convexity of tangent function

We will show that the tangent function is convex on the interval $[0,\pi /2)$.To do this, we will use the addition formula for the tangent and the fact thata continuous real function $f$ is convex (http://planetmath.org/ConvexFunction) if and only if $f((x+y)/2)\le (f(x)+f(y))/2$.

We start with the observation that, if and , then by thearithmetic-geometric mean inequality (http://planetmath.org/ArithmeticGeometricMeansInequality),

$-2xy$$\ge -x^2-y^2$

$1-2xy+x^2{y}^2$$\ge 1-x^2-y^2+x^2{y}^2$

${(1-xy)}^2$$\ge (1-x^2)(1-y^2),$

so

$$\frac{{(1-xy)}^2}{(1-x^2)(1-y^2)}\ge 1.$$

Let $u$ and $v$ be two numbers in the interval $[0,\pi /4)$. Set $x=\tan u$ and $y=\tan v$.Then and By the addition formula, we have

$\tan (2u)$$={\displaystyle \frac{2x}{1-x^2}}$

$\tan (u+v)$$={\displaystyle \frac{x+y}{1-xy}}$

$\tan (2v)$$={\displaystyle \frac{2y}{1-y^2}}.$

Hence,

${\displaystyle \frac{1}{2}}\left(\tan (2u)+\tan (2v)\right)$$={\displaystyle \frac{x+y-x^{2}y-x{y}^2}{(1-x^2)(1-y^2)}}$

$={\displaystyle \frac{(x+y)(1-xy)}{(1-x^2)(1-y^2)}}$

$={\displaystyle \frac{x+y}{1-xy}}{\displaystyle \frac{{(1-xy)}^2}{(1-x^2)(1-y^2)}}$

$\ge {\displaystyle \frac{x+y}{1-xy}}=\tan (u+v),$

so the tangent function is convex.