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