 # Understand the problem

Let $a,b,c\ge-1$ be real numbers with $a^3+b^3+c^3=1$.
Prove that $a+b+c+a^2+b^2+c^2\le4$, and determine the cases of equality.
##### Source of the problem
Austria MO 2016. Final Round, Problem 4
Inequality
##### Suggested Book
Challenges and Thrills of Pre-College Mathematics

Do you really need a hint? Try it first!

The idea is that you have to capture the symmetry in the equations and correspondingly find it. Observe that the inequality $a+b+c+a^2+b^2+c^2\le4$ with the constraint $a^3+b^3+c^3=1$ can be written as  $(a^3 – a ^2 – a +1) + (b^3 – b ^2 – b +1) + (c^3 – c ^2 – c +1) \ge 0$ using the constraint.
Now, $(a^3 – a ^2 – a +1) + (b^3 – b ^2 – b +1) + (c^3 – c ^2 – c +1) \ge 0$ demands you to look into the polynomial $P(x)=(x+1)(x-1)^2=x^3-x^2-x+1$. Thus, the problem reduces to show that if $a,b,c\ge-1$ are real numbers, then $P(a)+P(b)+P(c)\ge0$.
What if we can show that individually if $x\ge-1$ we always have $P(x)\ge0$? Then our problem will be solved right? We have $P(x)=(x+1)(x-1)^2=x^3-x^2-x+1$, observe that it automatically implies that if $x+1 \geq 0$ then we will have $P(x) \geq 0$.
Equality Cases: For equality we must have $P(a)=P(b)=P(c)=0$, and hence $a,b,c\in\{-1,+1\}$.
Hence equality holds if and only if one of the three variables is $-1$ and the other two are $+1$. QED

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