Thread: prove <= 8/27

1. prove <= 8/27

for a,b,c real positive numbers, show that

((a+b)(b+c)(c+a))/ ((a+b+c)^3) <= 8/27

i'm thinking of cuberoot both sides..what do u guys think?
or is there any theorem?

2. hmm i'm not sure it's true... this is the route i went down:

let d = max(a,b,c) and e = min(a,b,c)

then

{your LHS} <= (d+d)(d+d)(d+d)/(e+e+e)^3 = ((2d)^3)/((3e)^3)

=8d^3/27e^3 = (8/27) * (d/e)^3

But e <= d so the upper bound can be more than 8/27...

You sure its true?

Si

3. Originally Posted by nh149
for a,b,c real positive numbers, show that

((a+b)(b+c)(c+a))/ ((a+b+c)^3) <= 8/27
By AM–GM,

$2(a+b+c)$

$=\ (a+b)+(b+c)+(c+a)$

$\ge\ 3\sqrt[3]{(a+b)(b+c)(c+a)}$

The result follows by cubing both sides.