Inequality concerning segments of an interior point to vertices of a triangle 
Let $ABC$ be a triangle and it has an interior point $P$ inside. Show that: $$BC\cdot BP\cdot CP + CA\cdot CP\cdot AP + AB\cdot AP\cdot BP\geq AB\cdot BC\cdot CA$$  

If can, how?
Or does the triangle need specific conditions to match such inequality?
 A: Let $P$ it's an origin, $A(a)$, $B(b)$ and $C(c)$ in the Gauss plane.
Thus,
$$\frac{AP\cdot BP}{AC\cdot BC}+\frac{AP\cdot CP}{AB\cdot BC}+\frac{BP\cdot CP}{AB\cdot AC}=$$
$$=\left|\frac{ab}{(c-a)(b-c)}\right|+\left|\frac{ca}{(a-b)(b-c)}\right|+\left|\frac{bc}{(a-b)(c-a)}\right|\geq$$
$$\geq\left|\frac{ab}{(c-a)(b-c)}+\frac{ca}{(a-b)(b-c)}+\frac{bc}{(a-b)(c-a)}\right|=|-1|=1$$
and we are done!
A: I will give a geometric proof to the inequality.
Let $|BC|=a, |CA|=b, |AB|=c, |AP|=x, |BP|=y, |CP|=z$. So, O.P. is
$$ ayz + bzx + cxy \geq abc \tag{1}$$
Let's draw parallelograms $BCPD$ and $ABDE$. Thus $|AE|=|BD|=|CP|=b$ and they are parallel. So, $ACPE$ is a parallelogram and we get $|PE|=|CA|=b$. Also, e can write $|PD|=a, |DE|=c$.

Let's apply Ptolemy's inequality on the quadrilaterals $APBD$ and $APDE$:
$$ |AD|\cdot y + xz \geq ac \tag{2}$$
$$ az + xc \geq |AD|\cdot b \tag{3}$$
Now, let's multiply $(2)$ inequality by $b$ and let's multiply $(3)$ inequality by $y$. We find that
$$ |AD|\cdot b y + bzx \geq abc \tag{4}$$
$$ ayz + cxy \geq |AD|\cdot b y \tag{3}$$
and therefore, we get the $(1)$ equality:
$$ ayz + bzx + cxy \geq abc.$$
Note: Also, we can determine the equality condition. For the equality, $APBD$ and $APDE$ are cyclic quadrilaterals. So,
$\angle PAB = \angle PDB = \angle PCB = \alpha$,
$\angle PBA = \angle PDA = \angle PEA = \angle PCA = \beta $,
$\angle PAC = \angle EPA = \angle EDA = \angle DAB = \angle DPB = \angle PBC = \gamma $.
Hence, in the triangle $ABC$, $2(\alpha + \beta + \gamma) = 180^\circ $ and $\alpha + \beta + \gamma = 90^\circ $. Then we conclude that $ABC$ is an acute angle triangle and $P$ is the orthocenter.
