# Dan Sitaru's Inequality with Tangents II

### Solution 1

Let be $x=\cot A\cot B;\,$ $y=\cot B\cot C;\,$ $z=\cot C\cot A.\,$ It follows that $x+y+z=1.\,$ Further,

\displaystyle \begin{align} 45+\sum_{cycl}\tan A \tan B &= 45+\frac{1}{x}+\frac{1}{y}+\frac{1}{z}\\ &=\frac{xy+yz+zx+45xyz}{xyz}\\ &=\frac{\Bigr(1-(x+y+z)+xy+yz+zx-xyz\Bigr)+46xyz}{xyz}\\ &=\frac{(1-x)(1-y)(1-z)+46xyz}{xyz}\\ &\overbrace{\leq}^{AM-GM} \frac{\frac{1}{27}(1-x+1-y+1-z)^3+\frac{46}{27}(x+y+z)}{xyz}\\ &=\frac{\frac{1}{27}\cdot 8+\frac{46}{27}}{xyz}=\frac{2}{xyz}=\frac{2}{\displaystyle \prod_{cycl} \cot^2 A}\\ &=2\prod_{cycl} \tan^2 A. \end{align}

### Solution 2

Set $x=\tan A\gt 0,\,$ $y=\tan B\gt 0, z=\tan C\gt 0.\,$ Recollect that $x+y+z=xyz.\,$ The required inequality is equivalent to

$xy+yz+zx+45\le 2(x+y+z)^2,$

which, in turn, is equivalent to

$2(x^2+y^2+z^2) +3(xy+yz+zx)\ge 45.$

Since, by the rearrangement inequality, $x^2+y^2+z^2\ge xy+yz+zx,\,$ suffice it to prove that $5(xy+yz+zx)\ge 45,\,$ i.e., that $xy+yz+zx\ge 9,\,$ which follows from $\displaystyle \frac{1}{xy}+\frac{1}{yz}+\frac{1}{zx}=1,\,$ by the Cauchy-Schwarz inequality.

### Solution 3

\displaystyle \begin{align} 2\left(\prod_{cycl}\tan A\right)^2 &=2\left(\sum_{cycl}\tan A\right)^2\\ &\ge 6\sum_{cycl}\tan A\tan B, \end{align}

because $(x+y+z)^2\ge 3(xy+yz+zx).\,$ To continue, by the AM-GM inequality,

\displaystyle \begin{align} \sum_{cycl}\tan A\tan B &\ge 3\sqrt[3]{\tan^2A\tan^2B\tan^2C}\\ &=3\sqrt[3]{\left(\sum_{cycl}\tan A\right)^2}\\ &\ge 3\sqrt[3]{\left(3\sqrt{3}\right)^2}=9, \end{align}

because $\tan (x),\,$ being a convex function on $\displaystyle \left(0,\frac{\pi}{2}\right),\,$ $\displaystyle \sum_{cycl}\tan A\ge 3\tan\left(\frac{A+B+C}{3}\right).\,$ Finally.

\displaystyle \begin{align} 2\left(\prod_{cycl}\tan A\right)^2 &\ge 6\sum_{cycl}\tan A\tan B\\ &=\sum_{cycl}\tan A\tan B+5\sum_{cycl}\tan A\tan B\\ &\ge\sum_{cycl}\tan A\tan B+5\cdot 9. \end{align}

### Acknowledgment

The problem from the Romanian Mathematical Magazine has been kindly posted by Dan Sitaru at the CutTheKnotMath facebook page. Dan also has communicated his solution (Solution 1) in a latex file. Solution 2 is by Kevin Soto Palacios; Solution 3 is by Soumava Chakraborty.