Can there be two inequalities? $$x+a+\sqrt{x^2+a^2}>b$$
I can consider this inequality for $a,b,x>0$.
1) $\sqrt{x^2+a^2}>b-(x+a)$
$$x^2+a^2>b^2-2b(x+a)+(x+a)^2$$
$$0>b^2-2b(x+a)+2ax$$
$$x(2b-2a)>b^2-2ba$$
$$x>\frac{b^2-2ba}{(2b-2a)}$$
2)  $(x+a)-b>-\sqrt{x^2+a^2}$
$$x^2+a^2<b^2-2b(x+a)+(x+a)^2$$
$$0<b^2-2b(x+a)+2ax$$
$$x(2b-2a)<b^2-2ba$$
$$x<\frac{b^2-2ba}{(2b-2a)}$$
I get two inequalities. What is Im doing wrong? 

Supposing you want to solve for $x$ in $$\sqrt{x^2+a^2}>b-a-x,$$ note that since LHS is never negative, the inequality is automatically true whenever RHS is negative, that is, for all $b-a<x.$  Thus, suppose $b-a-x\ge 0,$ then you may square both sides to have $$x^2+a^2>(b-a-x)^2=b^2-2b(a+x)+(a+x)^2=b^2-2ab-2bx+a^2+2ax+x^2,$$ which simplifies to $$0>b^2-2ab-2bx+2ax=b^2-2ab-(2b-2a)x,$$ or $$(2b-2a)x>b^2-2ab=b(b-2a).$$ Now recall that we assumed that $b-a\ge x.$ Again we have three subsidiary cases, namely $a=b,a<b,a>b.$ You can check that in the first case we have no solutions. If we have $a<b,$ then we can divide both sides by $2b-2a$ without changing the order, to get $$x>\frac{b(b-2a)}{2(b-a)}.$$ In the final case this inequality is reversed.
