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The answer of Iosif helps one try to change variables as the solution verify $x^2+y^2=1$. This one is a bit long but oddly it has many powerful factorisation arguments.

So set $\sqrt{x^2+y^2}=a$ and $y^2=a^2-x^2$ with $-a\le x\le a$. The inequality is then: $$\sqrt{a^2+1-2x}(9xa^2-6a^2-3x^2-9x^2a+9xa-6a^3+6ax^2)\ge (x^2-a^2)(16x^2-16x+7)+a(9x+9xa^2-3a^2+3x^2-18x^2)-9x^2+18x^3-9x^4-7a^4-7x^4+14a^2x^2.$$ We check first that both sides are negative or zero: the left side factor is $l(x)=-3(a+1)(a-x)(2a-x)\le 0$. The right side is equal to: $f(x)=2x^3-x^2(2+2a^2+15a)+x(16a^2+9a+9a^3)-7a^2-3a^3-7a^4$, $f(a)=0$, $f(-a)\le 0$ then notice that $f'(x)\ge 0$, for $-a\le x\le a$, also we can write $f(x)=(x-a)(2x^2-x(2+2a^2+13a)+7a+3a^2+7a^3)$.

Simplifying by $(a-x)$ and squaring both sides we get the polynomial inequality $$(x+a)(4x^3+x^2(10(a-1)^2)+x(-5a^4-20a^3+47a^2-20a-5)+13a^5-30a^4+35a^3-30a^2+13a)=(x+a)p(x)\ge 0.$$ The inequality is true for $x=\pm a$ (by factorisation), for the third-degree polynomial $p$, the derivative has one negative and one positive root, so we need only to check $p(x)\ge 0$ when $0\le x\le a$, (for example, $-5a^4-20a^3+47a^2-20a-5 \le -5a^4-20a^3+50a^2-20a-5$ and this is $-5(x-1)^2(x^2+6x+1)\le 0$).

We take $p'(a)$, if $p'(a)\le 0$, $p$ is decreasing on $[0; a]$ and we are done, if $p'(a)>0$ we find the positive root $x_0$ of $p'(x)=0$ for $0\le x_0\le a$. Simplifying $p'(a)$ gives $p'(a)=19a^2-5a^4-5$. We only check $\dfrac{19-\sqrt{261}}{10}\le a^2\le \dfrac{19+\sqrt{261}}{10} $ or for simplicity $0.53\le a\le 1.88;$ we have $x_0=\dfrac{\sqrt{40a^4-40a^3+9a^2-40a+40}-5(a-1)^2}{6}.$ Replacing, $p(x_0)$ has the same sign as $$475a^6-348a^5-1560a^4+2920a^3-1560a^2-348a+475-(80a^4-80a^3+18a^2-80a+80)\sqrt{40a^4-40a^3+9a^2-40a+40}$$ written as $t(a)-w(a)\sqrt{40a^4-40a^3+9a^2-40a+40}.$ Check that $t(a)=475a^6-348a^5-1560a^4+2920a^3-1560a^2-348a+475\ge 475a^6-348a^5-1560a^4+2866a^3-1560a^2-348a+475\ge 0$.

The last polynomial is $(a-1)^2(475a^4+602a^3-831a^2+602a+475)$ and it is nonnegative for $a\le 1$ and $a\ge 1$. Also $w(a)\ge 0$ as $a^4-a^3-a+1\ge 0$.

Take finally $t^2-w^2(40a^4-40a^3+9a^2-40a+40)$. This equals $$-243(a-1)^2(a+1)^2(a^2-5a+1)^2(125a^4-550a^3+847a^2-550a+125). $$ Verify only that $a^4+1-4.4(a^3+a)+\frac{847}{125}a^2=q(a)\le 0$ when $0.53 \le a \le 1.88$. Approximating its roots or the ones of its derivative or also consider $q(a)\le a^4+1-4.4(a^3+a)+6.8a^2=0.2(a-1)^2(5a^2-12a+5)$. (Calculation are assisted by GeoGebra, online).

The answer of Iosif helps one try to change variables as the solution verify $x^2+y^2=1$. This one is a bit long but oddly it has many powerful factorisation arguments.

So set $\sqrt{x^2+y^2}=a$ and $y^2=a^2-x^2$ with $-a\le x\le a$. The inequality is then: $$\sqrt{a^2+1-2x}(9xa^2-6a^2-3x^2-9x^2a+9xa-6a^3+6ax^2)\ge (x^2-a^2)(16x^2-16x+7)+a(9x+9xa^2-3a^2+3x^2-18x^2)-9x^2+18x^3-9x^4-7a^4-7x^4+14a^2x^2.$$ We check first that both sides are negative or zero: the left side factor is $l(x)=-3(a+1)(a-x)(2a-x)\le 0$. The right side is equal to: $f(x)=2x^3-x^2(2+2a^2+15a)+x(16a^2+9a+9a^3)-7a^2-3a^3-7a^4$, $f(a)=0$, $f(-a)\le 0$ then notice that $f'(x)\ge 0$, for $-a\le x\le a$, also we can write $f(x)=(x-a)(2x^2-x(2+2a^2+13a)+7a+3a^2+7a^3)$.

