For the square roots to exist, we require $-1 \le x \le 1$. The inequality is false when $x \le 0$, so we assume $0 < x \le 1$. By multiplying above and below by $\sqrt{x+1} - \sqrt{1-x}$, the inequality becomes
$$\sqrt{x+1} - \sqrt{1-x} > \sqrt{\frac{4-2x}{7}}.$$
The left side is positive, since $x > 0$, so squaring both sides gives
$$2 - 2\sqrt{1 - x^2} > \frac{4 - 2x}{7}.$$
which gives $\frac{x+5}{7} > \sqrt{1-x^2}$. Squaring again gives $x^2+10x+25 > 49-49x^2$, so $25x^2+5x-12 > 0$. Factoring gives $(5x-3)(5x+4) > 0$ and since $x > 0$ this gives $x > \frac{3}{5}$. So the inequality is true if and only if $\frac{3}{5} < x \le 1$.