1962 IMO Problems/Problem 2
Problem
Determine all real numbers 
which satisfy the inequality:
Solution
Obviously we need 
for the outer square root to be defined, 
for the first inner square root to be defined,
and 
for the second inner square root to be defined. Solving these we get that the left hand side is defined for ![$x\in \left[ -1,1 \right]$](http://latex.artofproblemsolving.com/b/c/d/bcd1b79ebaa2f316e8a4640efe60a54f02a4bdcc.png)
.
Now obviously the function 
is continuous on ![$\left[ -1,1 \right]$](http://latex.artofproblemsolving.com/9/6/a/96a708526a0b1155ed2e4160fbada368658987d3.png)
, with 
and 
.
Moreover, as 
is a decreasing and 
an increasing function, both 
and 
are decreasing functions, and hence 
is a decreasing function. Therefore there is exactly one solution to 
.
We can now find this solution:
Solving the quadratic equation for , we get
![\[x_{1,2}=\dfrac{ 2^{11} \pm \sqrt{ 2^{22} - 2^{12}\cdot 897} }{2^{11}} = 1 \pm \dfrac{\sqrt{127}}{32}\]](http://latex.artofproblemsolving.com/3/0/6/3062ec05c3326e033a0888e9447c53feb71ecd1b.png)
The reason why we got two roots is that while solving the original equation we squared both sides twice, and this could have created additional solutions. In this case, obviously the root that is larger than 
is the additional solution, and 
is the root we need.
Hence the solutions to the given inequality are precisely the reals in the interval 
.
![[asy] import graph; size(300,300,IgnoreAspect); real f(real x) {return sqrt( sqrt(3-x) - sqrt(x+1) );}; real g(real x) {return 1/2;}; draw(graph(f,-1,1),blue,"$f(x)=\sqrt{\sqrt{3-x}-\sqrt{x+1}}$"); draw(graph(g,-1,1),red,"$g(x)=1/2$"); xaxis("$x$",BottomTop,Ticks); yaxis("$y$",LeftRight,Ticks); attach(legend(),point(E),20E,UnFill); [/asy]](http://latex.artofproblemsolving.com/b/4/2/b4208ee4ea18baca0541febbbfa64873d607434c.png)
See Also
| 1962 IMO (Problems) • Resources | ||
| Preceded by Problem 1 |
1 • 2 • 3 • 4 • 5 • 6 | Followed by Problem 3 |
| All IMO Problems and Solutions | ||
