<s>[INST] Prove that $AD\cdot BD \cdot CD \leq \dfrac{32}{27}$ where $ABC$ is a triangle of circumradius 1 and $D\in (BC)$. 
Let triangle $ABC$ of circumradius $1$ and $D$ a point on side $(BC)$.
  Prove that $$AD\cdot BD\cdot CD\leq \dfrac{32}{27}.$$

My idea. By letting $\alpha = \dfrac{BD}{BC}$ (of course $0<\alpha <1$) we get $BD=BC\cdot \alpha
, \enspace CD=BC\cdot(1-\alpha)\tag{1}$ and also
$$\overrightarrow{AD}=(1-\alpha)\cdot\overrightarrow{AB}+\alpha\cdot \overrightarrow{AC}.$$
By squaring this relation we have that $$AD^2=AB^2(1-\alpha)+AC^2\alpha+BC^2(\alpha^2-\alpha). \tag{$2$}$$
By law of sines we also have $AB=2\sin C$, $AC=2\sin B$ and $BC=2\sin A$.
Now combining with $(1)$ and $(2)$ we may rewrite the desired inequality as follows:
$$((1-\alpha)\sin^2C+\alpha\sin^2B+(\alpha^2-\alpha)\sin^2A)\cdot\alpha^2(1-\alpha)^2\sin^4A\leq \dfrac{2^4}{27^2}.$$
This is where I got stuck. Maybe we could also use the fact that $\sin A=\sin (\pi -B-C)=-\sin(B+C)=-(\sin B\cos C+\sin C\cos B)$ to get rid of $\sin A$?
Thank you in advance!
 [/INST] 
It is a really bad practice to ask a geometry problem without a picture. 
Let $OD = x.$ Then $BD\cdot CD = 1 - x^2.$ If $\angle AOD = \alpha,$ then:
$$AD^2 = 1+x^2-2x\cos\alpha.$$
So you need to prove:
$$(1-x^2)\sqrt{1+x^2-2x\cos\alpha}\leq\dfrac{32}{27}.$$ 
But this is just AM-GM:
$$(1-x^2)(1+x^2-2x\cos\alpha)^{\frac 12}\leq(1-x^2)(1+x)=4(1-x)\cdot\dfrac{1+x}{2}\cdot\dfrac{1+x}{2}\leq 4\cdot \left(\dfrac{2}{3}\right)^3 = \dfrac{32}{27}.$$
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