Extreme of $\cos A\cos B\cos C$ in a triangle without calculus. 
If $A,B,C$ are angles of a triangle, find the extreme value of $\cos A\cos B\cos C$.

I have tried using $A+B+C=\pi$, and applying all and any trig formulas, also AM-GM, but nothing  helps.
On this topic we learned also about Cauchy inequality, but I have no experience with it.
The answer according to Mathematica is when $A=B=C=60$.
Any ideas?

Assume without loss of generality that $\displaystyle A , B, C $ belongs to the first quadrant.By the identity  $\displaystyle sen^2x+cos^2x=1$, we can see that:
\begin{equation*}
    tanx+cotx=\frac{1}{senxcosx}
\end{equation*}
So:
\begin{equation}
    \frac{sen\frac{\beta}{2}sen\frac{\gamma}{2}}{sen\frac{\alpha}{2}cos\frac{\alpha}{2}}=\left(tan\frac{\alpha}{2}+cot\frac{\alpha}{2}\right)sen\frac{\beta}{2}sen\frac{\gamma}{2}
\end{equation}
\begin{equation}
    \frac{cos\frac{\beta}{2}cos\frac{\gamma}{2}}{sen\frac{\alpha}{2}cos\frac{\alpha}{2}}=\left(tan\frac{\alpha}{2}+cot\frac{\alpha}{2}\right)cos\frac{\beta}{2}cos\frac{\gamma}{2}
\end{equation}
Add so:
$\\ \\ \displaystyle \left(tan\frac{\alpha}{2}+cot\frac{\alpha}{2}\right)sen\frac{\beta}{2}sen\frac{\gamma}{2}+\left(tan\frac{\alpha}{2}+cot\frac{\alpha}{2}\right)cos\frac{\beta}{2}cos\frac{\gamma}{2}=\frac{cos\left(\frac{\beta}{2}-\frac{\gamma}{2}\right)}{sen\frac{\alpha}{2}cos\frac{\alpha}{2}}\leq \frac{1}{sen\frac{\alpha}{2}cos\frac{\alpha}{2}} \\ \\ $
We conclude that:
\begin{equation}
    \left(tan\frac{\alpha}{2}+cot\frac{\alpha}{2}\right)\left(cos\frac{\beta}{2}cos\frac{\gamma}{2}+sen\frac{\beta}{2}sen\frac{\gamma}{2}\right)\leq \frac{1}{sen\frac{\alpha}{2}cos\frac{\alpha}{2}}
\end{equation}
Apply Ravi transfomation, we get:
\begin{equation*}
    \left(\sqrt{\frac{xy}{z(x+y+z)}}+\sqrt{\frac{z(x+y+z)}{xy}}\right)\times
\end{equation*}
\begin{equation*}
    \left( \sqrt{\frac{x(x+y+z)}{(x+z)(x+y)}}\sqrt{\frac{y(x+y+z)}{(x+y)(y+z)}}+ \sqrt{\frac{yz}{(x+z)(x+y)}} \sqrt{\frac{xz}{(x+y)(y+z)}}\right)
\end{equation*}
\begin{equation}
    \leq \frac{(x+y)(y+z)}{\sqrt{xyz(x+y+z)}}
\end{equation}
Its easy see that:
\begin{equation}
    (xy+z(x+y+z))\left((x+y+z)\sqrt{xy}+z\sqrt{xy}\right)\leq (x+y)^2(y+z)\sqrt{(x+z)(y+z)}
\end{equation}
The above inequality is equivalent to:
\begin{equation}
    (x+z)(y+z)\sqrt{xy}\left(x+y+2z\right)\leq (x+y)^2(y+z)\sqrt{(x+z)(y+z)}
\end{equation}
Which with due cancellation reduces to the inequality below:
\begin{equation}
    (x+z)\sqrt{xy}\left(x+y+2z\right)\leq (x+y)^2\sqrt{(x+z)(y+z)}
\end{equation}
By symmetry we conclude the inequalities:
\begin{equation}
    (x+y)\sqrt{yz}\left(2x+y+z\right)\leq (y+z)^2\sqrt{(x+z)(x+y)}
\end{equation}
\begin{equation}
    (y+z)\sqrt{xz}\left(x+2y+z\right)\leq (x+z)^2\sqrt{(y+z)(x+y)}
\end{equation}
Multiplying, we get:
\begin{equation}
    xyz(2x+y+z)(x+2y+z)(x+y+2z)\leq [(x+z)(x+y)][(y+z)(x+y)][(x+z)(y+z)]
\end{equation}
