Prove that $e^x\geq(x+1)\sqrt {(x+1)^{\sin x}}\big{|}\cos x\big{|}$ for all $x> -1$ ( without working with derivatives )

Question

Prove that $$e^x\geq(x+1)\sqrt{(x+1)^{\sin x}}\big{|}\cos x\big{|}$$ for all $x>-1$ .

---

Some thoughts

I am trying to prove this inequality with high school techniques. I tried to use $e^x\geq x+1$. So it is sufficient to prove that $$x+1\geq(x+1)\sqrt{(x+1)^{\sin x}}\big{|}\cos x\big{|}$$

or $$\sqrt{(x+1)^{\sin x}}\big{|}\cos x\big{|}\leq 1$$

or $$(x+1)^{\sin x}\cos^2x\leq 1.$$

which is a wrong result as pointed out in the comments .

I'm stuck here, unable to make any further progress.

Case $x=\pi n-\frac {\pi}{2}$

$e^x$ is always positive. $\cos x=0$ gives $x=\pi n-\frac {\pi}{2}, n\geq 1$ and $n$ is an integer. Hence, the inequality is true for all $\pi n-\frac {\pi}{2}$.

Answer 1

Partial Answer: Note however the techniques used in this most likely won't be helpful for a full solution. I am posting what I could obtain from my try.

We'll prove this for $x \geq 1.5$. The expression given is

$$ e^x \geq (x+1)\sqrt{(x+1)^{\sin(x)}}|\cos(x)| = (x+1)^{1 + \frac{\sin(x)}{2}} |\cos(x)|$$

Taking $\log$

$$ x \geq \log\left((x+1)^{1 + \frac{\sin(x)}{2}} |\cos(x)|\right) = \left(1 + \frac{\sin(x)}{2}\right)(\log(x+1)) + \log|\cos(x)| $$

However, since $0\leq|\cos(x)|\leq 1$, we have $\log|\cos(x)| \leq 0, x \in R$, the given would hold true even if the following holds true.

$$ x \geq \left(1 + \frac{\sin(x)}{2}\right)(\log(x+1)) $$

Now consider $$ \frac{-1}{2} \leq \frac{\sin(x)}{2} \leq \frac{1}{2} $$ $$ \frac{1}{2} \leq \frac{\sin(x)}{2} + 1 \leq \frac{3}{2} $$ $$ \frac{\log(x+1)}{2} \leq \log(x+1)\left( \frac{\sin(x)}{2} + 1 \right) \leq \frac{3\log(x+1)}{2} $$

For any given $x$ the upper bound attained of the expression

$$ \log\left((x+1)^{1 + \frac{\sin(x)}{2}} |\cos(x)|\right) $$

is thus given by $ \frac{3\log(x+1)}{2} $

Thus, the given is true wherever $x \geq \frac{3\log(x+1)}{2} $, which certainly holds for all $x \geq 1.5$. More precisely it holds at around $x \geq 1.15$ but since the given is not what is asked, I preferred to keep a check-able by hand number.

Answer 2

Square both sides of the inequality to get the equivalent $$ e^{2x}\geq (1+x)^{2+\sin(x)}\cos^2(x) $$ and consider two cases.

Case 1

Let us first consider the case $-1 < x \leq \pi/2$, then $ \cos^2(x)\leq e^{-x^2}. $ (see How to see $\cos x \leq \exp(-x^2/2)$ on $x \in [0,\pi/2]$?). Thus it is sufficient to prove the inequality: $$ e^{2x}\geq (1+x)^{2+\sin(x)}e^{-x^2}, $$ or $$ e^{2x+x^2} \geq (1+x)^{2+\sin(x)}. $$ But $$(1+x)^{2+\sin(x)}\leq e^{2x+x\sin(x)}$$ using the inequality $(1+x)^r\leq e^{rx}$, since $2+\sin(x) > 0$. Thus it remains to prove $$ e^{2x+x^2} \geq e^{2x + x\sin(x)}\iff e^{x^2} \geq e^{x\sin(x)} \iff x^2\geq x\sin(x), $$ which is true.

Case 2 For $\pi/2 < x$ we can bound $\cos^2(x)\leq 1$, so it is sufficient to prove the inequality $$ e^{2x}\geq (1+x)^{2+\sin(x)} $$ But note that $(1+x)^{\sin(x)}\leq 1+x$, so it is sufficient to prove $$ e^{2x}\geq (1+x)^3, $$ which is true for $x > \pi/2$. Indeed note that $$ e^{2x/3 - \pi/3}\geq 1 + 2x/3 - \pi/3 \implies e^{2x/3}\geq e^{\pi/3}(1+ 2x/3 - \pi/3)=h(x). $$ But $h(\pi/2) = e^{\pi/3} > 1 + \pi/2$ and $h$ is a line with slope $2e^{\pi/3}/3 > 1$, implying that $h(x)\geq 1+x$ for $x \geq \pi/2$.

Alternatively, we can consider the derivative of the function $f(x) = e^{2x/3} - 1 - x$. Indeed, $f'(x) = \frac{2}{3}e^{\frac{2}{3}x} - 1$ is clearly an increasing function, and $f'(\pi/2) \approx 0.9 > 0$, implying that $f$ is increasing on the interval $(\pi/2,\infty)$. Also $f(\pi/2) \approx 0.279$, proving that $f$ is positive on $(\pi/2, \infty)$.