Mathematics Olympiad ~ Vinod Singh, Kolkata
$Problem$ #3 $$ $$
Find all real numbers $x$ for which \(\sqrt{3-x}-\sqrt{x+1} > \frac{1}{2}\) $$ $$
Let $f(x)$ \(=\sqrt{3-x}-\sqrt{x+1}\). First note that $f(x)$ is defined for \( -1 \leq x \leq 3 \) $$ $$
\( f'(x) = \frac{-1}{2\sqrt{3-x}} - \frac{1}{2\sqrt{x+1}} = - \big(\frac{1}{2\sqrt{3-x}} + \frac{1}{2\sqrt{x+1}}\big) < 0 \Rightarrow f(x)\) is strictly decreasing $$ $$
Now \( f(-1) = 2 > \frac{1}{2}\) and \( f(3) = -2 < \frac{1}{2}\) Since $f(x)$ is continuous, $\exists$ at least one x $\in$ ${(-1,3)}$ suct that $f(x) = \frac{1}{2}$ $$ $$
\( f(x) = \frac{1}{2} \Rightarrow \sqrt{3-x}-\sqrt{x+1} = \frac{1}{2} \Rightarrow 64x^2-128x+33 = 0 \Rightarrow x = 1 \pm \frac{\sqrt{31}}{8} \)$$ $$
but \(x = 1 + \frac{\sqrt{31}}{8}\) does not satisfy \(\sqrt{3-x}-\sqrt{x+1} = \frac{1}{2}\) Check yourself! So the only solution is \(x = 1 - \frac{\sqrt{31}}{8}\) $$ $$
Since $f(x)$ is strictly decreasing, the given inequality is true for \( x \in {[-1,1 - \frac{\sqrt{31}}{8}\big)}\) $$ $$
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