Prove $8(1-\cos^2a)(1-\cos^2b)(1-\cos^2c) \geq 27 \cos a\cos b\cos c$

Question

Let $a$, $b$ and $c$ be angles of an acute triangle. Prove that: $$8(1-\cos^2a)(1-\cos^2b)(1-\cos^2c) \geq 27 \cos a\cos b\cos c$$

I tried am-gm, rm-gm rm-am and a couple of other inequalities but I didn't get anywhere.

Answer 1

Let $u$, $v$ and $w$ be lengths-sides of the triangle such that

$\cos{a}=\frac{v^2+w^2-u^2}{2vw}$, $\cos{b}=\frac{u^2+w^2-v^2}{2uw}$, $\cos{c}=\frac{u^2+v^2-w^2}{2uv}$ and

let $u^2+v^2-w^2=r$, $u^2+w^2-v^2=q$, $v^2+w^2-u^2=p$.

Hence, we need to prove that $$8\prod_{cyc}\left(1-\frac{r^2}{(p+r)(q+r)}\right)\geq\frac{27pqr}{(p+q)(p+r)(q+r)}$$ or $$8(pq+pr+qr)^3\geq27pqr(p+q)(p+r)(q+r).$$ Let $pq=z$, $pr=y$ and $qr=x$.

Hence, we need to prove that $$8(x+y+z)^3\geq27(x+y)(x+z)(y+z),$$ which is AM-GM: $$\left(\frac{\sum\limits_{cyc}(x+y)}{3}\right)^3\geq\prod_{cyc}(x+y).$$

Done!

Answer 2

Let $$\ a,\ b,\ c \in\left(0, \dfrac{\pi}{2}\right),\quad a+b+c = \pi,\qquad(1) $$ and let us try to find minimum of function $$f(a,b) = 8\sin^2a\sin^2b\sin^2(\pi-a-b) - 27\cos a\cos b\cos(\pi-a-b),$$ in this area.

Equaling partial derivatives to zero, one can get $$\begin{cases} 16\sin a\sin^2b\sin(\pi-a-b)(\cos a \sin(\pi-a-b) - \sin(a)\cos(\pi-a-b))\\ - 27\cos b(-\sin a\cos(\pi-a-b) + \cos a\sin(\pi-a-b) = 0,\\ 16\sin^2a\sin b\sin(\pi-a-b)(\cos b \sin(\pi-a-b) - \sin(b)\cos(\pi-a-b))\\ - 27\cos a(-\sin b\cos(\pi-a-b) + \cos b\sin(\pi-a-b) = 0, \end{cases}$$ $$\begin{cases} -16\sin a\sin^2b\sin(\pi-a-b)\sin(\pi-2a-b) + 27\cos b(\sin(\pi-2a-b) = 0,\\ -16\sin^2a\sin b\sin(\pi-a-b)\sin(\pi-a-2b) + 27\cos a(\sin(\pi-a-2b) = 0, \end{cases}$$ $$\begin{cases} \genfrac{[}{.}{0}{0}{\sin(\pi-2a-b)=0}{16\sin a\sin^2b\sin(\pi-a-b) = 27\cos b }\\ \genfrac{[}{.}{0}{0}{\sin(\pi-a-2b)=0}{16\sin^2a\sin b\sin(\pi-a-b) = 27\cos a. }\\ \end{cases}$$ Let us consider all posible cases.

