Sum of perfect powers

Upon observation, the RHS of this inequality will be positive iff $x \in (0,1)$ . If $x \ge 1$ , we will have a contradiction whereby the positive LHS will be less than or equal to a non-positive quantity. Also, $0 < y < x < 1$ is another necessary condition for the inequality to be satisfied.

Let us now examine the behavior of $f(x)=x-x^3$ over $0 < x < 1.$ Taking $f'(x) = 1 - 3x^2 = 0 \Rightarrow x = \frac{1}{\sqrt{3}}$ and $f''(1/\sqrt{3}) = -\large \frac{6}{\sqrt{3}}$ , the maximum value of the inequality's RHS is $f(1/\sqrt{3}) = \large \frac{2}{3\sqrt{3}}$ . Taking the LHS into account, this yields: $0 < y+y^3 \le \frac{2}{3\sqrt{3}} \Rightarrow 0 < y \le 0.34414.$

Thus, the inequality requires $x \in [\frac{1}{\sqrt{3}} + k,1), y \in (0,0.34414-k]$ (where $k \in [0,0.34414)$ ), and the least integral value of $n$ that satisfies $x^2 + y^2 < n$ is $\boxed{n=1}.$