$x=1+\frac{\sqrt{31}}8$$
is extraneous.
Now consider
$$x=1-\frac{\sqrt{31}}8.$$
Then
$$\frac74-2x = \frac74-2\left(1-\frac{\sqrt{31}}8\right) = -\frac14+\frac{\sqrt{31}}4 = \frac{\sqrt{31}-1}{4}>0,$$
so no contradiction arises.
Substituting into the original equation confirms it works:
$$\sqrt{3-x}-\sqrt{x+1} = \frac12.$$
Thus the unique solution of the equality equation is
$$a=1-\frac{\sqrt{31}}8.$$
Since $f$ is strictly decreasing on $[-1,3]$, we have
$$f(x)>\frac12 \iff x<a.$$
Intersecting with the domain gives
$$\boxed{\left[-1,;1-\frac{\sqrt{31}}8\right)}.$$
Verification of Key Steps
The derivative computation is
$$f'(x) = \frac{d}{dx}\bigl(\sqrt{3-x}\bigr) - \frac{d}{dx}\bigl(\sqrt{x+1}\bigr).$$
Using the chain rule,
$$\frac{d}{dx}\bigl(\sqrt{3-x}\bigr) = \frac{-1}{2\sqrt{3-x}},$$
and
$$\frac{d}{dx}\bigl(\sqrt{x+1}\bigr) = \frac{1}{2\sqrt{x+1}}.$$
Therefore
$$f'(x) = -\frac1{2\sqrt{3-x}} -\frac1{2\sqrt{x+1}} <0$$
for all $x\in(-1,3)$. Hence $f$ is strictly decreasing.
At the endpoints,
$$f(-1)=\sqrt4-\sqrt0=2>\frac12,$$
while
$$f(3)=\sqrt0-\sqrt4=-2<\frac12.$$
Since $f$ is continuous and strictly decreasing, it crosses the value $\tfrac12$ exactly once, namely at
$$x=1-\frac{\sqrt{31}}8.$$
Therefore
$$f(x)>\frac12$$
holds exactly for
$$-1\le x<1-\frac{\sqrt{31}}8.$$
Alternative Approaches
A shorter approach avoids derivatives by rationalizing:
$$\sqrt{3-x}-\sqrt{x+1} = \frac{(3-x)-(x+1)}{\sqrt{3-x}+\sqrt{x+1}} = \frac{2-2x}{\sqrt{3-x}+\sqrt{x+1}}.$$
Since the denominator is positive on the domain, the expression decreases as $x$ increases. One may then solve the equality case exactly as above and conclude the same interval of solutions:
$$\boxed{\left[-1,;1-\frac{\sqrt{31}}8\right)}.$$