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Q3....................................................(just to reach min. char. limit

Aviral alok , 9 Years ago
Grade 12
anser 1 Answers
mycroft holmes
\because a\ge -1, (a+1)(a^2-2a+1)\ge 0 and hence we have
 
a^3 -a^2 -a +1 \ge 0 \Rightarrow a^3+1 \ge a^2+a
 
and so
 
\sum a^2+a \le \sum a^3+1 = 1+3 = 4
 
i.e. a+b+c+a^2+b^2+c^2 \le 4
 
Equality can occur when a,b,c \in \{1,-1 \} and since require that a^3+b^3+c^3 = 1 this means we can choose a=b=1 and c=-1 and permutations of this as the equality case.
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