<s>[INST] 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? 
 [/INST] You should use this fundamental fact about irrational inequalities:

On its domain, $\;\sqrt{A}>B\iff A>B^2\: \textbf{ or }\: B<0.$

</s>