$\newcommand{\res}{\operatorname{res}}$ Here is a complete proof of the inequality in question.

By symmetry, without loss of generality (wlog) $d$ is the maximum of $\{a,b,c,d\}$. By homogeneity, wlog $d=1$. So, the inequality reduces to \begin{equation*} r(a,b,c) :=\frac{\frac{a^2}{b}+\frac{b^2}{c}+c^2+\frac{1}{a}}{\sqrt[4]{a^4+b^4+c^4+1}}\ge2\sqrt2 \tag{1} \end{equation*} for $a,b,c$ in $(0,1]$. Consider the following polynomials in $a,b,c$: \begin{align*} D_ar(a,b,c)&:=\frac{\partial r(a,b,c)}{\partial a}\, a^2 b c \left(a^4+b^4+c^4+1\right)^{5/4} \\ D_br(a,b,c)&:=\frac{\partial r(a,b,c)}{\partial b}\, a b^2 c \left(a^4+b^4+c^4+1\right)^{5/4} \\ D_cr(a,b,c)&:=\frac{\partial r(a,b,c)}{\partial c}\, a b c^2 \left(a^4+b^4+c^4+1\right)^{5/4}, \\ p_1(a,b)&:=a^{16} b^4+2 a^{16} b^2+4 a^{13} b^3+3 a^{12} b^8+4 a^{12} b^6 \\ & +4 a^{12} b^4+4 a^{12} b^2+4 a^{10} b^7+2 a^{10} b^4+6 a^{10} b^2 \\ &+8 a^9 b^7+4 a^9 b^5+8 a^9 b^3+3 a^8 b^{12}+2 a^8 b^{10} \\ &+6 a^8 b^8+4 a^8 b^6+3 a^8 b^4+2 a^8 b^2+4 a^7 b^8 \\ &+4 a^7 b^3+4 a^6 b^{11}+4 a^6 b^8+4 a^6 b^7+6 a^6 b^6 \\ &+4 a^6 b^4+6 a^6 b^2+4 a^5 b^{11}+8 a^5 b^7+4 a^5 b^3 \\ &+a^4 b^{16}+4 a^4 b^{12}+3 a^4 b^8+2 a^4 b^4+4 a^3 b^{12} \\ &+4 a^3 b^8+4 a^3 b^7+4 a^3 b^3+2 a^2 b^{12}+4 a^2 b^8 \\ &+2 a^2 b^4+4 a^{13} b+4 a^9 b+a^{16}+a^{12}+b^8+b^4, \end{align*} which latter is manifestly $>0$.

Then \begin{equation*} p_{11}(a,b):=\frac1{a b^3 p_1(a,b)}\,\res_c(D_ar(a,b,c),D_br(a,b,c)) \end{equation*} is a polynomial in $a,b$ of respective degrees $35,25$, where $\res_c(D_ar(a,b,c),D_br(a,b,c))$ is the resultant with respect to $c$ of the polynomials $D_ar(a,b,c)$ and $D_br(a,b,c))$ in $a,b,c$. Similarly, \begin{equation*} p_{21}(a,b):=\frac1{a^4 b^3 p_1(a,b)}\,\res_c(D_br(a,b,c),D_cr(a,b,c)) \end{equation*} is a polynomial in $a,b$ of respective degrees $38,39$.

The key observation is that all the roots $(a,b)\in(0,1)\times(0,1)$ of the polynomials $p_{11}$ and $p_{21}$ satisfy the inequalities $b<a^2$ and $b\ge a^2$, respectively. (This was proved using the Mathematica command Reduce, which took about 6.5 sec for $p_{11}$ and 33 sec for $p_{21}$.) Thus, $p_{11}$ and $p_{21}$ have no common root $(a,b)\in(0,1)\times(0,1)$, and hence $r$ has no critical points in the cube $C:=(0,1)^3$.

It remains to consider the behavior of $r$ at/near the boundary of the cube $C$.

If $a\downarrow0$, then, by (1), $r(a,b,c)\to\infty$ uniformly in $(b,c)\in(0,1)\times(0,1)$.

If $b\downarrow0$, then, by (1), $r(a,b,c)\to\infty$ uniformly in $(a,c)\in(0,1)\times(0,1)$ unless $a\downarrow0$, which reduces the consideration to the previous paragraph.

If $c\downarrow0$, then, by (1), $r(a,b,c)\to\infty$ uniformly in $(a,b)\in(0,1)\times(0,1)$ unless $b\downarrow0$, which reduces the consideration to the previous paragraph.

It remains to consider $r(a,b,c)$ when at least one of the variables $a,b,c$ takes value $1$. In such cases, using again the Mathematica command Reduce, in splits of a second we get (1). $\Box$

Here is an image of a Mathematica notebook with details of calculations (click on the image to enlarge it):

$\newcommand{\res}{\operatorname{res}}$ Here is a complete proof of the inequality in question.

