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? 

You can't multiply or divide and keep the inequalities the the same unless you know the term you are multiplying by is positive.
In 1) $\sqrt{x^2 + a^2} > b-(x+a)$ does not mean
$(\sqrt{x^2 + a^2})^2 > [b-(x+a)]^2$ unless  $\sqrt{x^2 + a^2} > |b-(x+a)|$ and you don't know that.
And $x(2b-2a)>b^2-2ba$ does not mean 
$x > \frac {b^2 - 2ba}{2b-2a}$ unless you know that $2b - 2a > 0$.  And you do not know that.
And $(x+a)-b>-\sqrt{x^2+a^2}$ most certainly does NOT mean 
$[(x+a)-b]^2 > (-\sqrt{x^2+a^2})^2$.  That $-\sqrt{x^2 + a^2}$ is negative should have been a big tip off that if $(x+a) - b$ is positive we'd know absolutely nothing how the values compare and if $(x+a)-b < 0$ then this statement is completely false so $0 > (x+a)-b > -\sqrt{x^2 +a^2}\implies \sqrt{x^2 + a^2} > |(x+a)-b|\implies x^2 + a^2 > [(x+a)-b]^2$.
