Vrftsjtryk

How to prove the following inequality $forall tinmathbb{R}$:$$frac{2}{sqrt{2pi}}(t+frac{1}{t})e^{-t^2/2} leq frac{4}{t^2}$$

HINT

We have that

$$frac{2}{sqrt{2pi}}left(t+frac{1}{t}right)e^{-t^2/2} leq frac{4}{t^2} iff e^{t^2/2} geq frac{1}{2sqrt{2pi}}(t^3+t)$$

and then consider $f(t)=e^{t^2/2} -frac{1}{2sqrt{2pi}}(t^3+t)ge 0$, indeed we have

• $f'(t)=te^{t^2/2}-frac{3t^2+1}{2sqrt{2pi}}$


• $f''(t)=t^2e^{t^2/2}+e^{t^2/2}-frac{3t}{sqrt{2pi}}$

and we can show that $f'(t)$ has exactly one root at $x_0$ which is a positive minimum poinf for $f(t)$.

As an alternative, we can also use that $e^xge 1+x+frac12 x^2$ and consider the simpler

$$g(t)=1+frac12t^2+frac18t^4 -frac{1}{2sqrt{2pi}}(t^3+t)ge 0$$

(Since there are a powers of $2$)

Taking $log_2$ on both sides

$1-dfrac12-log_2 sqrt{pi}+log_2left(t+dfrac1tright)-dfrac{t^2}{2}leq2-2log_2t $ $implies dfrac{t^2}{2}-log_2left(t^3+tright)+log_2 sqrt{pi}+dfrac32geq0$

We show $f(t)=dfrac{t^2}{2}-log_2left(t^3+tright)+log_2 sqrt{pi}+dfrac32geq0 $. $f'(t)=t-dfrac{3t^2+1}{t^3+t}=0implies t^4-2t^2-1=0implies t^2-1=pmsqrt{2}implies t=pmsqrt{1pmsqrt{2}}.$

(We do not consider $pmsqrt{1-sqrt{2}}$ as t does not belong to $mathbb{C}$ and $-sqrt{1+sqrt{2}}:$ since $log_2$ cannot take negative values).

$f''(t)|_{t=sqrt{1+sqrt{2}}}=dfrac{t^6+5t^4+t^2+1}{(t^3+t)^2}>0$ and $f(t)|_{t=sqrt{1+sqrt{2}}}>0$. So the root is a point of positive minimum.
Hence the inequality holds.

We have $2sqrt{2pi} > 2sqrt{6.25} = 5$, so it is enough to prove
$5e^{t^2/2} > t + t^3$ (for all real $t$ - including $0$, for
which the original inequality doesn't make sense). This is obvious
if $t leqslant 0$, because the LHS is $> 0$ and the RHS is
$leqslant 0$. If $0 < t leqslant sqrt{2}$, then the LHS is $> 5$
and the RHS is $leqslant 3sqrt{2} < 5$, so the inequality is true. If
$sqrt{2} < t leqslant 2$, then $e^{t^2/2} > e > 2$, therefore
$5e^{t^2/2} > 10 geqslant t + t^3$, so the inequality is true
again. If $t > 2$, then from the series for $e^x$, $e^{t^2/2} > 1 + t^2/2 + t^4/8$, therefore $5e^{t^2/2} > t^2 + t^4/2 > t + t^3$. So the inequality holds
in all cases (and equality never holds).