This inequality $a+b^2+c^3+d^4\ge \frac{1}{a}+\frac{1}{b^2}+\frac{1}{c^3}+\frac{1}{d^4}$ 
let $0<a\le b\le c\le d$, and such $abcd=1$,show that
  $$a+b^2+c^3+d^4\ge \dfrac{1}{a}+\dfrac{1}{b^2}+\dfrac{1}{c^3}+\dfrac{1}{d^4}$$

it seems harder than This inequality $a+b^2+c^3\ge \frac{1}{a}+\frac{1}{b^2}+\frac{1}{c^3}$

it is trivial $d \ge 1$
there are two cases: $c\ge 1$ , or $c\le 1$
$c\ge 1$
let $x=\dfrac{1}{a},y=\dfrac{1}{b} \implies x\ge y\ge 1 ,xy=cd$
let $f(x)=x-\dfrac{1}{x}$, it is trivial $f(x)$ is mono increasing 
function
now we need to prove :
$f(c^3)+f(d^4) \ge f(x)+f(y^2)$
1)if $ c^3< x \cap c^3<y^2 \implies yc^3 <xy=cd \iff d> c^2y>y \cap x^2c^3 <x^2y^2=c^2d^2 \iff d^2>cx^2 \iff d>x $
$f(d^4)=f(d^2)(d^2+\dfrac{1}{d^2}) \ge 2 f(d^2) > f(x^2)+f(y^2)>f(x) +f(y^2)$
$f(c^3) \ge 0$  and this part is true.
2)if $x>y^2 \cap y^2 <c^3<x \implies c^3d^3=x^3y^3 <xd^3 \iff d^3>y^3x^2 \iff d^4>x^2 $
$f(c^3)+f(d^4) >f(y^2)+f(x^2)>f(x)+f(y^2)$
3)if $x<y^2 \cap x<c^3<y^2 \implies x^3y^3=c^3d^3 <y^2d^3\iff d^3>x^2y \ge y^3 \iff d>y$
$f(c^3)>f(x), f(d^4)>f(y^4)>f(y^2)$ it is true also.
4)it is trivial true when $c^3>x,c^3>y^2$
when $c \le 1 ,z=\dfrac{1}{c} \implies x \ge y \ge z \ge 1, d=xyz \ge 1$
we need to prove : $ f(d^4) \ge f(x)+f(y^2)+f(z^3)$
$f(z^4) \ge f(z^3),\iff f((xyz)^4) \ge f(x)+f(y^2)+f(z^4) \\ \iff (xy)^4((xy)^4-1)v^2+x^2y^2(x^2+xy^2-x^3y^2-x^2y^4)v+(xy)^4-1 \ge 0 ,v=z^4$
$(xy)^4((xy)^4-1)>0,$ if $\Delta \le 0 $ then it is true.
$\Delta \le 0 \\ \iff (x^2+xy^2-x^3y^2-x^2y^4)^2 \le 4((xy)^4-1)^2 \\ \iff x^2y^4+x^3y^2-x^2-xy^2 \le 2((xy)^4-1)  \\ \iff (2x^4-x^2)y^4-(x^3-x)y^2+x^2-2 \ge 0 $
$g(t=y^2)=(2x^4-x^2)t^2-(x^3-x)t+x^2-2$
note $\dfrac{(x^3-x)}{2(2x^4-x^2)}<1 \iff 4x^3 -x^2 -2x+1 \ge 0 \iff 2x^2(2x-1)+(x-1)^2 \ge 0$ is true,
$g_{min}=g(1)=2x^4-x^3+x-2=(x^2-1)(2x^2-x+2) \ge 0$
