Prove that for all real numbers $x\geq 0$ and $y \geq 0$ the following inequalities are true: $$\arctan(x+y)\leq\arctan(x)+\arctan(y) \qquad \tanh(x+y)\leq \tanh(x)+\tanh(y)$$

I tried to use both the addition formula and the geometrical approach, but I couldn't find anything leading me to the solution. Any hint/help is appreciated! Thanks in advance.

  • 2
    $\begingroup$ Let $X=\arctan(x), Y=\arctan(y), Z=\arctan(x+y)$. Then, we have, $$x+y=\tan(X)+\tan(Y)=(1+\tan(X)\tan(Y))\tan(X+Y)\geq \tan(X+Y)\\ \implies X+Y\leq\arctan(x+y)~\forall~x,y\geq 0$$ Here, we use the result that $\arctan$ is monotonically increasing on $[0,\infty)$. Can you try the other one now similarly? $\endgroup$ – Prasun Biswas Aug 21 '17 at 10:47
  • $\begingroup$ Sure thing! Thank you so much, I'll try it out. I'll be glad to give you a few points if you put this as an answer. $\endgroup$ – Alberto Andrenucci Aug 21 '17 at 10:50
  • 1
    $\begingroup$ Typo correction: $$x+y=\tan(X)+\tan(Y)=(1-\tan(X)\tan(Y))\tan(X+Y)\leq \tan(X+Y)\\ \implies \arctan(x+y)\leq X+Y~\forall~x,y\geq 0$$ $\endgroup$ – Prasun Biswas Aug 21 '17 at 10:55

For fixed $y \ge 0$ consider the function $$ h(x) = \arctan(x + y) - \arctan(x) - \arctan(y) $$ Then $h(0) = 0$ and $$ h'(x) = \frac{1}{1+(x+y)^2} - \frac{1}{1+x^2} \le 0 $$ for $x \ge 0$.

The same approach can be used to prove the second inequality.

More generally, any concave function $f: [0, \infty) \to \Bbb R$ with $f(0) \ge 0$ is subadditive.

  • $\begingroup$ Simple and easy to understand, thank you for your answer! $\endgroup$ – Alberto Andrenucci Aug 21 '17 at 14:45

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.