Challenge Basic Math Problem of the Week 10/10/2017

[PF PotW Robot]

Here is this week's basic math problem. We have several members who will check solutions, but we also welcome the community in general to step in. We also encourage finding different methods to the solution. If one has been found, see if there is another way. Using spoiler tags is optional. Occasionally there will be prizes for extraordinary or clever methods. Spoiler tags are optional.

Let that $a,\, b,\,c$ be three angles with $0<a,\,b,\,c<90^\circ$ that satisfy $sin a+\sin b+\sin c=1$

Prove that $\tan^2 a+\tan^2 b+\tan^2 c \ge \dfrac{3}{8}$

(PotW thanks to our friends at http://www.mathhelpboards.com/)

 

[Charles Link]

I have a partial proof of this. $\tan^2 {a}=\frac{\sin^2 {a}}{1-\sin^2 {a}}$ . Let $\sin{a}=x$. Similarly $y=\sin{b}$ and $z=\sin{c}$. Using Lagrange multiplier method to minimize $f(x,y,z)= \frac{x^2}{1-x^2}+\frac{y^2}{1-y^2}+\frac{z^2}{1-z^2}$ subject to $x+y+z-1=0$ results in $\frac{2x}{(1-x^2)^2}=\frac{2y}{(1-y^2)^2}=\frac{2z}{(1-z^2)^2}=-\lambda$ along with $x+y+z=1$. By symmetry of the 3 equations with the $\lambda$, the solution is $x=y=z$, so that by the constraint equation, $x=y=z=\frac{1}{3}$ is where a maximum or minimum will occur. The result is $f(\frac{1}{3},\frac{1}{3},\frac{1}{3})=\frac{3}{8}$. I haven't shown this to be the absolute minimum, and it would take additional effort to show that this is indeed the case.

 

[StoneTemplePython]

I liked @Charles Link 's response. It also got me thinking. Frequently when faced with a constrained multi-variable optimization problem, it can be tough to directly prove optimaliity. A much slicker approach, which sometimes works, is to use something like the tools in information theory (e.g. KL Divergence), or more generally: convexity. In particular, answers that involve all values identically matching tend to scream convexity for a solution. After suitable domain restrictions, trig functions are rich in convexity, so that's what I used here.

- - - - - - - - - - - -
edit:
to clean this up a bit and simplify notation, let

$\Theta = \frac{a+b+c}{3}$
- - - - - - - - - - - -
starter piece / lemma:

claim:
$\sin^2\big(\Theta \big) \geq \frac{1}{9}$

$\sin(x)$ is negative convex in $(0, \frac{\pi}{2})$, i.e. its second derivative is $-\sin(x)$, which is real non-positive over that interval. If we apply Jensen's Inequality, we have:

$\sin\big(\Theta \big) = \sin\big(\frac{a+b+c}{3}\big) \geq \frac{1}{3}\big(\sin(a) + \sin(b) + \sin(c)\big) = \frac{1}{3}\big(1) = \frac{1}{3}$
hence if we square both sides

$\sin\big(\Theta \big)^2 = \sin^2\big(\Theta \big) \geq \frac{1}{9}$

- - - - - - - - - - - -
now consider the original inequality we want to prove:

$\tan^2 a+\tan^2 b+\tan^2 c \geq \frac{3}{8}$

note that $\tan^2(x)$ is convex over our domain -- again verified via second derivative test, which is real, non-negative 2nd derivative.

Multiply each side by $\frac{1}{3}$ and apply Jensen's Inequality.

$\frac{1}{3}\big(\tan^2 a+\tan^2 b+\tan^2 c\big) \geq \tan^2\Big(\frac{1}{3}\big( a+b+ c \big)\Big) = \tan^2\Big(\Theta\Big) \geq \frac{1}{8}$

hence if we can prove

$\tan^2\Big(\Theta\Big) \geq \frac{1}{8}$

then we're home free.

