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?
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By the $AM-GM$ inequality, $$\frac{\cos A+\cos B+\cos C}{3}\geq \sqrt[3]{\cos A\cos B\cos C}$$ with equality only when $\cos A=\cos B=\cos C$
On
If $y=\cos A\cos B\cos C,$
$2y=\cos C[2\cos A\cos B]=\cos C\{\cos(A-B)+\cos(A+B)\}$
As $A+B=\pi-C,\cos(A+B)=-\cos C$
On rearrangement we have $$\cos^2C-\cos C\cos(A-B)+2y=0$$
As $C$ is real, so will be $\cos C$
$\implies$ the discriminant $$\cos^2(A-B)-8y\ge0\iff y\le\dfrac{\cos^2(A-B)}8\le\dfrac18$$
The equality occurs if $\cos^2(A-B)=1\iff\sin^2(A-B)=0$
$\implies A-B=n\pi$ where $n$ is any integer
As $0<A,B<\pi, n=0\iff A=B$ and consequently $$\cos^2C-\cos C+2\cdot\dfrac18=0\implies \cos C=\dfrac12\implies C=\dfrac\pi3$$
$\implies A=B=\dfrac{A+B}2=\dfrac{\pi-C}2=\dfrac\pi3=C$
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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
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We may assume $\alpha\leq\beta\leq\gamma$. The product $p:=\cos\alpha\cos\beta\cos\gamma$ is negative iff $\gamma>{\pi\over2}$, so that the minimum value $p_{\min}=-1$ is attained when $\alpha=\beta=0$, $\gamma=\pi$, i.e., for a degenerate triangle.
In the search of $p_{\max}$ we may assume $\gamma\leq{\pi\over2}$, hence $0\leq\cos\gamma\leq1$. From $$2\cos\alpha\cos\beta=\cos(\alpha-\beta)+\cos(\alpha+\beta)\leq 1+\cos(\alpha+\beta)$$ it follows that $$p\leq{1\over2}(1-\cos\gamma)\cos\gamma={1\over2}\left({1\over4}-\left(\cos\gamma-{1\over2}\right)^2\right)\leq{1\over8}\ .$$ This proves $p_{\max}={1\over8}$, and this value is attained when $\gamma={\pi\over3}$ and $\alpha=\beta$, i.e., for an equilateral tringle.
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If you don't like to my first solution, here is a more simple solution.Assume without loss of generality that A,B,C belongs to the first quadrant.And so, it's easy to see that: \begin{align*} x+y\geq 2\sqrt{xy} \tag{1} \end{align*}
\begin{align*} x+z\geq 2\sqrt{xz} \tag{2} \end{align*}
\begin{align*} y+z\geq 2\sqrt{yz} \tag{3} \end{align*}
Multiplying (1),(2),(3), we get:
$$ (x+y)(x+z)(y+z) \geq 8xyz$$ $$ \frac{1}{8} \geq \frac{xyz}{(x+y)(x+z)(y+z)}$$ $$ \displaystyle \frac{1}{8} \geq \sqrt{\frac{xy}{(x+z)(y+z)}}\sqrt{\frac{xz}{(x+y)(y+z)}}\sqrt{\frac{yz}{(x+y)(x+z)}} $$
By Ravi transformation and Law of cosines, we get $$ \frac{1}{8} \geq sen\frac{\alpha}{2} sen\frac{\beta}{2} sen\frac{\gamma}{2}$$ And this is equivalent to: $$ \frac{1}{8} \geq cos \left(\frac{\pi}{2}-\frac{\alpha}{2}\right) cos \left(\frac{\pi}{2}-\frac{\beta}{2}\right) cos \left(\frac{\pi}{2}-\frac{\gamma}{2}\right)$$ Take the substitution $\displaystyle \frac{\pi}{2}-\frac{\alpha}{2}=A$,$\displaystyle \frac{\pi}{2}-\frac{\beta}{2}=B$,$\displaystyle \frac{\pi}{2}-\frac{\gamma}{2}=C$, we get $\displaystyle A+B+C=\pi$ and: $$ \frac{1}{8} \geq cosAcosBcosC$$
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We know from geometry that for any triangle ABC the distance between its circumcenter $O$ and its orthocenter $H$ can be given by the following formula:
$$OH^2=R^2(1-8\cos A\cos B\cos C)$$ R being the circumradius.
Besides that, we know that orthocenter and circumcenter coincide only if the triangle is an equilateral one.
Therefore, $\cos A\cos B\cos C$ attains a maximum value of $\frac 18$ when $A=B=C=\pi/3$.
No Calculus needed.
On
Due to the inequality of means and the law of cosines, we know that:
\begin{equation*} 4a^4=((a^2+b^2-c^2)+(a^2+c^2-b^2))^2\geq 4(a^2+b^2-c^2)(a^2+c^2-b^2)=\\ 16a^2bc\cos\beta \cos\gamma \Rightarrow a^2\geq 4bc\cos\beta \cos\gamma \end{equation*}
\begin{equation*} 4b^4=((a^2+b^2-c^2)+(b^2+c^2-a^2))^2\geq 4(a^2+b^2-c^2)(b^2+c^2-a^2)=\\ 16ab^2c\cos\alpha \cos\gamma \Rightarrow b^2\geq 4ac\cos\alpha \cos\gamma \end{equation*}
\begin{equation*} 4c^4=((a^2+c^2-b^2)+(b^2+c^2-a^2))^2\geq 4(a^2+c^2-b^2)(b^2+c^2-a^2)=\\ 16abc^2\cos\alpha \cos\beta \Rightarrow c^2\geq 4ab\cos\alpha \cos\beta \end{equation*}
Multiplying the last three inequalities and extracting the root (assuming, without loss of generality, that alpha, beta and gamma are in the first quadrant) we will have the required inequality.
I use the Lagrange's multipliers theorem.
Let us define the functions $$g(A,B,C)=A+B+C-\pi,$$ $$f(A,B,C)=\cos A\cos B\cos C.$$ Then we have $$\mathrm{d}g(A,B,C)=\mathrm{d}A+\mathrm{d}B+\mathrm{d}C,$$ $$\mathrm{d}f(A,B,C)=-\sin A\cos B\cos C\mathrm{d}A-\cos A\sin B\cos C\mathrm{d}B-\cos A\cos B\sin C\mathrm{d}C$$ where $\left(\mathrm{d}A,\mathrm{d}B,\mathrm{d}C\right)$ is the coordinate forms on $\mathbb{R}^3$ (for example, $\mathrm{d}A[(1,2,3)]=1$).
Then we apply the Lagrange's multipliers theorem : we must cancel every determinants of the matrix $$\left(\begin{array}{cccccccc} 1 & 1 & 1 \\ -\sin A\cos B\cos C & -\cos A\sin B\cos C & -\cos A\cos B\sin C \end{array}\right).$$ This yields \begin{cases} -\cos A\sin B\cos C + \sin A\cos B\cos C = 0 \\ -\cos A\cos B\sin C + \sin A\cos B\cos C = 0 \\ -\cos A\cos B\sin C + \cos A\sin B\cos C=0 \end{cases} The first line gives us $\cos C = 0$ (and then $C=\pi/2$) or $$-\cos A\sin B + \sin A\cos B =0$$ that is $$\sin(A-B)=0$$ and then $A=B$. Do the same for the two other lines to get the condition $A=B=C$ (the other conditions are impossible, check that). Because in a triangle, we have $g(A,B,C)=0$, we finally find that $A=B=C=\pi/3$.