Question 1169194
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Prove that if a, b, and c are the sides of a triangle, then so are sqrt(a), sqrt(b), and sqrt(c). 
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&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;<U>Proof</U>



<pre>
Let assume that  

    {{{sqrt(a)}}} + {{{sqrt(b)}}} <= {{{sqrt(c)}}}.    (1)


where "a", "b" and "c" are the sides of a triangle.


I want to lead it to CONTRADICTION.


Indeed, square both sides of (1). You will get


    a + {{{2*sqrt(ab)}}} + b <= c,   or


    a + b <= c - {{{2*sqrt(ab)}}}.


Then even more so

    a + b < c.


But it contradicts to the <U>triangle inequality</U>  a + b > c.


Thus we proved that  a + b > c   <U>IMPLIES</U>   {{{sqrt(a)}}} + {{{sqrt(b)}}} > {{{sqrt(c)}}}.


It works for any combinations of sides of a triangle.


Thus we proved that the values  {{{sqrt(a)}}},  {{{sqrt(b)}}},  {{{sqrt(c)}}} satisfy all triangle inequalities,

if  "a",  "b"  and  "c"  are the sides of a triangle.


It implies that if  "a",  "b"  and  "c"  are the sides of a triangle, then  {{{sqrt(a)}}},  {{{sqrt(b)}}},  and  {{{sqrt(c)}}}  form a triangle, too.
</pre>

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