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$\newcommand\tH{\tilde H}$This problem is one of real algebraic geometry, which can be solved purely algorithmically. In Mathematica, such algorithms are implemented by Reduce and similar commands.

Here is a solution with Mathematica:

Here is a "more human" proof: Let $f(x,y,z)$ stand for the left-hand side of your desired inequality. We want to show that $f(x,y,z)\ge3/2$ if $x,y,z>0$ and $xyz=1$. Equivalently, we want to show that $$h(x,y):=\big(f(x,y,\tfrac1{xy})-\tfrac32\big) 2 x^3 y^3 (x + y) (1 + x^2 y) (1 + x y^2) \\ =2 x^{10} y^{10}+2 x^9 y^8+2 x^8 y^9+2 x^7 y^7-3 x^7 y^6-3 x^6 y^7-3 x^6 y^4+2 x^6 y^2-6 x^5 y^5+2 x^5 y^3+2 x^5-3 x^4 y^6-3 x^4 y^3+2 x^4 y+2 x^3 y^5-3 x^3 y^4+2 x^2 y^6+2 x y^4+2 y^5\overset{\text{(?)}}\ge0 \tag{1}\label{1}$$ for $x,y>0$. Now note that $$h(1+u,1+v)= 2 v^{10} u^{10}+20 v^9 u^{10}+90 v^8 u^{10}+240 v^7 u^{10}+420 v^6 u^{10}+504 v^5 u^{10}+420 v^4 u^{10}+240 v^3 u^{10}+90 v^2 u^{10}+20 v u^{10}+2 u^{10}+20 v^{10} u^9+200 v^9 u^9+902 v^8 u^9+2416 v^7 u^9+4256 v^6 u^9+5152 v^5 u^9+4340 v^4 u^9+2512 v^3 u^9+956 v^2 u^9+216 v u^9+22 u^9+90 v^{10} u^8+902 v^9 u^8+4086 v^8 u^8+11016 v^7 u^8+19572 v^6 u^8+23940 v^5 u^8+20412 v^4 u^8+11976 v^3 u^8+4626 v^2 u^8+1062 v u^8+110 u^8+240 v^{10} u^7+2416 v^9 u^7+11016 v^8 u^7+29954 v^7 u^7+53771 v^6 u^7+66552 v^5 u^7+57481 v^4 u^7+34186 v^3 u^7+13389 v^2 u^7+3116 v u^7+327 u^7+420 v^{10} u^6+4256 v^9 u^6+19572 v^8 u^6+53771 v^7 u^6+97664 v^6 u^6+122409 v^5 u^6+107083 v^4 u^6+64465 v^3 u^6+25520 v^2 u^6+5991 v u^6+633 u^6+504 v^{10} u^5+5152 v^9 u^5+23940 v^8 u^5+66552 v^7 u^5+122409 v^6 u^5+155352 v^5 u^5+137439 v^4 u^5+83450 v^3 u^5+33177 v^2 u^5+7782 v u^5+821 u^5+420 v^{10} u^4+4340 v^9 u^4+20412 v^8 u^4+57481 v^7 u^4+107083 v^6 u^4+137439 v^5 u^4+122540 v^4 u^4+74514 v^3 u^4+29382 v^2 u^4+6756 v u^4+703 u^4+240 v^{10} u^3+2512 v^9 u^3+11976 v^8 u^3+34186 v^7 u^3+64465 v^6 u^3+83450 v^5 u^3+74514 v^4 u^3+44802 v^3 u^3+17097 v^2 u^3+3690 v u^3+368 u^3+90 v^{10} u^2+956 v^9 u^2+4626 v^8 u^2+13389 v^7 u^2+25520 v^6 u^2+33177 v^5 u^2+29382 v^4 u^2+17097 v^3 u^2+6006 v^2 u^2+1077 v u^2+92 u^2+20 v^{10} u+216 v^9 u+1062 v^8 u+3116 v^7 u+5991 v^6 u+7782 v^5 u+6756 v^4 u+3690 v^3 u+1077 v^2 u+92 v u+2 v^{10}+22 v^9+110 v^8+327 v^7+633 v^6+821 v^5+703 v^4+368 v^3+92 v^2, $$ which is manifestly $\ge0$ for all real $u$ and $v$.

