# An Inequality for the Tangent to the Incircle

### Solution 1

The triangles $ABC$ and $ADE$ are similar, let $k$ be the coefficient of similarity, so that, e.g., $DE=ka$ and $[\Delta ADE]=k^2[\Delta ABC].$ Then, since the incircle of $\Delta ABC$ is an $A\text{-excircle}$ of $\Delta ADE,$ then, with $r$ being the inradius and $s$ the semiperimeter of $\Delta ABC,$ we have

$\displaystyle [\Delta ADE]=rk(s-a).$

Since also $[\Delta ABC]=sr,$ $\displaystyle k=\frac{s-a}{s}$ and, subsequently, $\displaystyle DE=\frac{a(s-a)}{s}.$

Introduce $x=s-a,$ $y=s-b,$ $z=s-c.$ The required inequality rewrites as

$\displaystyle \frac{x(x+y)}{x+y+z}\le\frac{x+y+z}{4},$

which is equivalent to $(x-y-z)^2\ge 0.$ Equality is attained only if $AB+AC=3\cdot BC.$

### Solution 2

WLOG, let the in-radius be unity. Let $\angle ABC = 2\alpha$ and $\angle ACB = 2\beta$.

\displaystyle \begin{align} BC&=\cot \alpha + \cot \beta \\ AB&=\cot \alpha + \cot (90^{\circ}-\alpha-\beta)=\cot \alpha + \tan (\alpha+\beta) \\ AC&=\cot \beta + \cot (90^{\circ}-\alpha-\beta) =\cot \beta + \tan (\alpha+\beta) \\ DE&=\cot (90^{\circ}-\alpha) + \cot (90^{\circ}-\beta)=\tan \alpha + \tan \beta. \end{align}

\displaystyle \begin{align} AB+BC+CE-8\cdot DE &=2\left[\cot \alpha + \cot \beta + \tan (\alpha+\beta)\right]-8\left[\tan \alpha + \tan \beta\right] \\ &=2\left[\frac{1}{\tan \alpha} + \frac{1}{\tan \beta} + \frac{\tan\alpha+ \tan \beta}{1-\tan \alpha \tan \beta}\right] -8\left[\tan \alpha + \tan \beta\right] \\ &=2\left[\tan \alpha + \tan \beta\right]\left[\frac{1}{\tan \alpha\tan \beta}+\frac{1}{1-\tan \alpha \tan \beta}-4\right] \\ &=\frac{2\tan(\alpha+\beta)}{\tan \alpha \tan \beta}\left(1-4\tan\alpha\tan\beta+4\tan^2\alpha\tan^2\beta\right) \\ &=\frac{2\tan(\alpha+\beta)}{\tan \alpha \tan\beta}\left(1-2\tan\alpha\tan\beta\right)^2\geq 0. \end{align}

Note, $\alpha$, $\beta$, and $(\alpha+\beta)$ are acute and their tangents are positive.

### Acknowledgment

This math folklore problem was kindly posted at the CutTheKnotMath facebook page by Leo Giugiuc. Solution 1 is Leo's. Solution 2 is by Amit Itagi.