AM GM only?

Using Titu's Lemma on the positive reals $a,b,c$ , we have,

$a^2+b^2+c^2\geq \frac{(a+b+c)^2}{1+1+1}\\ \implies (a+b+c)^2\leq 3\times 3=9 \implies a+b+c\leq 3$

and equality occurs when $a=b=c=1$ .

Now, using Titu's Lemma on $P$ , we have,

$P\geq \frac{(a+b+c)^2}{3(a+b+c)}=\frac{a+b+c}{3}\implies a+b+c\leq 3P$

From the two results obtained, we can tell that the minimum of $P$ is achieved when $3P$ provides the tightest upper bound for $(a+b+c)$ which is $3$ . Hence, we have,

$3\min(P)=3\implies \min(P)=1$

The minimum can be verified easily using the equality case for Titu's Lemma. The minimum occurs at $a=b=c=1$ which was the same for $(a+b+c)$ . Hence, the equality cases are matched and the minimum is confirmed to be $\boxed{1}$

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P.s- I will edit the solution a bit, if necessary, the next day. It's 5:00 AM now and I need to go to sleep.