Find the minimum value of $a^2+b^2+c^2+2abc$ when $a+b+c=3$ and $a,b,c\geq0$. 
Given $a,b,c\geq0$ such that $a+b+c=3$, find the minimum value of $$P=a^2+b^2+c^2+2abc.$$

It seems like the minimum value of $P$ is $5$ when $a=b=c=1$, but I can find at least one example where $P<5$.
My attempt:

Without loss of generality, I can suppose that $a\geq b\geq c$ and so $a\geq 1$. Therefore I have:
  $$P\geq a^2+b^2+c^2+2bc=a^2+(b+c)^2\geq \frac{(a+b+c)^2}{2}=\frac{9}{2}.$$

The issue with this is that equality doesn't occur with this method.
What's your take on the problem?

It occurs for $c=0$ and $a=b=\frac{3}{2}$ and we obtain a value $\frac{9}{2}$.
Also, you proved that it's a minimal value.
I like the following way.
We need to prove that
$$(a+b+c)(a^2+b^2+c^2)+6abc\geq\frac{1}{2}(a+b+c)^3$$ or
$$\sum_{cyc}(a^3-a^2b-a^2c+2abc)\geq0,$$ which is true by Schur.
