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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.

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    $\begingroup$ 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
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    $\begingroup$ 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
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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.

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

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