Range of $\sin \alpha\cdot \sin \beta+\sin \beta \cdot \sin \gamma+\sin \gamma \cdot \sin \alpha$ is

Question

If line makes an angle of $\alpha,\beta,\gamma$ with positive $x,y$ and $z$

axis respectively. Then Range of

$\sin \alpha\cdot \sin \beta+\sin \beta \cdot \sin \gamma+\sin \gamma \cdot \sin \alpha$ is

Try: If line makes an angle of $\alpha,\beta,\gamma$ with positive $x,y$ and

$z$ axis respectively. Then $\cos^2 \alpha+\cos^2 \beta+\cos^2 \gamma = 1$

means $\sin^2 \alpha+\sin^2 \beta+\sin^2 \gamma = 2.$

Using Cauchy Schwarz Inequality

$$(\sin^2 \alpha+\sin^2 \beta+\sin^2 \gamma )\cdot (\sin^2 \beta+\sin^2 \gamma+\sin^2 \alpha)\geq (\sin \alpha\cdot \sin \beta+\sin \beta \cdot \sin \gamma+\sin \gamma \cdot \sin \alpha)^2$$

So $$\sin \alpha\cdot \sin \beta+\sin \beta \cdot \sin \gamma+\sin \gamma \cdot \sin \alpha\leq 2$$

Now i did not understand how i find minimum value of $$\sin \alpha\cdot \sin \beta+\sin \beta \cdot \sin \gamma+\sin \gamma \cdot \sin \alpha$$

could some help me, Thanks

Answer 1

Let $O(0,0,0)$, $A(\sqrt{a},0,0),$ $B(0,\sqrt{b},0)$ and $C(0,0,\sqrt{c})$, where $a$, $b$ and $c$ are non-negatives such that $a^2+b^2+c^2\neq0.$

Thus, we can assume that $$\sin\alpha=\sqrt{\frac{b+c}{a+b+c}},$$ $$\sin\beta=\sqrt{\frac{a+c}{a+b+c}}$$ and $$\sin\gamma=\sqrt{\frac{a+b}{a+b+c}}.$$ Now, for $b=c=0$ we obtain: $$\sum_{cyc}\sin\alpha\sin\gamma=1.$$ We'll prove that it's a minimal value.

Indeed, we need to prove that $$\sum_{cyc}\sqrt{\frac{(a+b)(a+c)}{(a+b+c)^2}}\geq1$$ or $$\sum_{cyc}\sqrt{(a+b)(a+c)}\geq a+b+c,$$ which is obvious because $$\sqrt{(a+b)(a+c)}\geq\sqrt{a\cdot a}=a.$$

Answer 2

For the upper limit of the range, you've proved that $2$ is an upper bound on the range, but not that $2$ actually equals the upper bound. To show that $2$ is the upper limit of the range, you must exhibit an example (or otherwise prove) that $2$ can be attained. In this case, the upper bound can be attained with the line in the direction $\langle 1,1,1\rangle$. In this case, the angle with all axes is the same, so, by the equality condition for Cauchy-Schwarz, you get the value of $2$ (or, alternately, use the cross product to calculate that the sine of the angle is $\frac{\sqrt{2}}{\sqrt{3}}$ and calculate the desired value directly).

For the lower bound, this occurs when the line is in the direction of one of the axes, e.g., $\langle 1,0,0\rangle$. In this case, one of the sines is zero and the other two are $1$. This leads to a value of $1$. This can be proved with a bit of multivariate calculus, if you wish, or a little figuring.

Here's the sketch of how to argue that $1$ is the minimum value (without calculus): Observe that \begin{align} (\sin\alpha+\sin\beta+\sin\gamma)^2&=(\sin^2\alpha+\sin^2\beta+\sin^2\gamma)+2(\sin\alpha\sin\beta+\sin\alpha\sin\gamma+\sin\beta\sin\gamma)\\ &=2+2(\sin\alpha\sin\beta+\sin\alpha\sin\gamma+\sin\beta\sin\gamma). \end{align} Therefore, since everything is nonnegative, minimizing the desired quantity is the same as minimizing $\sin\alpha+\sin\beta+\sin\gamma$. Since each sine is between $0$ and $1$, we know that $\sin^2\alpha\leq\sin\alpha$, $\sin^2\beta\leq\sin\beta$, and $\sin^2\gamma\leq\sin\gamma$. Therefore, $$ 2=\sin^2\alpha+\sin^2\beta+\sin^2\gamma\leq \sin\alpha+\sin\beta+\sin\gamma. $$ The minimum is attained for the example above.