<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] HINT
I am going to deal with the case in which $b - x - a \geq 0$. Otherwise, any value of $x$ that satisfies $x \geq b - a$ is a solution.
\begin{align*}
& x + a + \sqrt{x^{2} + a^{2}} > b \Longleftrightarrow \sqrt{x^{2} + a^{2}} > b - x - a \Longleftrightarrow
\begin{cases}
x^{2} + a^{2} > (b - x - a)^{2}\\\\
b - x - a \geq 0
\end{cases}\\\\
& \begin{cases}
x^{2} + a^{2} > b^{2} + x^{2} + a^{2} - 2bx - 2ab + 2ax\\\\
b - x - a \geq 0
\end{cases} \Longleftrightarrow
\begin{cases}
b^{2} - 2bx - 2ab + 2ax < 0\\\\
b - x - a \geq 0
\end{cases}\\\\
& \begin{cases}
b^{2} + 2x(a - b) - 2ab < 0\\\\
b - x - a \geq 0
\end{cases} \Longleftrightarrow
\begin{cases}
2x(a-b) < 2ab - b^{2}\\\\
b - x - a \geq0
\end{cases}
\end{align*}
Then you consider the three possible situations: $a > b$, $a = b$ or $a < b$.
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