It also follows that the inequality in question is strict except for $(x,y,)=(1,1,1)$.

Thus, \eqref{1} follows. $\quad\Box$

$\newcommand\tH{\tilde H}$This problem is one of real algebraic geometry, which can be solved purely algorithmically. In Mathematica, such algorithms are implemented by Reduce and similar commands.

Here is a solution with Mathematica:

Here is a "more human" proof: Let $f(x,y,z)$ stand for the left-hand side of your desired inequality. We want to show that $f(x,y,z)\ge3/2$ if $x,y,z>0$ and $xyz=1$. Equivalently, we want to show that $$h(x,y):=\big(f(x,y,\tfrac1{xy})-\tfrac32\big) 2 x^3 y^3 (x + y) (1 + x^2 y) (1 + x y^2) \\ =2 x^{10} y^{10}+2 x^9 y^8+2 x^8 y^9+2 x^7 y^7-3 x^7 y^6-3 x^6 y^7-3 x^6 y^4+2 x^6 y^2-6 x^5 y^5+2 x^5 y^3+2 x^5-3 x^4 y^6-3 x^4 y^3+2 x^4 y+2 x^3 y^5-3 x^3 y^4+2 x^2 y^6+2 x y^4+2 y^5\overset{\text{(?)}}\ge0$$ for $x,y>0$.

Consider the partial derivatives $p(x,y):=h_x(x,y)$ and $q(x,y):=h_y(x,y)$. Let $r_1(y)$ and $r_2(x)$ denote the resultants of the polynomials $p(x,y)$ and $q(x,y)$ w.r. to $x$ and $y$, respectively. By symmetry, the real roots of $r_1(y)$ and $r_2(x)$ are the same: $z_1:=0$, $z_2:=1$, and a certain algebraic number $z_3\approx0.8180077783$. Therefore the critical points of $h$ in $(0,\infty)^2$ (if any) are of the form $(z_j,z_k)$ for $j,k\in\{2,3\}$. It is straightforward to check that $h(z_j,z_k)\ge0$ for $j,k\in\{2,3\}$.

So, it remains to check that the boundary values of $h$, near the boundary of the set $[0,\infty]^2$, are $\ge0$.

We have $h(0,y)=2 y^5\ge0$ for $y\ge0$ and $h(x,0)=2x^5\ge0$ for $x\ge0$. This does it for the boundary pieces $\{0\}\times[0,\infty)$ and $[0,\infty)\times\{0\}$.

By symmetry, it remains to show that $\liminf_{x,y\to\infty}h(x,y)\ge0$ and $\liminf_{y\to0,x\to\infty}h(x,y)\ge0$. Let $$H_{11}(x,y):=h(\tfrac1x,\tfrac1y)x^{10}y^{10} =2+2 x^{10} y^5+2 x^9 y^6+2 x^8 y^4-3 x^7 y^6+2 x^7 y^5+2 x^6 y^9-3 x^6 y^7-3 x^6 y^4+2 x^5 y^{10}+2 x^5 y^7-6 x^5 y^5+2 x^4 y^8-3 x^4 y^6-3 x^4 y^3-3 x^3 y^4+2 x^3 y^3+2 x^2 y+2 x y^2.$$ Obviously, $H_{11}(0+,0+)=2>0$. So, $\liminf_{x,y\to\infty}h(x,y)\ge0$.

