<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] Suppose that it is true for $n=3$:

$$
\forall 0 \le a \le b \le c \wedge abc = 1 :
a - \frac{1}{a}
  + b^2 - \frac{1}{b^2}
  + c^3 - \frac{1}{c^3} \ge 0.
$$

Let $d \ge c \ge 1$, then it is clear that

$$
\forall 0 \le a \le b \le c \le d \wedge abcd = d :
a - \frac{1}{a}
  + b^2 - \frac{1}{b^2}
  + c^3 - \frac{1}{c^3}
  + d^4 - \frac{1}{d^4} \ge 0.
$$

Let us now define
$$
a' = \frac{a}{\mu^{24}},
  b' = \frac{b}{\mu^{12}},
  c' = \frac{c}{\mu^{8}},
  d' = \frac{d}{\mu^{6}},
$$
using
$$
24 = 4!, 12 = 4!/2, 8 = 4!/3, 6 = 4!/4.
$$
We get
$$
a' b' c' d' =
\frac{a}{\mu^{24}} \frac{b}{\mu^{12}} \frac{c}{\mu^{8}} \frac{d}{\mu^{6}} =
\frac{abcd}{\mu^{50}} = \frac{d}{\mu^{50}},
$$
so
$$
a' b' c' d' =1,
$$
if
$$
\mu = \sqrt[50]{d}.
$$
As $d \ge 1$, it is clear that $\mu \ge 1$.
The inequality can be written as
$$
\frac{a}{\mu^{24}} - \frac{\mu^{24}}{a}
  + \frac{b^2}{\mu^{24}} - \frac{\mu^{24}}{b^2}
  + \frac{c^3}{\mu^{24}} - \frac{\mu^{24}}{c^3}
  + \frac{d^4}{\mu^{24}} - \frac{\mu^{24}}{d^4} \ge 0.
$$
Therefore
$$
\mu^{24} \left(
  \frac{a}{\mu^{48}} - \frac{1}{a}
  + \frac{b^2}{\mu^{48}} - \frac{1}{b^2}
  + \frac{c^3}{\mu^{48}} - \frac{1}{c^3}
  + \frac{d^4}{\mu^{48}} - \frac{1}{d^4}
\right) \ge 0.
$$
Whence
$$
\mu^{24} \left(
  a - \frac{1}{a}
  + b^2 - \frac{1}{b^2}
  + c^3 - \frac{1}{c^3}
  + d^4 - \frac{1}{d^4}
\right) -
\left( \mu^{12} - \frac{1}{\mu^{12}} \right)\big( a + b^2 + c^3 + d^4 \big)
\ge 0.
$$
Which can be written as
$$
a - \frac{1}{a}
  + b^2 - \frac{1}{b^2}
  + c^3 - \frac{1}{c^3}
  + d^4 - \frac{1}{d^4}
\ge 
\mu^{-24} \left( \mu^{12} - \frac{1}{\mu^{12}} \right)\big( a + b^2 + c^3 + d^4 \big).
$$
As $\mu \ge 1$, we obtain

$$
\forall 0 \le a \le b \le c \le d \wedge abcd = 1 :\\
a - \frac{1}{a}
  + b^2 - \frac{1}{b^2}
  + c^3 - \frac{1}{c^3}
  + d^4 - \frac{1}{d^4}
\ge 0.
$$

This method allows to show that the case $n+1$ is true, IF the case $n$ is true.
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