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?
Inequality concerning segments of an interior point to vertices of a triangle
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geometric-inequalities
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0I tried to relate radius of circumcircles of three triangles inside ABC but could not make any improvements in solving the problem. – 2017-02-17
1 Answers
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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!