Inequalities for elementary transcendental functions

For $0 \leq \theta \leq \frac{\pi }{2}$, show that

By using Part 1, or by any other method, show that if $\lambda < 1$, then

$\frac{x^{2}}{n^{2}} e^{-x}<e^{-x}-\left(1-\frac{x}{n}\right)^{n} \leqslant \frac{x^{2}}{n} e^{-x} \quad$ for $x \in [0, n], n>1$;

$\frac{x^{2}}{4 n} e^{-x} \leqslant \left(1+\frac{x}{n}\right)^{-n}-e^{-x} \leqslant \frac{x^{2}}{n} e^{-x}$ for $x \in [0, \sqrt{n}]$.

$(1+x) \ln^{2}(1+x)<x^{2} \quad$ for $x>-1, x \neq 0$;

$\frac{x}{1+\frac{2}{\pi } x}<\operatorname {arctg} x \quad$ for $x>0$;

$x^{2}<\ln \left(1+\operatorname {tg}^{2} x\right)<\sin x \operatorname {tg} x \quad$ for $x \in \left(0, \frac{\pi }{2}\right)$;

$3 x-x^{3}<2 \sin \left(\frac{\pi }{2} x\right) \quad$ for $x \in (0,1)$;

$x^{3}<\sin^{2} x \operatorname {tg} x \quad$ for $x \in \left(0, \frac{\pi }{2}\right)$.

Does the inequality $x^{3+\varepsilon }<\sin^{2+\varepsilon } x \operatorname {tg} x$ hold on $(0, \frac{\pi }{2})$ for some fixed $\varepsilon >0$?

$\left(1-x^{2}\right)\left(1-\frac{1}{4} x^{2}\right) \leqslant \cos \frac{\pi }{2} x \leqslant \left(1-x^{2}\right)\left(1-\frac{1}{5} x^{2}\right)$;

$\left(1-x^{2}\right)\left(1-\frac{2}{3} x^{2}\right) \leqslant \frac{\sin \pi x}{\pi x} \leqslant \left(1-x^{2}\right)\left(1-\frac{1}{2} x^{2}\right)$.

$\left(e^{x}-1\right) \ln (1+x)>x^{2} \quad$ for $x>0$;

$\operatorname {tg} x \operatorname {arctg} x>x^{2} \quad$ for $x \in \left(0, \frac{\pi }{2}\right)$;

$\left((1+x)^{p}-1\right)\left((1+x)^{1 / p}-1\right)>x^{2}$ for $x>0, p>0, p \neq 1$.

What happens for $p<0$?

$\left((1-x)^{p}-1\right)\left(1-(1+x)^{1 / p}\right)>x^{2}$ for $x \in (0,1), p<-1$.

What happens for $p \in (-1,0)$?

$\left|\left(\frac{\sin x}{x}\right)^{(n)}\right| \leqslant \frac{1}{n+1}$;

$\left|\left(\frac{1-\cos x}{x^{2}}\right)^{(n)}\right| \leqslant \frac{1}{(n+1)(n+2)}$;

$\left|\left(\frac{1}{1+x^{2}}\right)^{(n)}\right| \leqslant \frac{n!}{\left(1+x^{2}\right)^{\frac{n+1}{2}}}$.

$2 \leqslant (1+x)^{p}+(1-x)^{p} \leqslant 2^{p} \quad (p \geqslant 1)$;

$2\left(1+x^{p}\right) \leqslant (1+x)^{p}+(1-x)^{p} \leqslant 2^{p-1}\left(1+x^{p}\right) \quad (p \geqslant 2)$;

$\quad (1+x)^{p}+(1-x)^{p} \leqslant 2\left(1+(p-1) x^{2}\right)^{\frac{p}{2}} \quad (p \geqslant 2)$;

$\quad (1+x)^{p}+(1-x)^{p} \leqslant 2\left(1+x^{\frac{p}{p-1}}\right)^{p-1} \quad (p \geqslant 2)$.