$\tan^2\Big(\Theta\Big) = \frac{\sin^2\Big(\Theta \Big)}{1-\sin^2\Big(\Theta \Big)} \geq \frac{1}{8}$

$\sin^2\big(\Theta \big) \geq \frac{1}{8}\Big(1-\sin^2\big(\Theta \big)\Big) = \frac{1}{8} -\frac{1}{8}\sin^2\big(\Theta \big)$

$\frac{9}{8}\sin^2\big(\Theta \big) \geq \frac{1}{8}$

or equivalently

$\sin^2\Big(\Theta \Big) \geq \frac{1}{9}$

which is true by the initial lemma
- - - - - - - - - - - -
technical note:
$0 \lt \Big(1-\sin^2\big(\Theta \big)\Big)$

-- edit: if further justification is needed --
The above is equivalent to

$\sin^2\big(\Theta \big) \lt 1$

Note that the sine function has a magnitude at most of 1. And It cannot be equal to one over our domain (which $\Theta$ -- the average of a, b, and c-- must be in). Hence the magnitude of the sine function over our domain is less than one. This means that the squared value of the sine function (equivalently, the squared magnitude of the sine function) must be less than one over our domain.

 

[PetSounds]

You folks are way beyond me, but I think I cracked it nonetheless. Here's my high-school approach. The process I used was laborious, so I've streamlined a few steps in the LaTeX version. If there's a more elegant way to tackle this problem using only the basics, I'd love to see it—my way is ugly!

Let $p, q, r$ be real numbers with $-1 < p, q, r < 2$ and $p + q + r = 0$.

$\sin a = \frac {1+p}{3} \\ \sin b = \frac {1+q}{3} \\ \sin c = \frac{1 + r}{3}$

$\tan^2 a = \frac{1 + 2p + p^2}{8 - 2p - p^2} \\ \tan^2 b = \frac{1 + 2q + q^2}{8 - 2q - q^2} \\ \tan^2 c = \frac{1 + 2r + r^2}{8 - 2r - r^2}$

$\tan^2 a + \tan^2 b + \tan^2 c = \frac{(1 + 2p + p^2)(1 + 2q + q^2)(1 + 2r + r^2)}{(8 - 2p - p^2)(8 - 2q - q^2)(8 - 2r - r^2)}$

After multiplying, collecting like terms, and factoring (which I've omitted, because it would take me at least two hours to type up), the expression above reduces to this:

$\tan^2 a + \tan^2 b + \tan^2 c = \frac{3(64 + 16p^2 + 16q^2 + 16r^2 - 20pq - 20qr - 20pr - 10p^2q - 10p^2r - 10pq^2 - 10pr^2 - 10q^2r - 10qr^2 + 8pqr - 5p^2q^2 - 5p^2r^2 - 5q^2r^2 + 2p^2q^2r + 2p^2qr^2 + 2pq^2r^2 + p^2q^2r^2)}{8(64 - 8p^2 - 8q^2 - 8r^2 + 4pq + 4pr + 4qr + 2p^2q + 2p^2r + 2pq^2 + 2pr^2 + 2q^2r + 2qr^2 - pqr + p^2q^2 + p^2r^2 + q^2r^2 - \frac{p^2q^2r}{4} - \frac{1/4p^2qr^2}{4} - \frac{pq^2r^2}{4} - \frac {p^2q^2r^2}{8})}$

Now we show that the expression on the right hand side is greater than or equal to three-eighths. We will do this by showing that $64 + 16p^2 + 16q^2 + 16r^2 - 20pq - 20qr - 20pr - 10p^2q - 10p^2r - 10pq^2 - 10pr^2 - 10q^2r - 10qr^2 + 8pqr - 5p^2q^2 - 5p^2r^2 - 5q^2r^2 + 2p^2q^2r + 2p^2qr^2 + 2pq^2r^2 + p^2q^2r^2 \geq 64 - 8p^2 - 8q^2 - 8r^2 + 4pq + 4pr + 4qr + 2p^2q + 2p^2r + 2pq^2 + 2pr^2 + 2q^2r + 2qr^2 - pqr + p^2q^2 + p^2r^2 + q^2r^2 - \frac{p^2q^2r}{4} - \frac{p^2qr^2}{4} - \frac{pq^2r^2}{4} - \frac {p^2q^2r^2}{8}$

Suppose $64 + 16p^2 + 16q^2 + 16r^2 - 20pq - 20qr - 20pr - 10p^2q - 10p^2r - 10pq^2 - 10pr^2 - 10q^2r - 10qr^2 + 8pqr - 5p^2q^2 - 5p^2r^2 - 5q^2r^2 + 2p^2q^2r + 2p^2qr^2 + 2pq^2r^2 + p^2q^2r^2 < 64 - 8p^2 - 8q^2 - 8r^2 + 4pq + 4pr + 4qr + 2p^2q + 2p^2r + 2pq^2 + 2pr^2 + 2q^2r + 2qr^2 - pqr + p^2q^2 + p^2r^2 + q^2r^2 - \frac{p^2q^2r}{4} - \frac{p^2qr^2}{4} - \frac{pq^2r^2}{4} - \frac {p^2q^2r^2}{8}$.

