Given that $ a,b,c$ are positive numbers and $(a+b)(b+c)(c+a)=1$, find the maximum value of $P=ab+bc+ca$ I am stuck with this problem:

Given that $ a,b,c$ are positive numbers and $(a+b)(b+c)(c+a)=1$, find the maximum value of $P=ab+bc+ca$

I tried to use: $(a+b)(a+c)(b+c)=(a+b+c)(ab+bc+ca)-abc \implies  P=\frac{1+abc}{a+b+c}$ but I am stuck. Can anyone help me? Thank you!
I'm just 14 years old, so don't use derivative or something like that (my brother uses it but I cant understand).

We have $(a + b)(b + c)(c + a) = (a + b + c)(ab + bc + ca) - abc$ which results in
$$P = \frac{1 + abc}{a + b + c}.$$
Using AM-GM, we have
$$(a + b)(b + c)(c + a)
\ge 2\sqrt{ab}\cdot 2\sqrt{bc}\cdot 2\sqrt{ca} = 8abc$$
and
$$(a + b)(b + c)(c + a)\le \left(\frac{a + b + b + c + c + a}{3}\right)^3 = \frac{8}{27}(a + b + c)^3.$$
Thus, we have
$abc \le 1/8$ and $a + b + c\ge 3/2$.
Thus, we have
$$P = \frac{1 + abc}{a + b + c} \le \frac{1 + 1/8}{3/2} = 3/4.$$
Also, when $a = b = c = 1/2$,
we have $(a + b)(b + c)(c + a) = 1$ and $P = 3/4$.
Thus, the maximum of $P$ is $3/4$.
