Rearranging and squaring the expressiion we get $3-x > 1/4 + x + 1 + \sqrt(x+1)$ that is $\sqrt(x+1) <x+7/4$ sqaring this and rearranging we get $(x-1)^2>31/64$ which implies that $x >1+\sqrt(31)/8$ or $x<1-\sqrt(31)/8$ We ignore the first one as the interval for $x$ is $[-1,7/8)$ Hence the inequality holds for all reals between $-1$ and $1-\sqrt(31)/8$ excluding the upper limit and we are done. Ashwath

quite easy for an IMO problem!

Rewrite the inequality as $\sqrt{3-x}>\sqrt{x+1}+\frac{1}{2}$. Squaring both sides of the inequality gives $3-x>x+1+\sqrt{x+1}+\frac{1}{4}$. Multiplying both sides by $4$ makes our inequality equivalent to $12-4x>4x+4+4\sqrt{x+1}+1$. Now we subtract $4x+5$ from both sides to get $7-8x>4\sqrt{x+1}$. Square both sides to get rid of the radical and we have that $49-112x+64x^2>16x+16$. Bringing all terms to one side, $64x^2-128x+33>0$. Since $a>0$, we can conclude that this quadratic opens upwards. Now in our original inequality, we note that we can only have $x\in [-1,3]$ in order for the square root operations to both be satisfied. Applying the quadratic formula on $64x^2-128x+33=0$, we see that the roots are $\frac{8\pm\sqrt{31}}{8}$. It is easy to see that both these roots lie inside the range $[-1,3]$. So it follows that our answer is $\boxed{\left[-1,\frac{8-\sqrt{31}}{8}\right)}$.

ashwath.rabindranath wrote: quite easy for an IMO problem! IMO was lot more easier at that time.

redacted

PhysicsMonster_01 wrote:

forgive me for bumping a two year old thread, but I didn't get it. I got $64x^2-128x+33=0$, which yields $\frac{8\pm\sqrt{31}}{8}$. Now since both the roots are inside the range $[-1,3]$, won't the answer be $\left[-1,\frac{8-\sqrt{31}}{8}\right) \cup \left(\frac{8 + \sqrt{31}}{8}, 3 \right]$? See #4: Because $7-8x>4\sqrt{x+1}$, so we have $64x^2-128x+33>0$ and $7-8x>0$

Sorry for the bump, but I just encountered this problem and I would like to post a solution (if this is not allowed then just delete this post).

The function $f(x)=\sqrt{3-x}-\sqrt{x+1}$ is valid for $-1 \leq x \leq 3$. By little bruteforce, $f(-1)=2 > .5$ and $f(1-\frac{\sqrt{31}}{8}) =.5.$ So the ineq holds when $-1 \leq x < 1-\frac{\sqrt{31}}{8}.$

First, we must have $-1\leq x\leq3$ in order for $\sqrt{3-x}$ and $\sqrt{x+1}$ to be real. Then, $\sqrt{3-x}>\sqrt{x+1}+\tfrac{1}{2}$, and since both sides are nonnegative, we can square both sides to obtain an equivalent formulation: $$3-x>\left(\sqrt{x+1}+\frac{1}{2}\right)^2=x+\frac{5}{4}+\sqrt{x+1}\iff\frac{7}{4}-2x>\sqrt{x+1}.$$In order for the LHS to be nonnegative, we further restrict $x$ to the interval $[-1,\tfrac{7}{8}]$; then, squaring, we have $$4x^2-7x+\frac{49}{16}>x+1\iff4x^2-8x+\frac{33}{16}>0\iff64x^2-128x+33>0,$$and since the roots of the LHS are $\tfrac{8\pm\sqrt{31}}{8}$ by the Quadratic Formula, this inequality holds iff $x\in(-\infty,\tfrac{8-\sqrt{31}}{8})\cup(\tfrac{8+\sqrt{31}}{8},\infty)$. Intersection this union of intervals with $[-1,\tfrac{7}{8}]$ yields $\boxed{x\in[-1,\tfrac{8-\sqrt{31}}{8})}$, as requested. $\blacksquare$