Trigonometry Inequality

[anemone]

If X, Y and Z are in the interval $ (0, \frac{\pi}{2})$, prove that $cosX cosY cosZ+sinXsinYsinZ \leq 1$

Since I see no relevance between X, Y and Z, (but I know I can let X>Y>Z), I don't know how to begin and to be completely candid, I have no clue at all how to do it.

Any hint would be very much welcome.

Thanks.

 

[Opalg]

If X, Y and Z are in the interval $ (0, \frac{\pi}{2})$, prove that $\cos X \cos Y \cos Z+\sin X\sin Y\sin Z \leq 1$

One way would be to start with the Cauchy–Schwarz inequality: $$(\cos X \cos Y) \cos Z+(\sin X\sin Y)\sin Z \leqslant \sqrt{\cos^2X\cos^2Y + \sin^2X\sin^2Y}\sqrt{\cos^2Z+\sin^2Z}.$$ See if you can take it from there.

 

[anemone]

Oh...the Cauchy–Schwarz inequality!
OK. Let's see if I'm doing this right.

$(cosXcosY)cosZ+(sinXsinY)sinZ\leq \sqrt {cos^2Xcos^2Y+sin^2Xsin^2Y} \sqrt {cos^2Z+sin^2Z}$

$(cosXcosY)cosZ+(sinXsinY)sinZ\leq \sqrt {cos^2Xcos^2Y+sin^2Xsin^2Y} \sqrt {1}$

$(cosXcosY)cosZ+(sinXsinY)sinZ\leq \sqrt {(cosXcosY)^2+(sinXsinY)^2} \sqrt {1}$

$(cosXcosY)cosZ+(sinXsinY)sinZ\leq \sqrt {\frac{1}{2}[cos(X+Y)+cos(X-Y)])^2+(\frac{1}{2}[cos(X+Y)-cos(X-Y)])^2} $

$(cosXcosY)cosZ+(sinXsinY)sinZ\leq \sqrt {\frac{1}{4}(cos^2(X+Y)+cos^2(X-Y)+2cos(X+Y)cos(X-Y)+cos^2(X+Y)+cos^2(X-Y)-2cos(X+Y)cos(X-Y)}$

$(cosXcosY)cosZ+(sinXsinY)sinZ\leq \sqrt {\frac{1}{4}(2(cos^2(X+Y)+cos^2(X-Y))}$

$(cosXcosY)cosZ+(sinXsinY)sinZ\leq \sqrt {\frac{1}{2}(cos^2(X+Y)+cos^2(X-Y))}$

$(cosXcosY)cosZ+(sinXsinY)sinZ\leq \sqrt {\frac{1}{2}(\frac{cos2(X+Y)+1}{2}+\frac{cos2(X-Y)+1}{2})}$

$(cosXcosY)cosZ+(sinXsinY)sinZ\leq \sqrt {\frac{1}{2}(\frac{cos2(X+Y)+cos2(X-Y)}{2}+1)}$

$(cosXcosY)cosZ+(sinXsinY)sinZ\leq \sqrt {\frac{1}{2}(\frac{2cos2Xcos2Y}{2}+1)}$

$(cosXcosY)cosZ+(sinXsinY)sinZ\leq \sqrt {\frac{1}{2}(cos2Xcos2Y+1)}$

But we know that $-1 \leq cos2X\leq1$ and $-1 \leq cos2Y\leq1$ for $0 \leq2X\leq\pi$, so we can state that

$(cosXcosY)cosZ+(sinXsinY)sinZ\leq \sqrt {\frac{1}{2}((1)(1)+1)}$

$(cosXcosY)cosZ+(sinXsinY)sinZ\leq \sqrt {\frac{1}{2}(2)}$

$(cosXcosY)cosZ+(sinXsinY)sinZ\leq 1$ (Q.E.D.)

I cheat a bit because we're told that X, Y and Z are in the interval $ (0, \frac{\pi}{2})$ but I take it as they are lie in the interval $ [0, \frac{\pi}{2}]$...

 

[Alexmahone]

I had thought of an easier way.

Oh...the Cauchy–Schwarz inequality!
OK. Let's see if I'm doing this right.

$(cosXcosY)cosZ+(sinXsinY)sinZ\leq \sqrt {cos^2Xcos^2Y+sin^2Xsin^2Y} \sqrt {cos^2Z+sin^2Z}$

$(cosXcosY)cosZ+(sinXsinY)sinZ\leq \sqrt {cos^2Xcos^2Y+sin^2Xsin^2Y}$

Using the Cauchy–Schwarz inequality again,

$\cos^2X\cos^2Y+\sin^2X\sin^2Y\leq\sqrt{\cos^4X+ \sin^4X}\sqrt{\cos^4Y+\sin^4Y}$

$\cos^4X<\cos^2X$

$\sin^4X<\sin^2Y$

Adding, $\cos^4X+\sin^4X<1$

Similarly, $\cos^4Y+\sin^4Y<1$

So, $\cos^2X\cos^2Y+\sin^2X\sin^2Y\leq\sqrt{\cos^4X+ \sin^4X}\sqrt{\cos^4Y+\sin^4Y}<1$

So, $\sqrt{\cos^2X\cos^2Y+\sin^2X\sin^2Y}<1$

 

[anemone]

Oh..., one need to apply the Cauchy–Schwarz inequality twice to get the problem solved. Thanks, Alexmahone.

I think I've to polish up on the concept of Cauchy–Schwarz inequality.

BTW, do you think the condition that the problem set should read $ (0, \frac{\pi}{2}]$ rather than $ (0, \frac{\pi}{2})$ in order to have $cosX cosY cosZ+sinXsinYsinZ \leq 1$?

 

[Alexmahone]

BTW, do you think the condition that the problem set should read $ (0, \frac{\pi}{2}]$ rather than $ (0, \frac{\pi}{2})$ in order to have $cosX cosY cosZ+sinXsinYsinZ \leq 1$?

It should probably read $\left[0,\ \frac{\pi}{2}\right]$.

 

[Alexmahone]

Actually, there is a much easier way:

$\cos X\cos Y\cos Z<\cos X\cos Y$

$\sin X\sin Y\sin Z<\sin X\sin Y$

Adding,

$\cos X\cos Y\cos Z+\sin X\sin Y\sin Z<\cos X\cos Y+\sin X\sin Y=\cos(X-Y)<1$

 

[anemone]

Actually, there is a much easier way:

$\cos X\cos Y\cos Z<\cos X\cos Y$

$\sin X\sin Y\sin Z<\sin X\sin Y$

Adding,

$\cos X\cos Y\cos Z+\sin X\sin Y\sin Z<\cos X\cos Y+\sin X\sin Y=\cos(X-Y)<1$

Yes. That is a cool and intellectual approach.