# convexity of tangent function

We start with the observation that, if $0\leq x<1$ and $0\leq y<1$, then by the arithmetic-geometric mean inequality (http://planetmath.org/ArithmeticGeometricMeansInequality),

 $\displaystyle-2xy$ $\displaystyle\geq-x^{2}-y^{2}$ $\displaystyle 1-2xy+x^{2}y^{2}$ $\displaystyle\geq 1-x^{2}-y^{2}+x^{2}y^{2}$ $\displaystyle(1-xy)^{2}$ $\displaystyle\geq(1-x^{2})(1-y^{2}),$

so

 ${(1-xy)^{2}\over(1-x^{2})(1-y^{2})}\geq 1.$

Let $u$ and $v$ be two numbers in the interval $[0,\pi/4)$. Set $x=\tan u$ and $y=\tan v$. Then $0\leq x<1$ and $0\leq y<1.$ By the addition formula, we have

 $\displaystyle\tan(2u)$ $\displaystyle={2x\over 1-x^{2}}$ $\displaystyle\tan(u+v)$ $\displaystyle={x+y\over 1-xy}$ $\displaystyle\tan(2v)$ $\displaystyle={2y\over 1-y^{2}}.$

Hence,

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

so the tangent function is convex.

Title convexity of tangent function ConvexityOfTangentFunction 2013-03-22 17:00:12 2013-03-22 17:00:12 rspuzio (6075) rspuzio (6075) 14 rspuzio (6075) Result msc 26A09