Vrftjkry

Question: Given ${CD}^2=AD.BD$, then prove that, $sin A.sin B le sin^2{Cover2}$.

I got 2 approaches to the question. Both landing nowhere.

(i) Using sine rule I got that $sin A.sin B=sin alpha.sin beta$ but can't proceed.
(ii) On extending $CD$ further to $D'$ such that $CD=DD'={CD'over2}$ which makes $ABCD'$ a cyclic quadrilateral again ending where the first step ends.

Help me please.

Thanks for any hints or solution. Hope that question is decently put.

DeepSea's answer is nice, but it seems that you don't get it.

So, let me add more details.

You've already got
$$sin Asin B=sinalphasinbetatag1$$

Since
$$frac{(x+y)^2}{4}-xy=frac{x^2+2xy+y^2-4xy}{4}=frac{(x-y)^2}{4}ge 0$$
holds for any $x,yinmathbb R$, we see that
$$xylefrac{(x+y)^2}{4}$$
holds for any $x,yinmathbb R$.

So, we have
$$sinalphasinbetale frac{(sinalpha+sinbeta)^2}{4}tag2$$

Using the following sum-to-product identity
$$sinalpha+sinbeta=2sinbigg(frac{alpha+beta}{2}bigg)cosbigg(frac{alpha-beta}{2}bigg)$$
one gets
$$begin{align}frac{(sinalpha+sinbeta)^2}{4}&
=frac{1}{4}left(2sinbigg(frac{alpha+beta}{2}bigg)cosbigg(frac{alpha-beta}{2}bigg)right)^2
\\&=sin^2bigg(frac{alpha+beta}{2}bigg)cos^2bigg(frac{alpha-beta}{2}bigg)tag3end{align}$$

For any $thetainmathbb R$, we have $|costheta|le 1$, so $cos^2thetale 1$.

So, we have
$$cos^2bigg(frac{alpha-beta}{2}bigg)le 1$$
Multiplying the both sides by $sin^2(frac{alpha+beta}{2}) (gt 0)$ gives
$$sin^2bigg(frac{alpha+beta}{2}bigg)cos^2bigg(frac{alpha-beta}{2}bigg)le sin^2bigg(frac{alpha+beta}{2}bigg)=sin^2frac{C}{2}tag4$$

It follows from $(1)(2)(3)(4)$ that
$$sin Asin Blesin^2frac C2$$

Taking from where you left off,and using the well-known fact: $xy le dfrac{(x+y)^2}{4}$, we have : $ sin Acdot sin B = sin alpha cdot sin beta le dfrac{ (sin alpha + sin beta )^2}{4} = dfrac{left(2sin(frac{alpha+beta}{2})cdot cos(frac{alpha - beta}{2})right)^2}{4} = sin^2(frac{alpha+beta}{2})cdot cos^2(frac{alpha-beta}{2}) le sin^2 (frac{alpha+beta}{2}) = sin^2 (C/2)$ because $cos^2(frac{alpha - beta}{2}) le 1$. This establishes the claim.