Minimum of $\begin{aligned}\frac{a^2+b^2}{c^2}\end{aligned}$ in $\Delta ABC$

Question

In $\triangle ABC$, $\sin B=-\cos C$. Find the minimum of $\begin{aligned}\frac{a^2+b^2}{c^2}\end{aligned}$.

According to the law of sines, $\begin{aligned}\frac{a^2+b^2}{c^2}=\frac{\sin^2A+\sin^2B}{\sin^2C}\end{aligned}$.

Solution $1$

Let $\sin B=-\cos C=k$, then $\sin B=k>0$, so $\cos C=-k<0$, meaning that $\begin{aligned}C>\frac\pi2\end{aligned}$. Thus $\begin{aligned}B<\frac\pi2\end{aligned}$, so $\cos B=\sqrt{1-k^2}$.

Now $\sin A=\sin(B+C)=k(-k)+\sqrt{1-k^2}\times\sqrt{1-k^2}=1-2k^2$. So $$\frac{\sin^2A+\sin^2B}{\sin^2C}=4-4k^2+\frac{10}{1-k^2}-13\ge2\sqrt{40}-13=4\sqrt{10}-13.$$

Solution $2$

From $\sin B=-\cos C>0$ we have $$B=C-\frac{\pi}{2}, \sin B=\sin \left(C-\frac{\pi}{2}\right)=-\cos C , \sin A=\sin (B+C)=\sin \left(2 C-\frac{\pi}{2}\right)=-\cos 2 C$$. So \begin{aligned}\frac{\sin ^{2} A+\sin ^{2} B}{\sin ^{2} C}&=\frac{\cos ^{2} 2 C+\cos ^{2} C}{\sin ^{2} C} \\ &=\frac{\left(1-2 \sin ^{2} C\right)^{2}+\left(1-\sin ^{2} C\right)}{\sin ^{2} C} \\ &=\frac{2+4 \sin ^{4} C-5 \sin ^{2} C}{\sin ^{2} C}=\frac{2}{\sin ^{2} C}+4 \sin ^{2} C-5 \\ & \geqslant 2 \sqrt{\frac{2}{\sin ^{2} C} \cdot 4 \sin ^{2} C}-5=4 \sqrt{2}-5, \end{aligned}

Answer 1

Everything should be correct in solution $2$ because AM-GM can be used as $\sin^2 C$ is always non-negative.

For solution $1$ however, after $\sin A = 1 - 2k^2$ your steps don't follow: something must have been gone catastrophically wrong there. Using the values that we obtained for $\sin A, \sin B$ and that $\sin^2 C = 1 - \cos^2 C$:

$$\frac{\sin^2A+\sin^2B}{\sin^2C}= \frac{(1 - 2k^2)^2 + k^2}{1-k^2} = \frac{1 - 3k^2 + 4k^4}{1 - k^2}.$$

After doing some synthetic division on $\frac{4x^2 - 3x + 1}{x - 1}$ (or polynomial long division if you prefer):

$$ \begin{array}{c|cc} 1 & 4 & -3 & 1\\ \ & \ & 4 & 1 \\ \hline \ & 4 & 1 & 2 \\ \end{array} $$

and this implies $(4x^2 - 3x + 1) = (4x + 1)(x - 1) + 2$ or that $(4k^4 - 3k^2 + 1)$ $ = -(4k^2 + 1)(1 - k^2) + 2$. Thus we have:

$$\frac{1 - 3k^2 + 4k^4}{1 - k^2} = -(4k^2 + 1) + \frac{2}{1 - k^2} = 4 - 4k^2 + \frac{2}{1 - k^2} - 5.$$

instead of what you had there.

Now as $-1 ≤ \sin B ≤ 1$ and also $k \ne -1, 1$ due to the denominator, we can apply AM-GM since $1 - k^2$ will be non-negative.

This gives the minimum as $2 \sqrt{8} - 5 = 4 \sqrt{2} - 5$ like in your other answer. Thus solution $2$ is correct, but we can use AM-GM as well on your solution attempt $1$ with a bit more effort.

Addendum: the two functions which are in terms of $k$ or $\sin^2 C$ are remarkably similar. However, applying the transformation $x^2 \to 1 - x^2$ in both directions shows that the two functions do indeed have the same minimum, which is not obvious before polynomial division.

