<s>[INST] Prove inequality inside a triangle "Let $ABC$ be a triangle with centroid $G$. A line $PQ$ is drawed in the triangle such that it passes through $G$ and intersects the sides $AB$ and $AC$ in $P$ and $Q$ respectively."
Prove that:
$$\frac {PB \cdot QC}{PA\cdot QA} \le \frac {1}{4}$$ 
I made an aproximate drawing of this problem. Unfortunately i don't know how to start with this kind of problem and i need some hints. Thanks.

 [/INST] Let $M\in AC$ and $N\in AB$ such that $MN||BC$, and $G\in MN.$
Also, let $\measuredangle MGQ=\theta$.
Thus, in the standard notation $$MG=GN=\frac{1}{3}a$$ and  by law of sines we obtain:
$$\frac{PN}{\sin\theta}=\frac{\frac{1}{3}a}{\sin\beta}$$ or
$$PN=\frac{a\sin\theta}{3\sin\beta}.$$
Similarly, 
$$MQ=\frac{a\sin\theta}{3\sin\gamma}.$$
Id est, we need to prove that
$$\frac{\left(\frac{1}{3}c+\frac{a\sin\theta}{3\sin\beta}\right)\left(\frac{1}{3}b-\frac{a\sin\theta}{3\sin\gamma}\right)}{\left(\frac{2}{3}c-\frac{a\sin\theta}{3\sin\beta}\right)\left(\frac{2}{3}b+\frac{a\sin\theta}{3\sin\gamma}\right)}\leq\frac{1}{4}$$ or
$$\frac{(\sin\gamma\sin\beta+\sin\alpha\sin\theta)(\sin\gamma\sin\beta-\sin\alpha\sin\theta)}{(2\sin\gamma\sin\beta-\sin\alpha\sin\theta)(2\sin\gamma\sin\beta+\sin\alpha\sin\theta)}\leq\frac{1}{4}$$ or
$$\sin^2\alpha\sin^2\theta\geq0.$$
Done!
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