How to prove an inequality with $\tanh(x)$

Question

I want to prove that thre existts a constant $c>0$ such that

$$\tanh(\frac{t}{2}) - \frac{t}{2} < -c\frac{t}{2}$$

for all $t\ge 1$

Answer 1

Nothe that

$$\tanh \frac{t}2 - \frac{t}2 < -c\frac{t}2 \iff f(t)=\frac{\tanh \frac{t}2 - \frac{t}2 }{\frac{t}2 }<-c$$

which is true for some c>0, indeed

• $f(t)$ is continuos for $t \ge 1$
• $\tanh x<x \implies f(t)<\frac{\frac{t}2 - \frac{t}2 }{\frac{t}2 }=0 $
• $\lim_{t\to +\infty} f(t)=-1$

thus for Extreme value theorem $f(x)$ is bounded wih a maximum $M<0$ thus exists $c>0$ such that $0<-c<M\le f(t)$.

Answer 2

Consider the function $$ f(x)=(1-c)x-\tanh x $$ over $[1/2,\infty)$. Its derivative is $f'(x)=-c+\tanh^2x$. Since, for $x\ge1/2$, $\tanh^2x\ge\tanh^2(1/2)$, the derivative is positive whenever $c<\tanh^2(1/2)$, so the function is increasing.

Since $f(1/2)=(1-c)/2-\tanh(1/2)$, we just need to set $$ 1-c>2\tanh(1/2) $$ that is, $c<1-2\tanh(1/2)$. Thus any $c$ satisfying $$ 0<c<\min(1-2\tanh(1/2),\tanh^2(1/2)) $$ will do.

Note that $$ 1-2\tanh(1/2)=1-2\frac{e-1}{e+1}=\frac{e+1-2e+2}{e+1}=\frac{3-e}{e+1}>0 $$