Prove that $a + b + c + a^2 + b^2 + c^2 \leq4$ given that $a,b,c\geq-1$ and $a^3 + b^3 + c^3=1$ Let $a,b,c\geq-1$ be real numbers with $a^3 + b^3 + c^3=1$.
Prove that $$a + b + c + a^2 + b^2 + c^2 \leq4.$$ 
 A: Hint:
This inequality is "separable" and it is enough to note that for $x> -1$,
$$f(x) = x^3-x^2-x+1 = (1+x)(1-x)^2 \geqslant 0$$ 
A: Equality holds for $(a,b,c) = (-1,1,1)$ and permutations thereof. (But that's only the second part of your question....)
Now the interesting part:
Write $f(a,b,c) = a + b + c + a^2 + b^2 + c^2 $ 
By symmetry, it is clear that $f$, subject to the condition $a^3 + b^3 + c^3=1$, will have an extremum at $a=b=c$. That means $a=b=c=\sqrt[3]{1/3}$ and hence the extremal $f = 3 (\sqrt[3]{1/3} + (\sqrt[3]{1/3})^2) \simeq 3.52 < 4$.
There will be other extremal points which can be seen with a Lagrangian argument. We need to look for zero partial derivatives $\frac{\partial F}{\partial a}$, $\frac{\partial F}{\partial b}$, $\frac{\partial F}{\partial c}$  of $F = f + \lambda (a^3 +b^3 +c^3 -1)$. The partial derivatives give quadratic equations in the variables, i.e.
$ 1 + 2 a + 3 \lambda a^2 = 0 $ with solutions $a_\pm$ (liekwise for b and c). Hence the candidate solution $(a_+, b_+, c_+)$ has been inspected above but the other candidates have to be inspected as well. Let's get an overview by plotting $f$ with direct replacement of $c = (1 - a^3 - b^3)^{1/3}$.  The maximum values that $a$ and $b$ can attain are $\sqrt[3]3 \simeq 1.44$. Here is a graph:

So indeed we have to look for the (-1)-boundary only. So let a=-1, then 
$f= b+c +b^2 + c^2$ and inserting $-1 + b^3 + c^3 = 1$ gives $f = b+(2-b^3)^{1/3} +b^2 +(2-b^3)^{2/3}$ and this attains, in the range $[-1 , \sqrt[3]3]$, its maximum value (4) indeed at $b=1$. So we're done. $\qquad \Box$ 
