<s>[INST] 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 $a+b^2+c^3\ge \frac{1}{a}+\frac{1}{b^2}+\frac{1}{c^3}$
 [/INST] Necessarily $\color{green}{d\ge 1}$ and at least one between $a,b,c$ is less or equal than $1$. We have the equivalent inequality
$$a+b^2+c^3+d^4 - \frac{1}{a}-\frac{1}{b^2}-\frac{1}{c^3}-\frac{1}{d^4}\ge 0\qquad (*)$$
Arranging otherwise to compare the larger numbers involved with smaller ones
$$(d^4-\frac 1a)+(a-\frac {1}{d^4}) +c^3+b^2-\frac{1}{c^3}-\frac{1}{b^2}\ge 0\qquad (**)$$ 
$$d^4-\frac 1a=d^4-bcd=d(d^3-bc)$$ Since $d^2\ge bc$ and $d\ge 1$ one has $d^3-bc$ is positive; it follows  $$d^4-\frac 1a\ge 0\iff a-\frac{1}{d^4}\ge 0$$ therefore, if $b$ and $c$ are greater or equal than $1$ then the proof of $(**)$ is finished. Because of this we consider the restriction $\color{green}{0\lt a\le b\le c\lt 1\le d}$.
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Now, by the known  $A-M\ge H-M$ and the statement’s inequality, one has the system $$\begin{cases}LHS\ge \frac {16}{RHS}\ge 0\\LHS\ge RHS\ge 0\end{cases} \Rightarrow (LHS)^2\ge 16\iff LHS\ge 4$$ 
Hence we have to show $$a+b^2+c^3+d^4\ge 4\qquad (***)$$
One has $$\frac{a^4+b^4+c^4+d^4}{4}\ge \sqrt[4]{a^4b^4c^4d^4}=1$$
But (since $\color{green}{0\lt a\le b\le c\lt 1}$) $$a\ge a^4\\b^2\ge b^4\\c^3\ge c^4\\d^4=d^4$$
Consequently $$a+b^2+c^3+d^4\ge a^4+b^4+c^4+d^4\ge 4$$ as we wanted to prove.
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