Answer 2

The Law of Cosines gives:

$$a^2 = b^2 + c^2 - 2bc \cos(A)$$ $$b^2 = a^2 + c^2 - 2ac \cos(B)$$ $$c^2 = a^2 + b^2 - 2ab \cos(C)$$

Adding these three equations gives:

But $A = \pi - (B + C)$, so

$$\cos(A) = \cos(\pi - (B + C))$$ $$= \cos(\pi)\cos(B + C) + \sin(\pi)\sin(B + C)$$ $$= -\cos(B + C)$$ $$= -(\cos(B)\cos(C) - \sin(B)\sin(C))$$ $$= (\sin(B)\sin(C) - \cos(B)\cos(C)$$

We are given that $\cos(C) = -\sin(B)$. Since $\sin^2(C) = 1 - \cos^2(C) = 1 - \sin^2(B) = \cos^2(B)$, we have $\sin(B) = \pm \cos(B)$. So,

$$a^2 = b^2 + c^2 - 2bc (\pm \cos(B)\sin(B) + \cos(B)\sin(B))$$ $$b^2 = a^2 + c^2 - 2ac \cos(B)$$ $$c^2 = a^2 + b^2 + 2ab \sin(B)$$

If we chose $\sin(C) = -\cos(B)$, then $\triangle ABC$ is a right triangle with A being the right angle. We have $a^2 = b^2 + c^2$, So $\frac{a^2 + b^2}{c^2} = \frac{2b^2 + c^2}{c^2} = 2\frac{b^2}{c^2} + 1$. This approaches a minimum value of 1 when $b$ is as small as possible relative to $c$.

For the rest of this answer, I'll examine the case when $\sin(C) = +\cos(B)$. Then the first equation above simplifies to $a^2 = b^2 + c^2 - 4bc \cos(B)\sin(B)$.

From the last two equations above, we get $\cos(B) = \frac{a^2 + c^2 - b^2}{2ac}$ and $\sin(B) = \frac{c^2 - a^2 - b^2}{2ab}$. Thus,

$$a^2 = b^2 + c^2 - 4bc (\frac{a^2 + c^2 - b^2}{2ac}) (\frac{c^2 - a^2 - b^2}{2ab})$$

Multiplying through by $(2ac)(2ab) = 4a^2bc$ to get rid of the fractions gives:

$$4a^4bc = 4a^2b^3c + 4a^2bc^3 - 4bc (a^2 + c^2 - b^2) (c^2 - a^2 - b^2)$$ $$a^4 = a^2b^2 + a^2c^2 - (a^2 + c^2 - b^2) (c^2 - a^2 - b^2)$$ $$a^4 = a^2b^2 + a^2c^2 - (a^2c^2 - a^4 - a^2b^2 + c^4 - a^2c^2 - b^2c^2 - b^2c^2 + a^2b^2 + b^4)$$ $$a^4 = a^2b^2 + a^2c^2 - a^2c^2 + a^4 + a^2b^2 - c^4 + a^2c^2 + b^2c^2 + b^2c^2 - a^2b^2 - b^4$$ $$b^4 - 2b^2c^2 + c^4 = a^2(b^2 + c^2)$$ $$a^2 = \frac{(b^2 - c^2)^2}{b^2 + c^2}$$

Plugging this into the target expression gives:

$$q := \frac{a^2 + b^2}{c^2}$$ $$= \frac{\frac{(b^2 - c^2)^2}{b^2 + c^2} + b^2}{c^2}$$ $$= \frac{(b^2 - c^2)^2 + b^2(b^2 + c^2)}{c^2(b^2 + c^2)}$$ $$= \frac{2b^4 + c^4 - b^2c^2}{c^2(b^2 + c^2)}$$

Let $b = cx$. Then

$$q = \frac{2c^4x^4 + c^4 - c^4x^2}{c^2(c^2x^2 + c^2)}$$ $$= \frac{2c^4x^4 - c^4x^2 + c^4}{c^4x^2 + c^4}$$ $$= \frac{2x^4 - x^2 + 1}{x^2 + 1}$$

Let $u = x^2$. Then:

$$q = \frac{2u^2 - u + 1}{u + 1}$$

Setting the derivative equal to zero gives:

$$\frac{dq}{du} = \frac{(4u - 1)(u + 1) - (2u^2 - u + 1)(1)}{(u + 1)^2} = 0$$ $$(4u - 1)(u + 1) - (2u^2 - u + 1) = 0$$ $$4u^2 + 3u - 1 - 2u^2 + u - 1 = 0$$ $$2u^2 + 4u - 2 = 0$$ $$u^2 + 2u - 1 = 0$$ $$u = \frac{-2 \pm \sqrt{8}}{2}$$ $$u = -1 \pm \sqrt{2}$$

But since triangles can't have negative or imaginary side lengths, only the positive root makes sense. So $u = \sqrt{2} - 1$. Thus:

$$q = \frac{2(\sqrt{2} - 1)^2 - (\sqrt{2} - 1) + 1}{(\sqrt{2} - 1) + 1}$$ $$= \frac{2(2 - 2\sqrt{2} + 1) - \sqrt{2} + 1 + 1}{\sqrt{2}}$$ $$= \frac{8 - 5\sqrt{2}}{\sqrt{2}}$$ $$= \frac{8}{\sqrt 2} - 5$$ $$= 4 \sqrt 2 - 5$$

So, my solution agrees with your Solution 2.