# 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.

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

• Thank you. This is lovely! – RH_data_maths Nov 30 '18 at 10:57

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.