Question 1120745
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&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;<U>1. &nbsp;&nbsp;Prove  &nbsp;A >= G  &nbsp;holds if and only if &nbsp;&nbsp;a = b</U>.



<pre>
You have this chain of equivalent inequalities


{{{(a+b)/2}}} >= {{{sqrt(ab)}}}  <------>  a + b >= {{{2*sqrt(ab)}}}  <------>  {{{(sqrt(a))^2}}} + {{{(sqrt(b))^2}}}  >= {{{2*sqrt(a)*sqrt(b)}}}  <------>  {{{(sqrt(a))^2 - 2*sqrt(a)*sqrt(b) + (sqrt(b)^2)}}} >= 0  <------>  {{{(sqrt(a) - sqrt(b))^2}}} >= 0


It is true for all real non-negative  numbers  "a" and "b",   and the equality is hold if and only if   {{{sqrt(a)}}} - {{{sqrt(b)}}} = 0, 


which in turn is equivalent to  a = b.


You can read this chain of inequalities / (of arguments) from the right to the left - then you will get the proof you need.
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&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;<U>2. &nbsp;&nbsp;Prove  &nbsp;G >= H  &nbsp;holds if and only if &nbsp;&nbsp;a = b</U>.



<pre>
{{{sqrt(ab)}}} >= {{{(2ab)/(a+b)}}}  <------>  {{{a}}} + {{{b)}}} >= {{{(2ab)/sqrt(ab)}}}  <------>  {{{a}}} + {{{b)}}} >= {{{2*sqrt(ab)}}},


and the last inequality is exactly  <U>THE SAME</U>  as the proven in  #1.
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Solved.