For positive real no. a,b,c,d , then { (a+b+c)(a+c+d)}^1/3 >= (ab)^1/3 +(cd)^1/3

Answer 1

Here is a quick outline of the proof. A lot of the algebra is left for you...

Working backwards by first cubing both sides

The left hand side (LHS) becomes:

(a+c)^2 + ad + bc + bd + (ab + cd)

THe right hand side (RHS) becomes:

3 (abcd)^(1/3) [ (cd)^(1/3) + (ab)^1/3)] + (ab + cd)

(ab+cd) cancels out

Let v = MIN(a,b,c,d) and T = max(a,b,c,d)

Then to maximize/minimize, replaces each var with T or v respectively

minimizing the left side, LHS > 7v^2

maximizing the right side, RHS < 6T ^(4/3)

v < T

-v > -t

v > -v < -t

v^2 > -t^(4/3)

7v^2 > -t^(4/3)

So then LHS > 7v^2

which means

LHS - RHS > 7v^2 - RHS > 7v^2 - 6t^(4/3) > 0

Therefore, LHS > RHS, because LHS at its lowest is bigger than right hand side at its biggest.

cube root of both sides completes the proof