Question 983964

Let &nbsp;<B>a</B>&nbsp; and &nbsp;<B>b</B>&nbsp; be two real positive unequal numbers.


Their arithmetic mean is &nbsp;{{{(a + b)/2}}}. &nbsp;Their geometric mean is &nbsp;{{{sqrt(a*b)}}}. &nbsp;We need to prove that 


{{{sqrt(a*b)}}} < {{{(a + b)/2}}}.


Let us start with inequality


{{{(sqrt(a) - sqrt(b))^2}}} > 0. &nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;&nbsp;(1)


This inequality is true because the difference &nbsp;{{{sqrt(a) - sqrt(b)}}}&nbsp; is not zero and the square of a non-zero real number is positive. &nbsp;Now, &nbsp;expand &nbsp;(1)&nbsp; as 


{{{(sqrt(a) - sqrt(b))^2}}} = {{{(sqrt(a))^2}}} - {{{2*sqrt(a)*sqrt(b)}}} + {{{(sqrt(b))^2}}} = {{{a}}} - {{{2*sqrt(a)*sqrt(b)}}} + {{{b}}}.


Thus the original inequality &nbsp;(1)&nbsp; is equivalent to  


{{{a}}} - {{{2*sqrt(a)*sqrt(b)}}} + {{{b}}} > {{{0}}}, &nbsp;&nbsp;&nbsp;&nbsp;or


{{{a}}}  + {{{b}}} > {{{2*sqrt(a)*sqrt(b)}}}, &nbsp;&nbsp;&nbsp;&nbsp;or


{{{(a + b)/2}}} > {{{sqrt(a*b)}}}. 


It is exactly what has to be proved.