Proving an inequality including tanh functions For $k_2 \geq k_1 > 0$ and $d\geq 1$, I need to show that
$$ k_1\tanh(k_1d) + k_2\tanh(k_2d) - 2\sqrt{k_1k_2\tanh(k_1d)\tanh(k_2d)} \leq (k_2-k_1)\tanh((k_2-k_1)d). $$
I've started by letting $$ f(k) = \sqrt{k\tanh(kd)}$$ so that the problem can be written as $$ f(k_2) - f(k_1) \leq f(k_2 - f_1), $$
and then set $k_2 = k_1 + a$ for some $a\geq0$ so I need to show that
$$ g(k_1) = f(k_1 + a) - f(k_1) - f(a) \leq 0. $$
I've differentiated $g(k_1)$ w.r.t $k_1$ with the intent of showing it to be $\leq 0$ with a maximum value of $0$ when $k_1=0$. Numerically it seems feasible but I'm struggling to show it analytically. Explicitly, 
$$g'(k_1) = \dfrac{(k_1+a)d+\tanh((k_1+a)d)[1 - (k_1+a)d\tanh((k_1+a)d)]}{2\sqrt{(k_1+a)\tanh((k_1+a)d)}} - \dfrac{k_1d+\tanh(k_1d)[1 - k_1d\tanh(k_1d)]}{2\sqrt{k_1\tanh(k_1d)}} \leq 0.$$
Is this a good way of going about the proof? Any alternative ideas would be appreciated. 
 A: Let's see.
Writing $a$ for $k_1$
and
$b$ for $k_2$
we have,
with $a < b$,
$a\tanh(ad) + b\tanh(bd) - 2\sqrt{ab\tanh(ad)\tanh(bd)} 
\leq (b-a)\tanh((b-a)d)
$
or
$(\sqrt{b\tanh(ad)} - \sqrt{a\tanh(bd)})^2
\leq (b-a)\tanh((b-a)d)
$.
Taking the square root,
$\sqrt{b\tanh(bd)} - \sqrt{a\tanh(ad)}
\leq \sqrt{(b-a)\tanh((b-a)d)}
$.
Multiplying by $\sqrt{d}$
we get
$\sqrt{bd\tanh(ad)} - \sqrt{ad\tanh(bd)}
\leq \sqrt{(b-a)d\tanh((b-a)d)}
$.
Letting
$f(x) = \sqrt{x\tanh(x)}$
this is
$f(bd)-f(ad)
\le f((b-a)d)
$
or
$f(y)-f(x)
\le f(y-x)
$.
This isn't quite what you have,
but close.
Since
$f'(x)
=\dfrac{x\ sech(x)^2 + \tanh(x)}{2 \sqrt{x \tanh(x)}}
$,
for $x > 0$
we have
$f'(x) > 0$,
$f'(0^+) = 1$,
and
$f'(x) \to 0$
as $x \to \infty$.
Letting $y \to x$
and dividing by $y-x$,
since $f(0) = 0$,
this becomes
$f'(x) \le f'(0)$.
Looking at Wolfy,
I'm sure that this is true,
though I don't know how to prove it
right now.
Since $f'(0^+) = 1$,
we want
$x\ sech(x)^2 + \tanh(x)
\le 2 \sqrt{x \tanh(x)}
$
or
$x^2\ sech(x)^4+2x\ sech(x)\tanh(x)+\tanh^2(x)
\le 4 x \tanh(x)
$.
This seems to be true,
but no proof yet.
This is as far as I can go right now,
so I'll leave it at this.
A: Without loss of generality $d=1$, and you want to prove $f(x):=\sqrt{x\tanh x}\implies f(x+y)\le f(x)+f(y)$. We only need verify the case $x,\,y\ge 0$, since $f$ is even and $f(x+y)=f(|x+y|)\le f(|x|+|y|)$. Since $g(x):=\frac{f(x)}{x}=\sqrt{\frac{\tanh x}{x}}$ is decreasing on $\Bbb R^+$ (you may find it easier to check this property for $g^2$), $f(x+y)=(x+y)g(x+y)\le xg(x)+yg(y)=f(x)+f(y)$. 
