Simplifying absolute value expression with square roots I'm attempting to simplify the expression $|(|\sqrt2+ \sqrt3|-|\sqrt5-\sqrt7|)|$
Previously I've shown that if $0<a<b$ then $\sqrt{a}<\sqrt{b}$
Thus we have $\sqrt5-\sqrt7<0$ and therefore $|\sqrt5-\sqrt7|=\sqrt7-\sqrt5$
This means $|\sqrt2+\sqrt3|-|\sqrt5-\sqrt7|=\sqrt2+\sqrt3+\sqrt5-\sqrt7$
Now I've clearly simplified the expression, but I want to find out if it is possible to simplify further. As stated about $\sqrt5-\sqrt7<0$, and $\sqrt2+\sqrt3>0$. Intuitively I somehow see that this whole expression has to be greater than zero, but how do I prove it?
Thanks  
 A: You seem to be asking two questions: (1) can this be simplified further, and (2) is the expression positive or negative?


*

*No.  There isn't really anything that you can do to make this expression simpler.  Each radical expression has a different radicand, so there are no like terms to combine, and the radicals themselves are pretty simple.

*One possible approach, among many, is as follows:
\begin{align}
\sqrt{2} + \sqrt{3} + \sqrt{5} - \sqrt{7} \ge 0
     &\iff \sqrt{2} + \sqrt{3} + \sqrt{5} \ge \sqrt{7} \\
     &\iff \left( \sqrt{2} + \sqrt{3} + \sqrt{5} \right)^2 \ge 7. && (\text{$x \mapsto x^2$ is increasing on $[0,\infty)$})
\end{align}
Multiplying out the expression on the left, we get
\begin{align}
\left(\sqrt{2} + \sqrt{3} + \sqrt{5}\right)^2
     &= 2 + \sqrt{6} + \sqrt{10} + \sqrt{6} + 3 + \sqrt{15} + \sqrt{10} + \sqrt{15} + 5 \\
     &= 10 + 2\sqrt{6} + 2\sqrt{10} + 2\sqrt{15}.
\end{align}
Therefore we have
$$ \left( \sqrt{2} + \sqrt{3} + \sqrt{5} \right)^2 = 10 + \underbrace{2\sqrt{6} + 2\sqrt{10} + 2\sqrt{15}}_{\ge 0} \ge 10 > 7, $$
from which it follows that
$$ \sqrt{2} + \sqrt{3} + \sqrt{5} - \sqrt{7} > 0. $$
A: When $a > 0, b > 0$, it is easy to see that $\sqrt{a} + \sqrt{b} > \sqrt{a + b}$. Thus we have $\sqrt{2} + \sqrt{5} - \sqrt{7} > 0$. So you can prove something stronger - $$\sqrt{2} + \sqrt{3} + \sqrt{5} - \sqrt{7} > \sqrt{3} + 0 = \sqrt{3}.$$
A: we have $$\sqrt{2}+\sqrt{3}>0$$ and since $$|a-b|=|b-a|$$ our term is $$\sqrt{2}+\sqrt{3}+\sqrt{5}-\sqrt{7}$$ and we have
$$\sqrt{2}+\sqrt{3}+\sqrt{5}>\sqrt{7}$$
all Terms are positive and we get by squaring
$$2(\sqrt{6}+\sqrt{15}+\sqrt{10})>-3$$ which is true.
