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 $\pi n-\frac {\pi}{2}$.
Best Answer
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 $(\pi/2, \infty)$.