By symmetry, without loss of generality (wlog) $d$ is the maximum of $\{a,b,c,d\}$. By homogeneity, wlog $d=1$. So, the inequality reduces to \begin{equation*} r(a,b,c) :=\frac{\frac{a^2}{b}+\frac{b^2}{c}+c^2+\frac{1}{a}}{\sqrt[4]{a^4+b^4+c^4+1}}\ge2\sqrt2 \tag{1} \end{equation*} for $a,b,c$ in $(0,1]$. Consider the following polynomials in $a,b,c$: \begin{align*} D_ar(a,b,c)&:=\frac{\partial r(a,b,c)}{\partial a}\, a^2 b c \left(a^4+b^4+c^4+1\right)^{5/4} \\ D_br(a,b,c)&:=\frac{\partial r(a,b,c)}{\partial b}\, a b^2 c \left(a^4+b^4+c^4+1\right)^{5/4} \\ D_cr(a,b,c)&:=\frac{\partial r(a,b,c)}{\partial c}\, a b c^2 \left(a^4+b^4+c^4+1\right)^{5/4}, \\ p_1(a,b)&:=a^{16} b^4+2 a^{16} b^2+4 a^{13} b^3+3 a^{12} b^8+4 a^{12} b^6 \\ & +4 a^{12} b^4+4 a^{12} b^2+4 a^{10} b^7+2 a^{10} b^4+6 a^{10} b^2 \\ &+8 a^9 b^7+4 a^9 b^5+8 a^9 b^3+3 a^8 b^{12}+2 a^8 b^{10} \\ &+6 a^8 b^8+4 a^8 b^6+3 a^8 b^4+2 a^8 b^2+4 a^7 b^8 \\ &+4 a^7 b^3+4 a^6 b^{11}+4 a^6 b^8+4 a^6 b^7+6 a^6 b^6 \\ &+4 a^6 b^4+6 a^6 b^2+4 a^5 b^{11}+8 a^5 b^7+4 a^5 b^3 \\ &+a^4 b^{16}+4 a^4 b^{12}+3 a^4 b^8+2 a^4 b^4+4 a^3 b^{12} \\ &+4 a^3 b^8+4 a^3 b^7+4 a^3 b^3+2 a^2 b^{12}+4 a^2 b^8 \\ &+2 a^2 b^4+4 a^{13} b+4 a^9 b+a^{16}+a^{12}+b^8+b^4, \end{align*} which latter is manifestly $>0$.

Then \begin{equation*} p_{11}(a,b):=\frac1{a b^3 p_1(a,b)}\,\res_c(D_ar(a,b,c),D_br(a,b,c)) \end{equation*} is a polynomial in $a,b$ of respective degrees $35,25$, where $\res_c(D_ar(a,b,c),D_br(a,b,c))$ is the resultant with respect to $c$ of the polynomials $D_ar(a,b,c)$ and $D_br(a,b,c)$ in $a,b,c$. Similarly, \begin{equation*} p_{21}(a,b):=\frac1{a^4 b^3 p_1(a,b)}\,\res_c(D_br(a,b,c),D_cr(a,b,c)) \end{equation*} is a polynomial in $a,b$ of respective degrees $38,39$.

The key observation is that all the roots $(a,b)\in(0,1)\times(0,1)$ of the polynomials $p_{11}$ and $p_{21}$ satisfy the inequalities $b<a^2$ and $b\ge a^2$, respectively. (This was proved using the Mathematica command Reduce, which took about 6.5 sec for $p_{11}$ and 33 sec for $p_{21}$.) Thus, $p_{11}$ and $p_{21}$ have no common root $(a,b)\in(0,1)\times(0,1)$, and hence $r$ has no critical points in the cube $C:=(0,1)^3$.

It remains to consider the behavior of $r$ at/near the boundary of the cube $C$.

If $a\downarrow0$, then, by (1), $r(a,b,c)\to\infty$ uniformly in $(b,c)\in(0,1)\times(0,1)$.

If $b\downarrow0$, then, by (1), $r(a,b,c)\to\infty$ uniformly in $(a,c)\in(0,1)\times(0,1)$ unless $a\downarrow0$, which reduces the consideration to the previous paragraph.

If $c\downarrow0$, then, by (1), $r(a,b,c)\to\infty$ uniformly in $(a,b)\in(0,1)\times(0,1)$ unless $b\downarrow0$, which reduces the consideration to the previous paragraph.