Let $$H_{01}(x,y):=h(\tfrac1x,y)x^{10} =2 x^{10} y^5+2 x^9 y^4+2 x^8 y^6+2 x^7 y^5-3 x^7 y^4-3 x^6 y^6-3 x^6 y^3+2 x^6 y-6 x^5 y^5+2 x^5 y^3+2 x^5-3 x^4 y^7-3 x^4 y^4+2 x^4 y^2+2 x^3 y^7-3 x^3 y^6+2 x^2 y^9+2 x y^8+2 y^{10}.$$ Removing from the latter expression terms dominated by other terms whenever $x,y\to0$, we end up with having to show that $\liminf_{x,y\to0}\tH(x,y)\ge0$, where $$\tH(x,y):=(2+o(1)) x^5 + (2+o(1)) x^4 y^2 - (3+o(1)) x^3 y^6 + 2 x y^8 + 2 x^2 y^9 + 2 y^{10},$$ which follows because $$(2+o(1)) x^4 y^2 - (3+o(1)) x^3 y^6 + 2 x y^8 \\ \ge(2+o(1)) x^5 y^4 - (3+o(1)) x^3 y^6 + 2 x y^8 \\ =x y^8[(2+o(1)) s^4 - (3+o(1)) s^2 + 2 ]>0$$ for all small enough $x,y>0$, where $s:=x/y$. So, $\liminf_{y\to0,x\to\infty}h(x,y)\ge0$. $\quad\Box$

$\newcommand\tH{\tilde H}$This problem is one of real algebraic geometry, which can be solved purely algorithmically. In Mathematica, such algorithms are implemented by Reduce and similar commands.

Here is a solution with Mathematica:

Here is a "more human" proof: Let $f(x,y,z)$ stand for the left-hand side of your desired inequality. We want to show that $f(x,y,z)\ge3/2$ if $x,y,z>0$ and $xyz=1$. Equivalently, we want to show that $$h(x,y):=\big(f(x,y,\tfrac1{xy})-\tfrac32\big) 2 x^3 y^3 (x + y) (1 + x^2 y) (1 + x y^2) \\ =2 x^{10} y^{10}+2 x^9 y^8+2 x^8 y^9+2 x^7 y^7-3 x^7 y^6-3 x^6 y^7-3 x^6 y^4+2 x^6 y^2-6 x^5 y^5+2 x^5 y^3+2 x^5-3 x^4 y^6-3 x^4 y^3+2 x^4 y+2 x^3 y^5-3 x^3 y^4+2 x^2 y^6+2 x y^4+2 y^5\overset{\text{(?)}}\ge0$$ for $x,y>0$.

Consider the partial derivatives $p(x,y):=h_x(x,y)$ and $q(x,y):=h_y(x,y)$. Let $r_1(y)$ and $r_2(x)$ denote the resultants of the polynomials $p(x,y)$ and $q(x,y)$ w.r. to $x$ and $y$, respectively. By symmetry, the real roots of $r_1(y)$ and $r_2(x)$ are the same: $z_1:=0$, $z_2:=1$, and a certain algebraic number $z_3\approx0.8180077783$. Therefore the critical points of $h$ in $(0,\infty)^2$ (if any) are of the form $(z_j,z_k)$ for $j,k\in\{2,3\}$. It is straightforward to check that $h(z_j,z_k)\ge0$ for $j,k\in\{2,3\}$.

So, it remains to check that the boundary values of $h$, near the boundary of the set $[0,\infty]^2$, are $\ge0$.

We have $h(0,y)=2 y^5\ge0$ for $y\ge0$ and $h(x,0)=2x^5\ge0$ for $x\ge0$. This does it for the boundary pieces $\{0\}\times[0,\infty)$ and $[0,\infty)\times\{0\}$.

