Prove that $AD\cdot BD \cdot CD \leq \dfrac{32}{27}$ where $ABC$ is a triangle of circumradius 1 and $D\in (BC)$.

Question

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!

Answer 1

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: $$(1-x^2)(1+x^2-2x\cos\alpha)^{\frac 12}\leq(1-x^2)(1+x)=4(1-x)\cdot\dfrac{1+x}{2}\cdot\dfrac{1+x}{2}\leq 4\cdot \left(\dfrac{2}{3}\right)^3 = \dfrac{32}{27}.$$

Answer 2

Just some work, that was too big for a comment.

Well, when we have a $\triangle\text{ABC}$:

We know that:

$$ \begin{cases} \angle\alpha+\angle\beta+\angle\gamma=\pi\\ \\ \frac{\text{a}}{\sin\left(\angle\alpha\right)}=\frac{\text{b}}{\sin\left(\angle\beta\right)}=\frac{\text{c}}{\sin\left(\angle\gamma\right)}\\ \\ \text{a}^2=\text{b}^2+\text{c}^2-2\text{b}\text{c}\cos\left(\angle\alpha\right)\\ \\ \text{b}^2=\text{a}^2+\text{c}^2-2\text{a}\text{c}\cos\left(\angle\beta\right)\\ \\ \text{c}^2=\text{a}^2+\text{b}^2-2\text{a}\text{b}\cos\left(\angle\gamma\right) \end{cases}\tag1 $$

We also know that the circumradius of that triangle is given by:

$$\text{R}=\frac{\text{a}\text{b}\text{c}}{\sqrt{\left(\text{a}+\text{b}+\text{c}\right)\left(\text{b}+\text{c}-\text{a}\right)\left(\text{a}+\text{b}-\text{c}\right)}}\tag2$$

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So, when $\text{R}=1$ we know that:

$$\text{a}\text{b}\text{c}=\sqrt{\left(\text{a}+\text{b}+\text{c}\right)\left(\text{b}+\text{c}-\text{a}\right)\left(\text{a}+\text{b}-\text{c}\right)}\tag3$$

Which is the same as:

$$\text{a}^2\cdot\text{b}^2\cdot\text{c}^2=\left(\text{a}+\text{b}+\text{c}\right)\left(\text{b}+\text{c}-\text{a}\right)\left(\text{a}+\text{b}-\text{c}\right)\tag4$$

Answer 3

Let $M$ the second intersection of $AD$ and the circumcircle of triangle $ABC$. By power of a point, $BD \cdot DC=AD\cdot DM$. Let $x=AD$ and $y=DM$. The inequality can be rewritten as $$x^2y\leq \dfrac{32}{27}.$$

It is clear that $$(x-2y)^2(4x+y)\geq 0$$ which is equivalent to $$x^2y\leq (x+y)^3\cdot \dfrac{4}{27}$$ and combining with $x+y\leq 2$ (which is true as $x+y=AM$ is not greater than the diameter of the circle) we get the conclusion.