It remains to consider $r(a,b,c)$ when at least one of the variables $a,b,c$ takes value $1$. In such cases, using again the Mathematica command Reduce, in splits of a second we get (1). $\Box$

Here is an image of a Mathematica notebook with details of calculations (click on the image to enlarge it):

$\newcommand{\res}{\operatorname{res}}$ By symmetry, without loss of generality (wlog) $d$ is the maximum of $\{a,b,c,d\}$. By homogeneity, wlog $d=1$. So, the inequality reduces to \begin{equation*} r(a,b,c) :=\frac{\frac{a^2}{b}+\frac{b^2}{c}+c^2+\frac{1}{a}}{\sqrt[4]{a^4+b^4+c^4+1}}\ge2\sqrt2 \tag{1} \end{equation*} for $a,b,c$ in $(0,1]$. Consider the following polynomials in $a,b,c$: \begin{align*} D_ar(a,b,c)&:=\frac{\partial r(a,b,c)}{\partial a}\, a^2 b c \left(a^4+b^4+c^4+1\right)^{5/4} \\ D_br(a,b,c)&:=\frac{\partial r(a,b,c)}{\partial b}\, a b^2 c \left(a^4+b^4+c^4+1\right)^{5/4} \\ D_cr(a,b,c)&:=\frac{\partial r(a,b,c)}{\partial c}\, a b c^2 \left(a^4+b^4+c^4+1\right)^{5/4}, \\ p_1(a,b)&:=a^{16} b^4+2 a^{16} b^2+4 a^{13} b^3+3 a^{12} b^8+4 a^{12} b^6 \\ & +4 a^{12} b^4+4 a^{12} b^2+4 a^{10} b^7+2 a^{10} b^4+6 a^{10} b^2 \\ &+8 a^9 b^7+4 a^9 b^5+8 a^9 b^3+3 a^8 b^{12}+2 a^8 b^{10} \\ &+6 a^8 b^8+4 a^8 b^6+3 a^8 b^4+2 a^8 b^2+4 a^7 b^8 \\ &+4 a^7 b^3+4 a^6 b^{11}+4 a^6 b^8+4 a^6 b^7+6 a^6 b^6 \\ &+4 a^6 b^4+6 a^6 b^2+4 a^5 b^{11}+8 a^5 b^7+4 a^5 b^3 \\ &+a^4 b^{16}+4 a^4 b^{12}+3 a^4 b^8+2 a^4 b^4+4 a^3 b^{12} \\ &+4 a^3 b^8+4 a^3 b^7+4 a^3 b^3+2 a^2 b^{12}+4 a^2 b^8 \\ &+2 a^2 b^4+4 a^{13} b+4 a^9 b+a^{16}+a^{12}+b^8+b^4, \end{align*} which latter is manifestly $>0$.

Then \begin{equation*} p_{11}(a,b):=\frac1{a b^3 p_1(a,b)}\,\res_c(D_ar(a,b,c),D_br(a,b,c)) \end{equation*} is a polynomial in $a,b$ of respective degrees $35,25$, where $\res_c(D_ar(a,b,c),D_br(a,b,c))$ is the resultant with respect to $c$ of the polynomials $D_ar(a,b,c)$ and $D_br(a,b,c))$ in $a,b,c$. Similarly, \begin{equation*} p_{21}(a,b):=\frac1{a^4 b^3 p_1(a,b)}\,\res_c(D_br(a,b,c),D_cr(a,b,c)) \end{equation*} is a polynomial in $a,b$ of respective degrees $38,39$.

The key observation is that all the roots $(a,b)\in(0,1)\times(0,1)$ of the polynomials $p_{11}$ and $p_{21}$ satisfy the inequalities $b<a^2$ and $b\ge a^2$, respectively. (This was proved using the Mathematica command Reduce, which took about 6.5 sec for $p_{11}$ and 34 sec for $p_{21}$.) Thus, $p_{11}$ and $p_{21}$ have no common root $(a,b)\in(0,1)\times(0,1)$, and hence $r$ has no critical points in the cube $C:=(0,1)^3$.

It remains to consider the behavior of $r$ at/near the boundary of the cube $C$.

If $a\downarrow0$, then, by (1), $r(a,b,c)\to\infty$ uniformly in $(b,c)\in(0,1)\times(0,1)$.

If $b\downarrow0$, then, by (1), $r(a,b,c)\to\infty$ uniformly in $(a,c)\in(0,1)\times(0,1)$ unless $a\downarrow0$, which reduces the consideration to the previous paragraph.

If $c\downarrow0$, then, by (1), $r(a,b,c)\to\infty$ uniformly in $(a,b)\in(0,1)\times(0,1)$ unless $b\downarrow0$, which reduces the consideration to the previous paragraph.

It remains to consider $r(a,b,c)$ when at least one of the variables $a,b,c$ takes value $1$. In such cases, using again the Mathematica command Reduce, in splits of a second we get (1). $\Box$

$\newcommand{\res}{\operatorname{res}}$ Here is a complete proof of the inequality in question.

