2010/023) prove that a^4+b^4+c^4> or =abc(a+b+c)

we know

(a^2-b^2) ^2 >= 0

or a^4+b^4 >= 2a^b^2

similarly

b^4 + c^4 >= 2 b^2 c^2

c^4+ a^4 >= 2 c^2 a^2
adding all 3 above and deviding by 2

a^4+b^4+c^4 >= (a^2b^2+b^2c^2 + c^2 a^2) ... 1


now a^2+c^2 >= 2ac

multiply by b^2 on both sides

b^2(a^2+c^2) >= 2b^2ac -- 2

similarly
a^2(b^2+c^2) >= 2a^2bc ... 3

and c^2(a^2+b^2) >= 2c^2ab .. 4

adding (2) (3) and (4) we get

2(a^2b^2+b^2c^2+c^2a^2) >= 2abc(b+a+c)

or (a^2b^2+b^2c^2+c^2a^2) >= abc(a+b+c) ...5

from 1 and 5 we get a^4+b^4+c^4>=abc(a+b+c)

proved