# Prove that $e^x \cos (\sqrt{x^2+1}) \leq 1$

I would like to prove that :

$$\forall x \in [0,1], e^x\cos(\sqrt{x^2+1}) \leq 1$$

When plotting the graph this inequality is not sharp at all and we even have : $$\forall x \in [0,1], e^x \cos(\sqrt{x^2+1}) \leq 0.8$$

I tried several things such has :

Calculating the derivative and try to apply the mean value theorem to get an upper bound, but the derivative is hard to manipulate and it doesn’t seem I am getting something.

Moreover trying something on convexity but once again this is difficult due to the horrible looking of the derivative.

I am very interested in sharper upper bound, even if I can’t manage to prove the inequality for $$1$$...

• “sharp”? ${}{}{}$ – gen-z ready to perish Nov 7 '18 at 22:41
• @ChaseRyanTaylor It means that the inequality achieves equality somewhere, creating (somewhat incorrectly) the impression that the inequality cannot be "improved". – Theo Bendit Nov 7 '18 at 22:45
• The maximum value is about $0.722$ – Yuriy S Nov 8 '18 at 1:07

## 4 Answers

We can rewrite the inequality to be proved as

$$e^{-x}-\cos(\sqrt{1+x^2})\ge0$$

for $$0\le x\le1$$.

By truncating Taylor series and using a crude estimate to keep things quadratic, we have

$$e^{-x}\ge1-x+{x^2\over2}-{x^3\over6}=1-x+{x^2\over2}\left(1-{x\over3}\right)\ge1-x+{x^2\over2}\left(1-{1\over3}\right)=1-x+{x^2\over3}$$

and (since $$\sqrt{1+x^2}\le\sqrt2$$ for $$0\le x\le1$$ and the alternating terms in the Taylor series for $$\cos\sqrt2$$ are monotonically decreasing)

$$\cos(\sqrt{1+x^2})\le1-{1+x^2\over2}+{(1+x^2)^2\over24}={13-10x^2+x^4\over24}\le{13-10x^2+1\over24}={7-5x^2\over12}$$

It follows that

$$e^{-x}-\cos(\sqrt{1+x^2})\ge1-x+{x^2\over3}-{7-5x^2\over12}={9x^2-12x+5\over12}={(3x-2)^2+1\over12}\gt0$$

Remark: Part of what makes this work is that the inequality, as the OP observed, is not sharp, so there is a fair amount of room for crude estimates to keep things simple.

$$e^x$$ is increasing on $$[0,1]$$, and $$\cos\sqrt{x^2+1}$$ is decreasing. Therefore:

• if $$x\in[0,\frac12]$$, then $$e^x\le e^\frac12$$ and $$\cos\sqrt{x^2+1}\le\cos 1$$, so $$e^x\cos\sqrt{x^2+1}\le e^\frac12\cos 1 < 0.9$$

• if $$x\in[\frac12,\frac34]$$, then $$e^x\le e^\frac34$$ and $$\cos\sqrt{x^2+1}\le\cos\sqrt\frac54$$, so $$e^x\cos\sqrt{x^2+1}\le e^\frac34\cos \sqrt\frac54 < 0.93$$

• if $$x\in[\frac34,1]$$, then $$e^x\le e$$ and $$\cos\sqrt{x^2+1}\le\cos\sqrt{(\frac34)^2+1}=\cos\frac54$$, so $$e^x\cos\sqrt{x^2+1}\le e\cos\frac54 < 0.86$$

This proves that your function is less than $$0.93$$ on $$[0,1]$$. And you can get as close to the true maximum as you like using this method, by dividing $$[0,1]$$ into ever more intervals.

• That’s clever. I didn’t think about dividing $[0,1]$ into smaller part. I’ve seen that $e^1 \cdot \cos(1) >1$ unfortunately... The only problem with this kind of proofs is that we obvisouly need some calculator to continue the proof. – Interesting problems Nov 8 '18 at 7:42

We can reduce to a polynomial inequality using that

• $$e^x\le 1+x+x^2 \quad 0\le x \le 1\quad$$ proved here

• $$\cos x< 1-\frac25 x^2 \quad 1\le x \le \sqrt 2 \quad$$ to be proved

therefore

$$e^x \cos (\sqrt{x^2+1})< (1+x+x^2)\left(1-\frac25 (x^2+1)\right)\stackrel{?}<0.90<1$$

which can be easily checked by derivatives and IVT.

