$(a + x)^{1/2} + (a - x)^{1/2} \gt a$ for any real $a\gt 0$.
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Obviously $-a\lt x\lt a$ (Comment of Shawn). Squaring both sides... $(a+x)+(a-x)+2\sqrt{a^2 - x^2} = 2a + 2\sqrt{a^2-x^2} > a^2$ or $2\sqrt{a^2-x^2} > a(a-2)$ (When $a\geq2$) $4(a^2-x^2) > a^2 (a^2 - 4a + 4)$ or $a^4 - 4a^3 + 4x^2 < 0$ or $x^2 < a^3 - a^4/4$ or $-a \sqrt{a - a^2/4} < x < a \sqrt{a - a^2/4}$. (When $a\leq2$) (Comment of Arturo Magidin) Since $a(a-2)\lt0$, $-a\lt x \lt a$ is enough. |
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Squaring both sides, we obtain the equivalent inequality $$2\sqrt{a^2-x^2} >a^2-2a$$ If $a^2-2a<0$, that is, if $0<a<2$, the inequality automatically holds for all $x$ at which the left-hand side is defined, that is, for all $x$ such that $|x| \le a$. So now consider the case $a\ge 2$. Then the inequality $2\sqrt{a^2-x^2}>a^2-2a$ is equivalent to $4(a^2-x^2)>(a^2-2a)^2$. This simplifies to $$4x^2 <4a^3-a^4$$ or equivalently $|x|<(a/2)\sqrt{4a-a^2}$. Note that in particular there are no solutions if $a \ge 4$. |
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