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?
Best Answer
Your attempt already contains very good observations, and is quite close to a complete solution. All that remains is an analysis of when both inequalities are equalities. You use the following two inequalities: \begin{eqnarray*} a^2+b^2+c^2+2abc&\geq&a^2+b^2+c^2+2bc\tag{1}\\ a^2+(b+c)^2&\geq&\frac{(a+b+c)^2}{2}\tag{2} \end{eqnarray*} The first inequality is an equality if and only if $abc=bc$, i.e. if and only if either $a=1$ or $bc=0$.
The second inequality is an equality if and only if $a=b+c$. Then from $a+b+c=3$ it follows that $a=\tfrac32$. Then for the first inequality to be an equality we must have $bc=0$, and hence from $b\geq c$ it follows that then $c=0$. Because $a+b+c=0$ it finally follows that $b=\tfrac32$, and so $$(a,b,c)=(\tfrac32,\tfrac32,0).$$
This shows that the minimum of $P$ is indeed $\tfrac92$, and that it is attained precisely at the points $$(a,b,c)=(0,\tfrac32,\tfrac32),\qquad(a,b,c)=(\tfrac32,0,\tfrac32),\qquad(a,b,c)=(\tfrac32,\tfrac32,0).$$
Original answer, by brute force and elementary methods:
Plugging in $c=3-a-b$, we want the minimum of $$P=2a^2+2b^2+9-6a-6b+8ab-2a^2b-2ab^2,$$ with the restriction that $a,b\geq0$ and $a+b\leq3$. At the boundary points we either have $a=0$ or $b=0$ or $a+b=3$, and hence correspondingly either \begin{eqnarray*} P&=&2b^2-6b+9,\\ P&=&2a^2-6a+9,\\ P&=&2a^2+2(3-a)^2+9-6a-6(3-a)+8a(3-a)+2a^2(3-a)+2a(3-a)^2\\ &=&2a^2-6a+9, \end{eqnarray*} where in each case $0\leq a,b\leq3$. From here the local minima on the boundary are easily determined to be at $$(a,b)=(0,\tfrac32),\qquad(a,b)=(\tfrac32,0),\qquad(a,b)=(\tfrac32,\tfrac32)$$ each with value $\tfrac92$. For the extrema of $P$ on the interior we compute the derivatives of $P$ w.r.t. $a$ and $b$, which shows that $$4a-6+8b-4ab-2b^2=0\qquad\text{ and }\qquad 4b-6+8a-2a^2-4ab=0,$$ at any interior extremum $(a,b)$ of $P$. Taking the difference shows that $$0=2(a^2-b^2)-4(a-b)=2(a-b)(a+b-2)$$ so either $a=b$ or $a+b=2$. If $a=b$ then the quadratics above both become $$0=4a-6+8b-4a^2-2a^2=6(a-1)^2,$$ which shows that $a=b=1$, and then $P=5$, which is not minimal. If $a+b=2$ then $$0=4(2-a)-6+8a-2a^2-4a(2-a)=2(a-1)^2,$$ which again yields $a=b=1$, again not yielding a minimum.
In conclusion the minimum of $P$ is $\tfrac92$, and it is attained precisely at the points $$(a,b,c)=(0,\tfrac32,\tfrac32),\qquad(a,b,c)=(\tfrac32,0,\tfrac32),\qquad(a,b,c)=(\tfrac32,\tfrac32,0).$$