Case $1$ $$\begin{cases} \sin(\pi-2a-b)=0\\ \sin(\pi-a-2b)=0 \end{cases}\qquad\rightarrow \begin{cases} 2a+b = \pi\\ a+2b = \pi, \end{cases}$$ $$a=b=\pi/3,\quad f(a,b) = 0.$$

Case $2$ $$\begin{cases} \sin(\pi-2a-b)=0\\ 16\sin^2a\sin b\sin(\pi-a-b) = 27\cos a, \end{cases}\rightarrow \begin{cases} b = \pi-2a\\ 16\sin^3a\sin2a = 27\cos a, \end{cases}$$ $$\sin^4a = \frac{27}{32},\quad \sin^2a >5/6,\quad \cos2a = 1-2\sin^2a<-\dfrac23.\qquad(2)$$ So $$f(a,\pi-2a) = 8\sin^4a\sin^22a + 27\cos2a\cos^2a = 16\sin^5a\sin2a\cos a + 27\cos2a\cos^2a$$ $$ = 27\sin^2a\cos^2a + 27\cos2a\cos^2a = 27\cos^2a(\sin^2a + \cos2a),$$ $$f(a, \pi-2a) > 27\cos^2a\cdot\dfrac16 >0.$$

Case $3$ $$\begin{cases} \sin(\pi-a-2b)=0\\ 16\sin^2a\sin b\sin(\pi-a-b) = 27\cos b, \end{cases}$$ proves similarly to case $2$.

Case $4$ $$\begin{cases} 16\sin a\sin^2b\sin(\pi-a-b) = 27\cos b\\ 16\sin^2a\sin b\sin(\pi-a-b) = 27\cos a. \end{cases}\quad(3).$$

Subtraction of these equations with non-zero factors $\cos a$ and $\cos b$ leads to equation $$\sin a\sin b\sin(\pi-a-b)(\sin b\cos a - \sin a\cos b) = 0,$$ $$\sin a\sin b\sin(\pi-a-b)\sin(a-b) = 0,$$ $$\sin(a-b) = 0,\quad a=b.$$ Substitution to $(3)$ gives $$16\sin^3a\sin2a = 27\cos a,$$ $$32\sin^4a = 27,$$ and then using $(2)$ $$f(a,a) = 8\sin^4a\sin^22a + 27\cos2a\cos^2a = 16\sin^5a\sin2a\cos a + 27\cos2a\cos^2a$$ $$ = 27\sin^2a\cos^2a + 27\cos2a\cos^2a = 27\cos^2a(\sin^2a + \cos2a),$$ $$f(a, a) > 27\cos^2a\cdot\dfrac16 >0.$$

That means that in the OP conditions $$\boxed{8(1-\cos^2a)(1-\cos^2b)(1-\cos^2c) \geq 27 \cos a\cos b\cos c}$$

Answer 3

Here's a proof that accounts for all acute-angled triangles whose angles lie in $ \left(\tan^{-1}{\sqrt{2}}, \dfrac{\pi}{2}\right)$.

Note that $f(x) = \ln \sin(x) \tan(x)$ is convex over $ \left( \tan^{-1}{\sqrt{2}}, \dfrac{\pi}{2}\right)$ since $f''(x) = \dfrac{\tan^{2}(x) -2}{\sin^{2}(x)} > 0 \quad \forall x\in \left(\tan^{-1}{\sqrt{2}}, \dfrac{\pi}{2}\right)$

Hence we can apply Jensen's inequality to $f(x)$ and $a, b,c \in \left(\tan^{-1}{\sqrt{2}}, \dfrac{\pi}{2}\right)$ to get:

$f\left(\dfrac{a+b+c}{3}\right)\leq \dfrac{f(a) + f(b)+ f(c)}{3}$

$\Rightarrow \ln \dfrac{3}{2} \leq \dfrac{\ln \dfrac{\sin^{2}(a)\,\sin^{2}(b)\,\sin^{2}(c)}{\cos(a)\,\cos(b)\,\cos(c)}}{3}$

$\Rightarrow \dfrac{27}{8} \leq \dfrac{\sin^{2}(a)\,\sin^{2}(b)\,\sin^{2}(c)}{\cos(a)\,\cos(b)\,\cos(c)}$, as required.

Can someone figure out an extension of this argument that accounts for those acute-angled triangles with one angle in $ \left(0, \tan^{-1}{\sqrt{2}}\right)$?