No, you are wrong. You said, you can find numbers a,b,c which fulfil conditions for any x, but you must find x, which for any possible a,b,c (a+b+c=3) is filfuled. for example you said x=-50. it is wrong but I can find numbers a,b,c which isnt filfuled (a^3+b^3+c^3>=3).. for example a=-50 b=26 c=27
it says for arbitrary values THAT FULFILL THE GIVEN CONDITIONS. you said yourself that arbitrary means "any." if it means "any" why does it matter if i pick them or not? i found arbitrary values that fulfill all conditions, but for everyone i find, i can find smaller and smaller x's, thus there is no minimum value for the x.
in math, to disprove something, all you need is ONE counter example, and you are allowed to choose it if you can find one. here i found infinitely many counter-examples. it is easy to satisfy the equations and inequalities, but there is no minimum for the x value as the problem claimed. since i can find arbitrary a,b, and c values that fulfill all requirements, but require a smaller x value than any that i chose before. if you look at my post, you will realize that i could choose the a,b,c first and then find the x. a,b, and c are arbitrary, so i can choose any a,b,c i want, as long as they fulfill the required conditions. (your example does not fulfill the conditions by the way).
my example doesnt fulfil a^3+b^3+c^3>=3 because chosen x is wrong... you must first find x and then for every posibility a,b,c which fulfil a+b+c=3, this a^3+b^3+c^3>=3 is true... I repeat fo every combination of a,b,c , not one I found this ONE exapmle to disprove your chosen x (-50) for example x=0, can you find a,b,c, which doesnt filful equation and inequality?