We have:
a3+b3=(a+b)(a2+ab+b2)=2a3+b3=(a+b)(a2+ab+b2)=2 (*)
Other way:
a2+ab+b2=a2+2.12b+(12b)2+34b2a2+ab+b2=a2+2.12b+(12b)2+34b2
=(a+12b)2+34b2=(a+12b)2+34b2
Because (a+12b)2≥0(a+12b)2≥0 with ∀a,b∀a,b
34b2≥034b2≥0 with ∀b∀b
So (a+12b)2+34b2≥0(a+12b)2+34b2≥0 with ∀a,b∀a,b
or a2+ab+b2≥0a2+ab+b2≥0 with ∀a,b∀a,b (**)
From (*) and (**) we have a+b≤2a+b≤2
Your ex is so hard to do :)
