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.


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.


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 $g^2$), $f(x+y)=(x+y)g(x+y)\le xg(x)+yg(y)=f(x)+f(y)$.

  • $\begingroup$ Thank you. This is lovely! $\endgroup$ Nov 30 '18 at 10:57

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