By symmetry, it remains to show that $\liminf_{x,y\to\infty}h(x,y)\ge0$ and $\liminf_{y\to0,x\to\infty}h(x,y)\ge0$. Let $$H_{11}(x,y):=h(\tfrac1x,\tfrac1y)x^{10}y^{10} =2+2 x^{10} y^5+2 x^9 y^6+2 x^8 y^4-3 x^7 y^6+2 x^7 y^5+2 x^6 y^9-3 x^6 y^7-3 x^6 y^4+2 x^5 y^{10}+2 x^5 y^7-6 x^5 y^5+2 x^4 y^8-3 x^4 y^6-3 x^4 y^3-3 x^3 y^4+2 x^3 y^3+2 x^2 y+2 x y^2.$$ Obviously, $H_{11}(0+,0+)=2>0$. So, $\liminf_{x,y\to\infty}h(x,y)\ge0$.

Let $$H_{01}(x,y):=h(\tfrac1x,y)x^{10} =2 x^{10} y^5+2 x^9 y^4+2 x^8 y^6+2 x^7 y^5-3 x^7 y^4-3 x^6 y^6-3 x^6 y^3+2 x^6 y-6 x^5 y^5+2 x^5 y^3+2 x^5-3 x^4 y^7-3 x^4 y^4+2 x^4 y^2+2 x^3 y^7-3 x^3 y^6+2 x^2 y^9+2 x y^8+2 y^{10}.$$ Removing from the latter expression terms dominated by other terms whenever $x,y\to0$, we end up with having to show that $\liminf_{x,y\to0}\tH(x,y)\ge0$, where $$\tH(x,y):=(2+o(1)) x^5 + (2+o(1)) x^4 y^2 - (3+o(1)) x^3 y^6 + 2 x y^8 + 2 x^2 y^9 + 2 y^{10},$$ which follows because $$(2+o(1)) x^4 y^2 - (3+o(1)) x^3 y^6 + 2 x y^8 \\ \ge(2+o(1)) x^5 y^4 - (3+o(1)) x^3 y^6 + 2 x y^8 \\ =x y^8[(2+o(1)) s^4 - (3+o(1)) s^2 + 2 ]>0$$ for all small enough $x,y>0$, where $s:=x/y$. So, $\liminf_{y\to0,x\to\infty}h(x,y)\ge0$. $\quad\Box$

$\newcommand\tH{\tilde H}$This problem is one of real algebraic geometry, which can be solved purely algorithmically. In Mathematica, such algorithms are implemented by Reduce and similar commands.

Here is a solution with Mathematica:

