Just adding more steps for others to understand
\Assume a = x/z, b = y/z
Is a^4 + b^4 > 1
1. a^2 + b^2 >1
(a^2+b^2)^2 > 1, squaring a positive doesn't change the inequality.
a^4 + b^4 + 2a^2b^2 > 1 ---(1)
(a^2-b^2) >= 0 (any square is +ve; this is the same used in deriving AM >= GM inqualitty, gm = geometric mean and am = arithmetic mean)
a^4 + b^4 -2a^2b^2 >= 0 -- (2)
Add (1) and (2): a^4 + b^4 >= 1/2 (insufficient)
2. a + b > 1 when z is +ve
a^2+b^2 + 2ab > 1 (square the above) --(3)
(a-b)^ 2>= 0
a^2 + b^2 -2ab >= 0 --- (4)
Add (3) and (4): a^2+b^2 > 1/2
Now square it again and use (a^2 - b^2)^2 and sum them up, you will get a^4+b^4 >= 1/8 (insufficient)
------------------------------------
There is a faster way to do all these things
a^2+b^2 >= 2ab (arithmetic mean >= geometric mean)
Now we need to get the minimum value of 2ab when a + b > 1: how to go about?
just assume a = b and solve a+b =1, ya end up with a = b = 1/2
Substitute these in the above inequality.
2ab = 2(1/2)(1/2) = 1/2
a^2 + b^2 > =1/2
a^4+b^4 >= 2a^2b^2 (arith mean >= geometric mean)
What is a^2 when a^2 = b^2 and a^2 + b^2 = 1/2
a^2 = b^2 = 1/4
therfore a^4 + b^4 >= 2(1/4(1/4)
a^4+b^4 >= 1/8
