Question 461625
<pre>

Since the inequality we are to prove for the letters
representing unequal positive numbers:
{{{drawing(175,60,0,10,-10,7, locate(1,3,
expr(1/4)*(a+b+c+d)>(abcd)^(1/4) ) )}}}

and one of the other parts is

{{{drawing(175,60,0,10,-10,7, locate(1,3,
expr(1/3)*(a+b+c)>(abc)^(1/3) ) )}}}

Then we might suppose that the first thing we need to
prove is that for any different positive numbers a and b 

{{{drawing(175,60,0,10,-10,7, locate(1,3,
expr(1/2)*(a+b)>(ab)^(1/2) ) )}}}

The technique I will use is to assume that it is false,
and then reach a contradiction.  

So for contradiction we will assume that for {{{0>a<>b>0}}},  
 
{{{drawing(175,60,0,10,-10,7, locate(1,3,
expr(1/2)*(a+b)<=(ab)^(1/2) ) )}}}

Multiply both sides by 2

{{{drawing(175,60,0,10,-10,7, locate(1,3,
a+b<=2(ab)^(1/2) ) )}}}

Square both sides:

{{{drawing(175,60,0,10,-10,7, locate(1,3,
(a+b)^2<=4ab ) )}}}

{{{drawing(175,60,0,10,-10,7, locate(1,3,
a^2+2ab+b^2<=4ab ) )}}}

Subtract 4ab from both sides:

{{{drawing(175,60,0,10,-10,7, locate(1,3,
a^2-2ab+b^2<=0 ) )}}}

Factor the left side:

{{{drawing(175,60,0,10,-10,7, locate(1,3,
(a-b)^2<=0 ) )}}}

This is false since the left side is positive,
since a and b are different.

So we have reached a contradiction and therefore
we have proved that 

{{{drawing(175,60,0,10,-10,7, locate(1,3,
expr(1/2)*(a+b)>(ab)^(1/2) ) )}}}

Note that the two sides are equal if a=b.

We have also proved by letting a=c and b=d.
that for {{{0>c<>d>0}}} 

{{{drawing(175,60,0,10,-10,7, locate(1,3,
expr(1/2)*(c+d)>(cd)^(1/2) ) )}}}

Down below we will need this for positive x,y: 

{{{drawing(175,60,0,10,-10,7, locate(1,3,
expr(1/2)*(x+y)>=(xy)^(1/2) ) )}}}

with equality holding if and only if x=y.

We do this to avoid getting letters confused.


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

Next we will prove that for a,b,c,d > 0, all different,

{{{drawing(175,60,0,10,-10,7, locate(1,3,
expr(1/4)*(a+b+c+d)>(abcd)^(1/4) ) )}}}

We start with

{{{drawing(175,60,0,10,-10,7, locate(1,3,
expr(1/2)*(a+b)>(ab)^(1/2) ) )}}}
{{{drawing(175,60,0,10,-10,7, locate(1,3,
expr(1/2)*(c+d)>(cd)^(1/2) ) )}}}

Multiply both sides by 2, and we have:

{{{drawing(175,60,0,10,-10,7, locate(1,3,
a+b>2(ab)^(1/2) ) )}}}
{{{drawing(175,60,0,10,-10,7, locate(1,3,
c+d>2(cd)^(1/2) ) )}}}

Add the two inequalities:

{{{drawing(250,60,0,10,-10,7, locate(1,3,
a+b+c+d>2(ab)^(1/2)+2(cd)^(1/2) ) )}}}

Factor out 2 on the right

{{{drawing(250,60,0,10,-10,7, locate(1,3,
a+b+c+d>2((ab)^(1/2)+(cd)^(1/2)) ) )}}}

Make the left side into the left side of what
we have to prove by multiplying thru hy 1/4

{{{drawing(300,60,0,10,-10,7, locate(1,3,
expr(1/4)(a+b+c+d)>expr(1/2)((ab)^(1/2)+(cd)^(1/2)) ) )}}}

Now we will show that the right side is greater than or equal
to {{{drawing(100,60,0,10,-10,7, locate(1,3, (abcd)^(1/4)) )}}}

We recall from above that for positive x, y,

{{{drawing(175,60,0,10,-10,7, locate(1,3,
expr(1/2)*(x+y)>=(xy)^(1/2) ) )}}}

with equality holding if and only if x=y

We let x = {{{drawing(100,60,0,10,-10,7, locate(1,3, (ab)^(1/2)) )}}}
and y = {{{drawing(100,60,0,10,-10,7, locate(1,3, (cd)^(1/2)) )}}}

{{{drawing(400,60,0,10,-10,7, locate(1,9,
expr(1/2)*((ab)^(1/2)+(cd)^(1/2))>=((ab)^(1/2)(cd)^(1/2))^(1/2) ) )}}}

{{{drawing(400,60,0,10,-10,7, locate(1,3,
expr(1/2)*((ab)^(1/2)+(cd)^(1/2))>=(abcd)^(1/4) ) )}}}

[Notice that we had to include the possibility of equality 
since even though a,b,c,d are all distinct, that does not
guarantee that {{{drawing(100,60,0,10,-10,7, locate(1,3, (ab)^(1/2)) )}}}
 and {{{drawing(100,60,0,10,-10,7, locate(1,3, (cd)^(1/2)) )}}}
 are distinct, e.g. if a=1,b=6,c=2,d=3, they are not distinct.]

So we have proved:

{{{drawing(400,60,0,10,-10,7, locate(1,3,
expr(1/4)(a+b+c+d)>expr(1/2)((ab)^(1/2)+(cd)^(1/2))>=(abcd)^(1/4) ) )}}}

which is what we had to proved.

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For different positive a,b,c, we need to prove

{{{drawing(175,60,0,10,-10,7, locate(1,3,
expr(1/3)*(a+b+c)>(abc)^(1/3) ) )}}}

Sorry, I haven't figured out how to do this one yet.

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The generalization.  If all a<sub>i</sub> are 
positive numbers, then

{{{expr(1/n)sum(a[i],i=1,n)>=(matrix(3,2,n,"",PI,a[i],i=1,""))^(1/n)}}}

and equality holds only when all the a<sub>i</sub> are equal.

Edwin</pre>