Here is a "more human" proof: Let $f(x,y,z)$ stand for the left-hand side of your desired inequality. We want to show that $f(x,y,z)\ge3/2$ if $x,y,z>0$ and $xyz=1$. Equivalently, we want to show that $$h(x,y):=\big(f(x,y,\tfrac1{xy})-\tfrac32\big) 2 x^3 y^3 (x + y) (1 + x^2 y) (1 + x y^2) \\ =2 x^{10} y^{10}+2 x^9 y^8+2 x^8 y^9+2 x^7 y^7-3 x^7 y^6-3 x^6 y^7-3 x^6 y^4+2 x^6 y^2-6 x^5 y^5+2 x^5 y^3+2 x^5-3 x^4 y^6-3 x^4 y^3+2 x^4 y+2 x^3 y^5-3 x^3 y^4+2 x^2 y^6+2 x y^4+2 y^5\overset{\text{(?)}}\ge0 \tag{1}\label{1}$$ for $x,y>0$. Now note that $$h(1+u,1+v)= 2 v^{10} u^{10}+20 v^9 u^{10}+90 v^8 u^{10}+240 v^7 u^{10}+420 v^6 u^{10}+504 v^5 u^{10}+420 v^4 u^{10}+240 v^3 u^{10}+90 v^2 u^{10}+20 v u^{10}+2 u^{10}+20 v^{10} u^9+200 v^9 u^9+902 v^8 u^9+2416 v^7 u^9+4256 v^6 u^9+5152 v^5 u^9+4340 v^4 u^9+2512 v^3 u^9+956 v^2 u^9+216 v u^9+22 u^9+90 v^{10} u^8+902 v^9 u^8+4086 v^8 u^8+11016 v^7 u^8+19572 v^6 u^8+23940 v^5 u^8+20412 v^4 u^8+11976 v^3 u^8+4626 v^2 u^8+1062 v u^8+110 u^8+240 v^{10} u^7+2416 v^9 u^7+11016 v^8 u^7+29954 v^7 u^7+53771 v^6 u^7+66552 v^5 u^7+57481 v^4 u^7+34186 v^3 u^7+13389 v^2 u^7+3116 v u^7+327 u^7+420 v^{10} u^6+4256 v^9 u^6+19572 v^8 u^6+53771 v^7 u^6+97664 v^6 u^6+122409 v^5 u^6+107083 v^4 u^6+64465 v^3 u^6+25520 v^2 u^6+5991 v u^6+633 u^6+504 v^{10} u^5+5152 v^9 u^5+23940 v^8 u^5+66552 v^7 u^5+122409 v^6 u^5+155352 v^5 u^5+137439 v^4 u^5+83450 v^3 u^5+33177 v^2 u^5+7782 v u^5+821 u^5+420 v^{10} u^4+4340 v^9 u^4+20412 v^8 u^4+57481 v^7 u^4+107083 v^6 u^4+137439 v^5 u^4+122540 v^4 u^4+74514 v^3 u^4+29382 v^2 u^4+6756 v u^4+703 u^4+240 v^{10} u^3+2512 v^9 u^3+11976 v^8 u^3+34186 v^7 u^3+64465 v^6 u^3+83450 v^5 u^3+74514 v^4 u^3+44802 v^3 u^3+17097 v^2 u^3+3690 v u^3+368 u^3+90 v^{10} u^2+956 v^9 u^2+4626 v^8 u^2+13389 v^7 u^2+25520 v^6 u^2+33177 v^5 u^2+29382 v^4 u^2+17097 v^3 u^2+6006 v^2 u^2+1077 v u^2+92 u^2+20 v^{10} u+216 v^9 u+1062 v^8 u+3116 v^7 u+5991 v^6 u+7782 v^5 u+6756 v^4 u+3690 v^3 u+1077 v^2 u+92 v u+2 v^{10}+22 v^9+110 v^8+327 v^7+633 v^6+821 v^5+703 v^4+368 v^3+92 v^2, $$ which is manifestly $\ge0$ for all real $u$ and $v$.

It also follows that the inequality in question is strict except for $(x,y,)=(1,1,1)$.

Thus, \eqref{1} follows. $\quad\Box$

$\newcommand\tH{\tilde H}$This problem is one of real algebraic geometry, which can be solved purely algorithmically. In Mathematica, such algorithms are implemented by Reduce and similar commands.

Here is a solution with Mathematica:

Here is a "more human" proof: Let $f(x,y,z)$ stand for the left-hand side of your desired inequality. We want to show that $f(x,y,z)\ge3/2$ if $x,y,z>0$ and $xyz=1$. Equivalently, we want to show that $$h(x,y):=\big(f(x,y,\tfrac1{xy})-\tfrac32\big) 2 x^3 y^3 (x + y) (1 + x^2 y) (1 + x y^2)\ge3/2 \\ =2 x^{10} y^{10}+2 x^9 y^8+2 x^8 y^9+2 x^7 y^7-3 x^7 y^6-3 x^6 y^7-3 x^6 y^4+2 x^6 y^2-6 x^5 y^5+2 x^5 y^3+2 x^5-3 x^4 y^6-3 x^4 y^3+2 x^4 y+2 x^3 y^5-3 x^3 y^4+2 x^2 y^6+2 x y^4+2 y^5\overset{\text{(?)}}\ge0$$ for $x,y>0$.

