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

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.


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

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


$e^x$ 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.

Tags:

Inequality

Calculus

Real Analysis

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