By symmetry, without loss of generality (wlog) $d$ is the maximum of $\{a,b,c,d\}$. By homogeneity, wlog $d=1$. So, the inequality reduces to \begin{equation*} r(a,b,c) :=\frac{\frac{a^2}{b}+\frac{b^2}{c}+c^2+\frac{1}{a}}{\sqrt[4]{a^4+b^4+c^4+1}}\ge2\sqrt2 \tag{1} \end{equation*} for $a,b,c$ in $(0,1]$. Consider the following polynomials in $a,b,c$: \begin{align*} D_ar(a,b,c)&:=\frac{\partial r(a,b,c)}{\partial a}\, a^2 b c \left(a^4+b^4+c^4+1\right)^{5/4} \\ D_br(a,b,c)&:=\frac{\partial r(a,b,c)}{\partial b}\, a b^2 c \left(a^4+b^4+c^4+1\right)^{5/4} \\ D_cr(a,b,c)&:=\frac{\partial r(a,b,c)}{\partial c}\, a b c^2 \left(a^4+b^4+c^4+1\right)^{5/4}, \\ p_1(a,b)&:=a^{16} b^4+2 a^{16} b^2+4 a^{13} b^3+3 a^{12} b^8+4 a^{12} b^6 \\ & +4 a^{12} b^4+4 a^{12} b^2+4 a^{10} b^7+2 a^{10} b^4+6 a^{10} b^2 \\ &+8 a^9 b^7+4 a^9 b^5+8 a^9 b^3+3 a^8 b^{12}+2 a^8 b^{10} \\ &+6 a^8 b^8+4 a^8 b^6+3 a^8 b^4+2 a^8 b^2+4 a^7 b^8 \\ &+4 a^7 b^3+4 a^6 b^{11}+4 a^6 b^8+4 a^6 b^7+6 a^6 b^6 \\ &+4 a^6 b^4+6 a^6 b^2+4 a^5 b^{11}+8 a^5 b^7+4 a^5 b^3 \\ &+a^4 b^{16}+4 a^4 b^{12}+3 a^4 b^8+2 a^4 b^4+4 a^3 b^{12} \\ &+4 a^3 b^8+4 a^3 b^7+4 a^3 b^3+2 a^2 b^{12}+4 a^2 b^8 \\ &+2 a^2 b^4+4 a^{13} b+4 a^9 b+a^{16}+a^{12}+b^8+b^4, \end{align*} which latter is manifestly $>0$.

Then \begin{equation*} p_{11}(a,b):=\frac1{a b^3 p_1(a,b)}\,\res_c(D_ar(a,b,c),D_br(a,b,c)) \end{equation*} is a polynomial in $a,b$ of respective degrees $35,25$, where $\res_c(D_ar(a,b,c),D_br(a,b,c))$ is the resultant with respect to $c$ of the polynomials $D_ar(a,b,c)$ and $D_br(a,b,c))$ in $a,b,c$. Similarly, \begin{equation*} p_{21}(a,b):=\frac1{a^4 b^3 p_1(a,b)}\,\res_c(D_br(a,b,c),D_cr(a,b,c)) \end{equation*} is a polynomial in $a,b$ of respective degrees $38,39$.

The key observation is that all the roots $(a,b)\in(0,1)\times(0,1)$ of the polynomials $p_{11}$ and $p_{21}$ satisfy the inequalities $b<a^2$ and $b\ge a^2$, respectively. (This was proved using the Mathematica command Reduce, which took about 6.5 sec for $p_{11}$ and 33 sec for $p_{21}$.) Thus, $p_{11}$ and $p_{21}$ have no common root $(a,b)\in(0,1)\times(0,1)$, and hence $r$ has no critical points in the cube $C:=(0,1)^3$.

It remains to consider the behavior of $r$ at/near the boundary of the cube $C$.

If $a\downarrow0$, then, by (1), $r(a,b,c)\to\infty$ uniformly in $(b,c)\in(0,1)\times(0,1)$.

If $b\downarrow0$, then, by (1), $r(a,b,c)\to\infty$ uniformly in $(a,c)\in(0,1)\times(0,1)$ unless $a\downarrow0$, which reduces the consideration to the previous paragraph.

If $c\downarrow0$, then, by (1), $r(a,b,c)\to\infty$ uniformly in $(a,b)\in(0,1)\times(0,1)$ unless $b\downarrow0$, which reduces the consideration to the previous paragraph.

It remains to consider $r(a,b,c)$ when at least one of the variables $a,b,c$ takes value $1$. In such cases, using again the Mathematica command Reduce, in splits of a second we get (1). $\Box$

Here is an image of a Mathematica notebook with details of calculations (click on the image to enlarge it):