2011/105) Prove a^4 + b^4 + c^4 >= abc(a+b+c)

this can be done in 2 steps

we know by AM GM inequality

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

adding the 3 above and deviding by 2 we get

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

now we repeat the process of a^2b^2 , b^2 c^2 and c^2 a^2 to get as below

a^2 b^2 + b^2 c^2 > = 2 b^2ac
b^2c^2 + c^2 a^2 >= 2 c^2ab
c^2a^2 + a^2 b^2 >= 2 a^2bc

adding the above and deviding by 2 we get

a^2b^2 + b^2 c^2 + c^2 a^2 >= (b^2ac+c^2ab+a^2bc) or abc(b+c+a) ,.2

from (1) and (2) it follows

a^4 + b^4 + c^4 >= abc(a+b+c)