Supose $\displaystyle xy+xz+yz=1$, note that x,y and z will be cotangents of angles of an acute triangle, hence it follows
$\\ \displaystyle [(x+z)(x+y)][(y+z)(x+y)][(x+z)(y+z)]=$
$\\ \displaystyle [x^2+xy+xz+yz][y^2+xy+xz+yz][z^2+xy+xz+yz]=$
$\\ \displaystyle [x^2+1][y^2+1][z^2+1]=[cot^2\alpha'+1][cot^2\beta'+1][cot^2\gamma'+1]=csc^2\alpha' csc^2\beta' csc^2\gamma' \\ \\$
From where we can see that:
\begin{equation*}
    \cot\alpha \cot\beta \cot\beta \left(2\cot\alpha+\cot\beta+\cot\beta\right)\left(\cot\alpha+2\cot\beta+\cot\beta\right)\left(\cot\alpha+\cot\beta+2\cot\beta\right) \leq 
\end{equation*}
\begin{equation}
  csc^2\alpha' csc^2\beta' csc^2\gamma'
\end{equation}
Extracting the cube root on both sides of the inequality, we have:
\newpage
\begin{equation*}
    (\cot\alpha \cot\beta \cot\beta)^{\frac{1}{3}} \left(2\cot\alpha+\cot\beta+\cot\beta\right)^{\frac{1}{3}}\left(\cot\alpha+2\cot\beta+\cot\beta\right)^{\frac{1}{3}}\left(\cot\alpha+\cot\beta+2\cot\beta\right)^{\frac{1}{3}}  
\end{equation*}
\begin{equation}
  \leq csc^{\frac{2}{3}}\alpha' csc^{\frac{2}{3}}\beta' csc^{\frac{2}{3}}\gamma'
\end{equation}
Look at the expression:
$$\\ \\ \sqrt[3]{(2\times a+b+c)(a+2\times b+c)(a+b+2\times c)}$$
So:
$$ \displaystyle a+a+b+c\geq 4\sqrt[4]{a^2bc}$$
$$  \displaystyle a+b+b+c\geq  4\sqrt[4]{ab^2c}$$
$$ \displaystyle a+b+c+c\geq  4\sqrt[4]{abc^2}$$
And see, taking the product:
$$\\ \\ (2\times a+b+c)(a+2\times b+c)(a+b+2\times c)\geq 64\sqrt[4]{4a^4b^4c^4}=64abc$$
And so:
$$\\ \\ (2\times a+b+c)(a+2\times b+c)(a+b+2\times c)\geq 64abc$$
Extracting the cube root, we have:
$$\\ \\ \sqrt[3]{(2\times a+b+c)(a+2\times b+c)(a+b+2\times c)}\geq4\sqrt[3]{abc} \\ \ \ $$
Make the replacement $\displaystyle a=\cot(\alpha),b=\cot(\beta),c=\cot(\gamma)$, and so
$$\\ \\ \sqrt[3]{(  2\cot(\alpha)+\cot(\beta)+\cot(\gamma)   )(\cot(\alpha)+2\cot(\beta)+\cot(\gamma))(\cot(\alpha)+\cot(\beta)+2\cot(\gamma))}\geq$$
$$4\sqrt[3]{\cot(\alpha)\cot(\beta)\cot(\gamma)} \\ \ \ $$
Multiplying by
$\displaystyle (\cot\alpha \cot\beta \cot\beta)^{\frac{1}{3}}$, we get:
\newpage
\begin{equation*}
     4(\cot\alpha \cot\beta \cot\beta)^{\frac{2}{3}} \leq 
\end{equation*}
\begin{equation}
      (\cot\alpha \cot\beta \cot\beta)^{\frac{1}{3}}\left(2\cot\alpha+\cot\beta+\cot\beta\right)^{\frac{1}{3}}\left(\cot\alpha+2\cot\beta+\cot\beta\right)^{\frac{1}{3}}\left(\cot\alpha+\cot\beta+2\cot\beta\right)^{\frac{1}{3}} 
\end{equation}
We arrive by transitivity to the inequality below:
\begin{equation}
     4(\cot\alpha \cot\beta \cot\beta)^{\frac{2}{3}} \leq  \csc^{\frac{2}{3}}\alpha \csc^{\frac{2}{3}}\beta \csc^{\frac{2}{3}}
\end{equation}
We can assume without loss of generality that the angles are in the first quadrant, so we can extract the root and preserve the sign of inequality, because by this hypothesis all terms are positive... our inequality is equivalent to the inequality required in the problem.Done.For more solution enter to the link https://www.overleaf.com/read/qyrxbsjhhjst
