<s>[INST] Find all Integral solutions to $x+y+z=3$, $x^3+y^3+z^3=3$. Suppose that $x^3+y^3+z^3=3$ and $x+y+z=3$. 
What are all integral solutions of this equation? 
I can only find $x=y=z=1$.
 [/INST] Using the Power Mean Inequality we have that:
$$ 1=\frac{1}{3}(x^3 + y^3 + z^3)  \leq \left[\frac{1}{3}(x+y+z)\right]^3 =1$$
Equality only holds when $x=y=z=1$.  

Let's replace $z \mapsto -z$ and try to solve the system of equations for $x,y,z \geq 0$:
$$ x^3 + y^3 = 3 + z^3 \text{ and } x + y = z + 3 $$
We can still use the power mean inequality (in only two of the varaibles) and use the linear relation
$$ \frac{1}{3}z^3 < \frac{1}{3}(3 + z^3 ) = \frac{1}{3}(x^3 + y^3 )  \leq \left[\frac{1}{3}(x+y)\right]^3 = \left[\frac{1}{3}(z+3)\right]^3
< \frac{1}{24}(z+3)^3 $$
Maybe if I take cube roots the left and right sides are easier to compare.  I also rounded $3^3 \approx 27$
$$ z <  \frac{1}{2}(z + 3)$$
This means $0 \leq z < 3$.  Can it be that $0^3 + 3 = 3, 1^3 + 3 = 4, 2^3 + 3 = 11$ are the sums of two cubes?

If two of the 3 numbers are negative,  the starting point should be:
$$ x^3 + y^3 = -3 + z^3 \text{ and } x + y = z - 3 $$
Then using the power mean inequality we can attempt to deduce a range on $z$
$$  \frac{1}{3}(-3 + z^3 ) < \frac{1}{24}(z-3)^3 $$
This one is harder to estimate.  After fiddling around, we see that $z \leq 0$:
$$   z <  \frac{1}{2}\sqrt[3]{(z-3)^3 + 24} 
< \frac{1}{2}z $$
So there are definitely no solutiuons of this kind.
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