Question 980601
<pre>
The other tutor is right, so for the statement to be true, we must
change the conclusion from {{{x < y}}} to {{{x <= y}}}

But that's not all we have to change.

The statement is also not true if p=-1, q=-9

For in that case  {{{ x= sqrt((-1)(-9))=sqrt(9)=3 }}}
and {{{ y = (1/2) ((-1)^""+(-9))=(1/2)(-10)=-5}}} and x>y

The statement is also not true if p=0, q=-2

For in that case  {{{ x= sqrt((0)(-2))=sqrt(0)=0 }}}
and {{{ y = (1/2) ((0)^""+(-2))=(1/2)(-2)=-1}}} and x>y

Also by symmetry it is not true if p=-2, q=0

---------------------------------------------

So we must change your statement to something that is always true:

If {{{ x= sqrt(pq) }}}, {{{ y = (1/2) (p+q)}}}, {{{p >= 0}}}, {{{q >= 0}}}, prove {{{x<=y}}}

{{{sqrt(pq)=sqrt(p)sqrt(q)}}}

So what we are to prove becomes

If {{{ x= sqrt(p)sqrt(q) }}} and {{{ y = (1/2) (p+q)}}}. Prove {{{x<=y}}}

To prove {{{x<=y}}} is the same as proving {{{x-y<=0}}}

Let {{{r=sqrt(p)}}} and {{{s=sqrt(q)}}}

Then {{{p=r^2}}} and {{{q=s^2}}}

Then what we are to prove becomes:

If {{{ x= rs }}} and {{{ y = (1/2) (r^2+s^2)}}}. Prove {{{x-y<=0}}}

So we consider 

{{{x-y=rs-(1/2)(r^2+s^2) = rs -(r^2+s^2)/2 = 2rs/2 -(r^2+s^2)/2 = 
(2rs-(r^2+s^2))/2 = ""}}}
{{{(2rs-r^2-s^2)/2 = (-r^2+2rs-s^2)/2 = (-(r^2-2rs+s^2))/2 =
(-(r-s)^2)/2 = -(r-s)^2/2 <= 0 }}}

With the inclusion in the premise than p and q are non-negative and
in the conclusion that {{{p<=q}}} we have proved the statement. 

Edwin</pre>