Consider the partial derivatives $p(x,y):=h_x(x,y)$ and $q(x,y):=h_y(x,y)$. Let $r_1(y)$ and $r_2(x)$ denote the resultants of the polynomials $p(x,y)$ and $q(x,y)$ w.r. to $x$ and $y$, respectively. By symmetry, the real roots of $r_1(y)$ and $r_2(x)$ are the same: $z_1:=0$, $z_2:=1$, and a certain algebraic number $z_3\approx0.8180077783$. Therefore the critical points of $h$ in $(0,\infty)^2$ (if any) are of the form $(z_j,z_k)$ for $j,k\in\{2,3\}$. It is straightforward to check that $h(z_j,z_k)\ge0$ for $j,k\in\{2,3\}$.

So, it remains to check that the boundary values of $h$, near the boundary of the set $[0,\infty]^2$, are $\ge0$.

We have $h(0,y)=2 y^5\ge0$ for $y\ge0$ and $h(x,0)=2x^5\ge0$ for $x\ge0$. This does it for the boundary pieces $\{0\}\times[0,\infty)$ and $[0,\infty)\times\{0\}$.

By symmetry, it remains to show that $\liminf_{x,y\to\infty}h(x,y)\ge0$ and $\liminf_{y\to0,x\to\infty}h(x,y)\ge0$. Let $$H_{11}(x,y):=h(\tfrac1x,\tfrac1y)x^{10}y^{10} =2+2 x^{10} y^5+2 x^9 y^6+2 x^8 y^4-3 x^7 y^6+2 x^7 y^5+2 x^6 y^9-3 x^6 y^7-3 x^6 y^4+2 x^5 y^{10}+2 x^5 y^7-6 x^5 y^5+2 x^4 y^8-3 x^4 y^6-3 x^4 y^3-3 x^3 y^4+2 x^3 y^3+2 x^2 y+2 x y^2.$$ Obviously, $H_{11}(0+,0+)=2>0$. So, $\liminf_{x,y\to\infty}h(x,y)\ge0$.

Let $$H_{01}(x,y):=h(\tfrac1x,y)x^{10} =2 x^{10} y^5+2 x^9 y^4+2 x^8 y^6+2 x^7 y^5-3 x^7 y^4-3 x^6 y^6-3 x^6 y^3+2 x^6 y-6 x^5 y^5+2 x^5 y^3+2 x^5-3 x^4 y^7-3 x^4 y^4+2 x^4 y^2+2 x^3 y^7-3 x^3 y^6+2 x^2 y^9+2 x y^8+2 y^{10}.$$ Removing from the latter expression terms dominated by other terms whenever $x,y\to0$, we end up with having to show that $\liminf_{x,y\to0}\tH(x,y)\ge0$, where $$\tH(x,y):=(2+o(1)) x^5 + (2+o(1)) x^4 y^2 - (3+o(1)) x^3 y^6 + 2 x y^8 + 2 x^2 y^9 + 2 y^{10},$$ which follows because $$(2+o(1)) x^4 y^2 - (3+o(1)) x^3 y^6 + 2 x y^8 \\ \ge(2+o(1)) x^5 y^4 - (3+o(1)) x^3 y^6 + 2 x y^8 \\ =x y^8[(2+o(1)) s^4 - (3+o(1)) s^2 + 2 ]>0$$ for all small enough $x,y>0$, where $s:=x/y$. So, $\liminf_{y\to0,x\to\infty}h(x,y)\ge0$. $\quad\Box$

$\newcommand\tH{\tilde H}$This problem is one of real algebraic geometry, which can be solved purely algorithmically. In Mathematica, such algorithms are implemented by Reduce and similar commands.

Here is a solution with Mathematica:

Here is a "more human" proof: Let $f(x,y,z)$ stand for the left-hand side of your desired inequality. We want to show that $f(x,y,z)\ge3/2$ if $x,y,z>0$ and $xyz=1$. Equivalently, we want to show that $$h(x,y):=\big(f(x,y,\tfrac1{xy})-\tfrac32\big) 2 x^3 y^3 (x + y) (1 + x^2 y) (1 + x y^2) \\ =2 x^{10} y^{10}+2 x^9 y^8+2 x^8 y^9+2 x^7 y^7-3 x^7 y^6-3 x^6 y^7-3 x^6 y^4+2 x^6 y^2-6 x^5 y^5+2 x^5 y^3+2 x^5-3 x^4 y^6-3 x^4 y^3+2 x^4 y+2 x^3 y^5-3 x^3 y^4+2 x^2 y^6+2 x y^4+2 y^5\overset{\text{(?)}}\ge0$$ for $x,y>0$.

