An Inequality with Cot, Cos, and Sin

Solution

For any $\alpha,\;$ with $\sin\alpha\gt 0,\;$ $\displaystyle\cos\alpha=\frac{\cot\alpha}{\sqrt{1+\cot^2\alpha}}\;$ and $\displaystyle\sin\alpha=\frac{1}{\sqrt{1+\cot^2\alpha}}.\;$ Also, in any triangle $ABC,\;$ $\displaystyle\sum_{cycl}\cot A\,\cot B=1\;$ and $\displaystyle\sum_{cycl}\cos^2A+2\cos A\;\cos B\;\cos C=1.$

Denote $\displaystyle\sqrt{3}\cot A=x,\;$ $\displaystyle\sqrt{3}\cot B=y,\;$ $\displaystyle\sqrt{3}\cot C=z.\;$ Then, $x,y,z\ge 0\;$ and $xy+yz+zx=3.$

Also, $\displaystyle\cos A=\frac{x}{\sqrt{3+x^2}},\;$ $\displaystyle\sin A=\frac{\sqrt{3}}{\sqrt{3+x^2}},\;$ etc. On the other hand,

\displaystyle\begin{align}\prod_{cycl}(3+x^2)&=\prod_{cycl}\left(\sum_{cycl}xy+x^2\right)\\ &=\left[\prod_{cycl}(x+y)\right]^2, \end{align}

implying

$\sqrt{(3+x^2)(3+y^2)(3+z^2)}=(x+y)(y+z)(z+x).$

It follows that

\displaystyle\begin{align} \cos^2 A+\cos^2B+\cos2^C &= 1-2\cos A\cos B\cos C\\ &=1-\frac{2xyz}{(x+y)(y+z)(z+x)}\\ &=\frac{(x+y+z)(xy+yz+zx)-3xyz}{(x+y)(y+z)(z+x)}\\ &=\frac{3(x+y+z)-3xyz}{(x+y)(y+z)(z+x)}\\ \end{align}

and, similarly,

$\displaystyle \sin^2 A+\sin^2B+\sin^2C = \frac{6(x+y+z)}{(x+y)(y+z)(z+x)}.$

Thus, the required inequality is equivalent to

$\displaystyle \frac{x^2y^2+y^2z^2+z^2x^2}{9}\ge\frac{x+y+z-xyz}{2(x+y+z)},$

which, in turn is equivalent to

(*)

$\displaystyle\frac{9-2xyz(x+y+z)}{9}\ge\frac{x+y+z-xyz}{2(x+y+z)}.$

Now, since $(x+y+z)^2\ge 3(xy+yz+zx),\;$ $x+y+z=3\;$ implies the existence of $t\ge 1\;$ such that $x+y+z=3\displaystyle\left(\frac{t^2+1}{2t}\right).$

Thus, we conclude that $\displaystyle xyz\le\frac{t^2-1}{2t^3}.\;$ Now (*) is equivalent to $\displaystyle\frac{3t(t^2+1)}{2(t^4+t^2+1)}\ge xyz.\;$ Hence, suffice it to prove that $\displaystyle\frac{3t(t^2+1)}{2(t^4+t^2+1)}\ge\frac{3t^2-1}{2t^3}\;$ which reduces to $(t^2-1)^2\ge 0\;$ and is, therefore, true.

Acknowledgment

Leo Giugiuc has kindly communicated to me, along with his solution, a problem by Nguyen Viet Hung.