How can I prove that the perimeter is at most 60? Problem: Let $\Delta$ be a triangle in the plane. Let $P$ be the perimeter of the triangle and $A$ be the area. Let $a,b,c$ be the length of the sides and suppose they are positive integers. Suppose finally that $A=P$. How can I prove that $P \leq 60$?
My attempt: I tried using Erone's formula getting:
$$a+b+c = \sqrt{\frac P 2 \left(\frac{P}{2}-a\right)\left(\frac P 2-b\right)\left(\frac P 2 - c\right)}$$
from which I obtained:
$$4P=-P^3 + 4 (a^2b+a^2c+ab^2+b^2c+ac^2+bc^2+abc)$$
and I noticed that $abc \leq \left(\frac{P}{2}\right)^2$ but I am not able to continue. Any help will be appreciated.
Remark: as suggested in the comments, defining $u=a+b-c$, $v=a+c-b$ and $w=b+c-a$ we get $16(u+v+w)=uvw$.
Noticing that $uvw$ should be even and that none of $u,v,w$ can be odd we can write $u=2k$, $v=2h$, $w=2n$ in such a way we deduce $4(k+h+n)=khn$.

$$hnk=4(h+n+k), h,n,k\in\mathbb{N}$$
$$(hk-4)n=4(h+k)$$
WLOG $h\geq n \geq k > 0$. Then
$$(hk-4)k \leq 4(h+k) \Rightarrow hk^2-8k \leq 4h \Rightarrow hk^2-8h\leq 4h\Rightarrow hk^2\leq12h\Rightarrow k\leq 3$$
$$n=\frac{4(h+k)}{hk-4}$$
At $k=1$:
$$n=\frac{4h+4}{h-4}=4+\frac{20}{h-4}$$
$$h+n+k=h+5+\frac{20}{h-4}=9+(h-4)+\frac{20}{h-4}$$
$$(h-4)|20 \Rightarrow (h-4)+\frac{20}{h-4}\leq 21 \Rightarrow h+n+k\leq 30$$
At $k=2$:
$$n=\frac{4h+8}{2h-4}=2+\frac{8}{h-2}$$
$$h+n+k=h+4+\frac{8}{h-2}=6+(h-2)+\frac{8}{h-2}$$
$$(h-2)|8 \Rightarrow (h-2)+\frac{8}{h-2}\leq 9 \Rightarrow h+n+k\leq 15$$
At $k=3$:
$$n=\frac{4h+12}{3h-4}=1+\frac{h+16}{3h-4}$$
$$n\geq k=3 \Rightarrow \frac{h+16}{3h-4}\geq 2\Rightarrow h+16\geq 6h-8 \Rightarrow h\leq \frac{24}{5}<5$$
$$h\leq 4\Rightarrow h+n+k\leq 4+4+3=11$$
Then $h+n+k\leq 30$ in all possible cases.
Perimeter of triangle $$P=a+b+c=u+v+w=2(h+k+n)\leq 60$$