Consider the partial derivatives $p(x,y):=h_x(x,y)$ and $q(x,y):=h_y(x,y)$. Let $r_1(y)$ and $r_2(x)$ denote the resultants of the polynomials $p(x,y)$ and $q(x,y)$ w.r. to $x$ and $y$, respectively. By symmetry, the real roots of $r_1(y)$ and $r_2(x)$ are the same: $z_1:=0$, $z_2:=1$, and a certain algebraic number $z_3\approx0.8180077783$. Therefore the critical points of $h$ in $(0,\infty)^2$ (if any) are of the form $(z_j,z_k)$ for $j,k\in\{2,3\}$. It is straightforward to check that $h(z_j,z_k)\ge0$ for $j,k\in\{2,3\}$.

So, it remains to check that the boundary values of $h$, near the boundary of the set $[0,\infty]^2$, are $\ge0$.

We have $h(0,y)=2 y^5\ge0$ for $y\ge0$ and $h(x,0)=2x^5\ge0$ for $x\ge0$. This does it for the boundary pieces $\{0\}\times[0,\infty)$ and $[0,\infty)\times\{0\}$.

By symmetry, it remains to show that $\liminf_{x,y\to\infty}h(x,y)\ge0$ and $\liminf_{y\to0,x\to\infty}h(x,y)\ge0$. Let $$H_{11}(x,y):=h(\tfrac1x,\tfrac1y)x^{10}y^{10} =2+2 x^{10} y^5+2 x^9 y^6+2 x^8 y^4-3 x^7 y^6+2 x^7 y^5+2 x^6 y^9-3 x^6 y^7-3 x^6 y^4+2 x^5 y^{10}+2 x^5 y^7-6 x^5 y^5+2 x^4 y^8-3 x^4 y^6-3 x^4 y^3-3 x^3 y^4+2 x^3 y^3+2 x^2 y+2 x y^2.$$ Obviously, $H_{11}(0+,0+)=2>0$. So, $\liminf_{x,y\to\infty}h(x,y)\ge0$.

Let $$H_{01}(x,y):=h(\tfrac1x,y)x^{10} =2 x^{10} y^5+2 x^9 y^4+2 x^8 y^6+2 x^7 y^5-3 x^7 y^4-3 x^6 y^6-3 x^6 y^3+2 x^6 y-6 x^5 y^5+2 x^5 y^3+2 x^5-3 x^4 y^7-3 x^4 y^4+2 x^4 y^2+2 x^3 y^7-3 x^3 y^6+2 x^2 y^9+2 x y^8+2 y^{10}.$$ Removing from the latter expression terms dominated by other terms whenever $x,y\to0$, we end up with having to show that $\liminf_{x,y\to0}\tH(x,y)\ge0$, where $$\tH(x,y):=(2+o(1)) x^5 + (2+o(1)) x^4 y^2 - (3+o(1)) x^3 y^6 + 2 x y^8 + 2 x^2 y^9 + 2 y^{10},$$ which follows because $$(2+o(1)) x^4 y^2 - (3+o(1)) x^3 y^6 + 2 x y^8 \\ \ge(2+o(1)) x^5 y^4 - (3+o(1)) x^3 y^6 + 2 x y^8 \\ =x y^8[(2+o(1)) s^4 - (3+o(1)) s^2 + 2 ]>0$$ for all small enough $x,y>0$, where $s:=x/y$. So, $\liminf_{y\to0,x\to\infty}h(x,y)\ge0$. $\quad\Box$