Question 590269
<pre>
{{{sqrt(2)}}} < {{{sqrt(5)}}}  because 2 < 5
{{{sqrt(3)}}} < {{{sqrt(7)}}}  because 3 < 7

Therefore

{{{sqrt(2)}}} + {{{sqrt(3)}}} < {{{sqrt(5)}}} + {{{sqrt(7)}}}  

because unequals in a certain order added to unequals in the same order
always gives unequals in that same order.

Edwin</pre>