Grade 12Algebra

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

Question image for Q3................................................

Aviral alok

9 Years agoGrade 12

1 Answer

mycroft holmes

Approved Tutor Answer9 Years ago

$\because a\ge -1, (a+1)(a^2-2a+1)\ge 0$ 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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Q3....................................................(just to reach min. char. limit

Aviral alok

1 Answer

mycroft holmes

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