$\displaystyle F(t)=\frac{t}{1+t}$
Prove that $F(s+t)\le F(s)+F(t)$ for $s, t\ge 0$.
I really don't like the brute-force method of summing 3 unlike fractions. Is there an easier way of doing this?
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$F$ is concave and $F(0) \ge 0 $. That implies that $F$ is subadditive on $\Bbb R_{+}$ – Martin R Sep 05 '18 at 17:33
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Essentially a duplicate of Prove that $\dfrac{|x+y|}{1+|x+y|}\leq\dfrac{|x|}{1+|x|}+\dfrac{|y|}{1+|y|}$ for any $x,y$. – Martin R Sep 05 '18 at 17:35
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Since $s,t\geq 0$, $$ F(s+t) = \frac{s+t}{1+s+t} = \frac{s}{1+s+t} + \frac{t}{1+s+t} \leq \frac{s}{1+s} + \frac{t}{1+t} = F(s) + F(t). $$
Rigel
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Hint Let $s \geq 0$ be arbitrary but fixed and define $$ H(t) = F(s+t)-F(s)-F(t). $$ Show that $H'(t)\leq 0,$ so that $H$ is non-increasing, and use that, consequently, $$ H(t) \leq H(0) = 0, $$ for all $t \geq 0.$
MSDG
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Yes. Since $$ F'(t) = \left(1 - \frac 1{1+t}\right)' = \frac 1{(1+t)^2}, F''(t) = - \frac 2{(1+t)^3} < 0, $$ $F'(t)$ is decreasing. WLOG, suppose $0 \leqslant s \leqslant t$, then by MVT, $$ F(s+t) - F(t) = s F'(a) \leqslant sF'(b) = F(s) - F(0) = F(s), $$ where $0 < b < s \leqslant t < a < s+t$.
xbh
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