Let define

• $$f(x)= (1+x+x^2)\left(1-\frac25 (x^2+1)\right)-\frac9{10}$$
• $$f(0)<0$$
• $$f(1)<0$$

and

• $$g(x)=f'(x)= -\frac85x^3-\frac65x^2+\frac25 x+3$$
• $$g(0)>0$$
• $$g(1)<0$$

and

• $$h(x)=g'(x)=-\frac{24}5x^2-\frac{12}5x+\frac25$$
• $$h(0)>0$$
• $$h(1)<0$$
• $$h'(x)=-\frac{48}5x-\frac{12}5<0$$

then

• $$h(x)$$ is strictly decreasing and has exactly one root on that interval
• $$g(x)$$ has a local maximum at that point and exactly one root on that interval that is $$x_0 \approx 0.622$$
• $$f(x_0)<0$$ is a maximum and therefore $$f(x)$$ is always negative on the interval that is

$$f(x)= (1+x+x^2)\left(1-\frac25 (x^2+1)\right)-\frac9{10}<0 \\\implies e^x \cos (\sqrt{x^2+1})< (1+x+x^2)\left(1-\frac25 (x^2+1)\right)<\frac9{10}<1$$

To prove $$\cos x< 1-\frac25 x^2 \quad 1\le x \le \sqrt 2 \quad$$ let consider

• $$f(x)=\cos x-1+\frac25 x^2$$
• $$f(1)<0$$
• $$f(\sqrt 2)<0$$

and

• $$g(x)=f'(x)=-\sin x+\frac45 x$$
• $$g(1)<0$$
• $$g(\sqrt 2)>0$$
• $$g'(x)=-\cos x+\frac45>0$$

therefore

• $$g(x)$$ is strictly increasing and has exactly one root on that interval
• $$f(x)$$ has a local minimum and is negative on that interval that is

$$f(x)=\cos x-1+\frac25 x^2< 0 \implies \cos x< 1-\frac25 x^2 \quad 1\le x \le \sqrt 2$$

• Thank you for the links about the two inequality you use. It can be helpful to have them in mind in the future :) – Interesting problems Nov 8 '18 at 7:40
• You are welcome, I'm also expanding the answer with the full solution. – gimusi Nov 8 '18 at 7:41
• Great explanation thank you ! – Interesting problems Nov 8 '18 at 8:21
• That’s a simple but powerful way to prove inequalities by a chain of derivatives. You are welcome! Bye – gimusi Nov 8 '18 at 8:22

Not a proof, but some considerations which could be developed into a proof.

Let's change the variable to $$x=\sinh t$$

Now the function becomes:

$$f(t)=e^{\sinh t} \cos (\cosh t), \quad t \in [0, \ln (1+\sqrt{2})]$$

Note that $$\ln (1+\sqrt{2}) < \ln 2.5 < 1$$

This function is easy to differentiate:

$$f'(t)=e^{\sinh t} (\cosh t \cos (\cosh t) - \sinh t \sin (\cosh t))$$

The extrema obey the equation:

$$\tanh t \tan (\cosh t)=1$$

Remember that: $$\cosh t \geq 1 > \frac{\pi}{4}$$

Considering the ranges of both tangents, it's not hard to guess that there should only be a single solution in our interval.

To approximate the value, we can use bisection method:

$$g(t)=\tanh t \tan (\cosh t)-1$$

$$g(0) = -1 <0 \\ g(\ln (1+\sqrt{2})=\frac{\tan \sqrt{2}}{\sqrt{2}}-1 >0$$

Because for the range $$a \in [0, \pi/2]$$ we have $$\tan a > a$$.

Now we check the middle of the interval:

$$g(1/2)=\tanh \frac{1}{2} \tan \left(\cosh \frac{1}{2} \right)-1=\frac{e-1}{e+1} \tan \frac{e+1}{2 \sqrt{e}}-1=-0.0264\dots$$

We had to use calculators here, but with some care and knowledge of the first few digits of $$e$$ we can at least prove the the value is negative.

Which means that the extremum point $$t_0 \in (1/2,1)$$. However, we don't need to check the middle of this new interval, because $$g(1/2)$$ is much smaller than $$g(0)$$ and it's quite clear that $$t_0$$ is close to $$1/2$$. Moreover, we know* that the function $$f(t)$$ is growing on the interval $$(0,1/2)$$. So we can write:

$$f(t)< f(1/2) = e^{\sinh \frac{1}{2}} \cos \left( \cosh \frac{1}{2} \right)=0.722 \dots$$

This is again done with a calculator, however we can fix that by picking some number close to $$1/2$$ which gives us a more simple expression.

$$^*$$ It remains to be shown that $$t_0$$ is a maximum, but it can be done in principle by checking a few suitable values of $$f(t)$$ around it.