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? 
 A: You should use this fundamental fact about irrational inequalities:

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

A: 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 $0 > (x+a)-b > -\sqrt{x^2 +a^2}\implies \sqrt{x^2 + a^2} > |(x+a)-b|\implies x^2 + a^2 > [(x+a)-b]^2$.
A: 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 $a < b$.
A: 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.