$24p^2 + 24q^2 + 24r^2 + 9pqr + \frac{9p^2q^2r}{4} + \frac{9p^2qr^2}{4} + \frac{9pq^2r^2}{4} + \frac{9p^2q^2r^2}{8} < 24pq + 24pr + 24qr + 12p^2q + 12p^2r + 12pq^2 + 12pr^2 + 12q^2r + 12qr^2 + 6p^2q^2 + 6p^2r^2 + 6q^2r^2 \\ 48p^2 + 24q^2 + 24r^2 + 9pqr + \frac{9p^2q^2r}{4} + \frac{9p^2qr^2}{4} + \frac{9pq^2r^2}{4} + \frac{9p^2q^2r^2}{8} < 24p^2 + 24pq + 24pr + 24qr + 12p^2q + 12p^2r + 12pq^2 + 12pr^2 + 12q^2r + 12qr^2 + 6p^2q^2 + 6p^2r^2 + 6q^2r^2 = 24p(p + q + r) + 24qr + 12p^2q + 12p^2r + 12pq^2 + 12pr^2 + 12q^2r + 12qr^2 + 6p^2q^2 + 6p^2r^2 + 6q^2r^2 = 24qr + 12p^2q + 12p^2r + 12pq^2 + 12pr^2 + 12q^2r + 12qr^2 + 6p^2q^2 + 6p^2r^2 + 3q^2r^2 + 3q^2r^2$

It's easy to show that the above statement is false by "pairing off" terms on each side, which implies that $64 + 16p^2 + 16q^2 + 16r^2 - 20pq - 20qr - 20pr - 10p^2q - 10p^2r - 10pq^2 - 10pr^2 - 10q^2r - 10qr^2 + 8pqr - 5p^2q^2 - 5p^2r^2 - 5q^2r^2 + 2p^2q^2r + 2p^2qr^2 + 2pq^2r^2 + p^2q^2r^2 \geq 64 - 8p^2 - 8q^2 - 8r^2 + 4pq + 4pr + 4qr + 2p^2q + 2p^2r + 2pq^2 + 2pr^2 + 2q^2r + 2qr^2 - pqr + p^2q^2 + p^2r^2 + q^2r^2 - \frac{p^2q^2r}{4} - \frac{p^2qr^2}{4} - \frac{pq^2r^2}{4} - \frac {p^2q^2r^2}{8}$. It follows that $\tan^2 a + \tan^2 b + \tan^2 c \geq \frac 3 8$.

I'll probably clean up/flesh out this solution tomorrow—it's close to 1 AM here now. I got engrossed and lost track of time

 

[mfb]

Whoa :D

Here is how I would solve it:
Define x=sin(a), y=sin(b), z=sin(c). Define $t(x)=\frac{x^2}{1-x^2}$. We want to minimize t(x)+t(y)+t(z) where $0 \leq x,y,z < 1$. In this range the second derivative of t is positive, therefore $2 t(\frac{x+y}{2}) < t(x) + t(y)$ if $x\neq y$. In other words x=y=z is the only option for a minimum. The original constraint leads to 1/3 for them, and 3t(1/3)=3/8 is the minimum.

Edit: Fixed definition of t.

 

[PetSounds]

Whoa :D

Here is how I would solve it:
Define x=sin(a), y=sin(b), z=sin(c). Define $t(x)=\frac{x}{1+x^2}$. We want to minimize t(x)+t(y)+t(z) where $0 \leq x,y,z < 1$. In this range the second derivative of t is positive, therefore $2 t(\frac{x+y}{2}) < t(x) + t(y)$ if $x\neq y$. In other words x=y=z is the only option for a minimum. The original constraint leads to 1/3 for them, and 3t(1/3)=3/8 is the minimum.

I knew there was an simpler way Just one question: how did you arrive at $t(x)=\frac{x}{1+x^2}$? I can only come up with
$\sin a = x \\ \sin^2 a = x^2 \\ \cos^2 a = 1 - x^2 \\ \tan^2 a = \frac{x^2}{1-x^2}$

 

[mfb]

I messed up its definition. Should be t(x) = x²/(1-x²) of course to mimic the tan².