Simplifying by $(a-x)$ and squaring both sides we get the polynomial inequality $$(x+a)(4x^3+x^2(10(a-1)^2)+x(-5a^4-20a^3+47a^2-20a-5)+13a^5-30a^4+35a^3-30a^2+13a)=(x+a)p(x)\ge 0.$$ The inequality is true for $x=\pm a$ (by factorisation), for the third-degree polynomial $p$, the derivative has one negative and one positive root, so we need only to check $p(x)\ge 0$ when $0\le x\le a$, (for example, $-5a^4-20a^3+47a^2-20a-5 \le -5a^4-20a^3+50a^2-20a-5$ and this is $-5(x-1)^2(x^2+6x+1)\le 0$).

We take $p'(a)$, if $p'(a)\le 0$, $p$ is decreasing on $[0; a]$ and we are done, if $p'(a)>0$ we find the positive root $x_0$ of $p'(x)=0$ for $0\le x_0\le a$. Simplifying $p'(a)$ gives $p'(a)=19a^2-5a^4-5$. We only check $\dfrac{19-\sqrt{261}}{10}\le a^2\le \dfrac{19+\sqrt{261}}{10} $ or for simplicity $0.53\le a\le 1.88;$ we have $x_0=\dfrac{\sqrt{40a^4-40a^3+9a^2-40a+40}-5(a-1)^2}{6}.$ Replacing, $p(x_0)$ has the same sign as $$475a^6-348a^5-1560a^4+2920a^3-1560a^2-348a+475-(80a^4-80a^3+18a^2-80a+80)\sqrt{40a^4-40a^3+9a^2-40a+40}$$ written as $t(a)-w(a)\sqrt{40a^4-40a^3+9a^2-40a+40}.$ Check that $$t(a)=475a^6-348a^5-1560a^4+2920a^3-1560a^2-348a+475\ge 475a^6-348a^5-1560a^4+2866a^3-1560a^2-348a+475\ge 0.$$

The last polynomial is $(a-1)^2(475a^4+602a^3-831a^2+602a+475)$ and it is nonnegative for $a\le 1$ and $a\ge 1$. Also $w(a)\ge 0$ as $a^4-a^3-a+1\ge 0$.

Take finally $t^2-w^2(40a^4-40a^3+9a^2-40a+40)$. This equals $$-243(a-1)^2(a+1)^2(a^2-5a+1)^2(125a^4-550a^3+847a^2-550a+125). $$ Verify only that $a^4+1-4.4(a^3+a)+\frac{847}{125}a^2=q(a)\le 0$ when $0.53 \le a \le 1.88$. Approximating its roots or the ones of its derivative or also consider $q(a)\le a^4+1-4.4(a^3+a)+6.8a^2=0.2(a-1)^2(5a^2-12a+5)$. (Calculation are assisted by GeoGebra, online).

The answer of Iosif helps one try to change variables as the solution verify $x^2+y^2=1$. This one is a bit long but oddly it has many powerful factorisation arguments.

So set $\sqrt{x^2+y^2}=a$ and $y^2=a^2-x^2$ with $-a\le x\le a$. The inequality is then: $$\sqrt{a^2+1-2x}(9xa^2-6a^2-3x^2-9x^2a+9xa-6a^3+6ax^2)\ge (x^2-a^2)(16x^2-16x+7)+a(9x+9xa^2-3a^2+3x^2-18x^2)-9x^2+18x^3-9x^4-7a^4-7x^4+14a^2x^2.$$ We check first that both sides are negative or zero: the left side factor is $l(x)=-3(a+1)(a-x)(2a-x)\le 0$. The right side is equal to: $f(x)=2x^3-x^2(2+2a^2+15a)+x(16a^2+9a+9a^3)-7a^2-3a^3-7a^4$, $f(a)=0$, $f(-a)\le 0$ then notice that $f'(x)\ge 0$, for $-a\le x\le a$, also we can write $f(x)=(x-a)(2x^2-x(2+2a^2+13a)+7a+3a^2+7a^3)$.

Simplifying by $(a-x)$ and squaring both sides we get the polynomial inequality $$(x+a)(4x^3+x^2(10(a-1)^2)+x(-5a^4-20a^3+47a^2-20a-5)+13a^5-30a^4+35a^3-30a^2+13a)=(x+a)p(x)\ge 0.$$ The inequality is true for $x=\pm a$ (by factorisation), for the third degree polynomial $p$, the derivative has one negative and one positive root so we need only to check $p(x)\ge 0$ when $0\le x\le a$,  (for example $-5a^4-20a^3+47a^2-20a-5\le -5a^4-20a^3+50a^2-20a-5=-5(x-1)^2(x^2+6x+1)\le 0$).

We take $p'(a)$, if $p'(a)\le 0$, $p$ is decreasing on $[0;a]$ and we are done, if $p'(a)>0$ we find the positive root $x_0$ of $p'(x)=0$ for $0\le x_0\le a$. Simplifying $p'(a)$ gives $p'(a)=19a^2-5a^4-5$. We only check $\dfrac{19-\sqrt{261}}{10}\le a^2\le \dfrac{19+\sqrt{261}}{10} $ or for simplicity $0.53\le a\le 1.88;$ we have $x_0=\dfrac{\sqrt{40a^4-40a^3+9a^2-40a+40}-5(a-1)^2}{6}.$ Replacing, $p(x_0)$ has the same sign as $$475a^6-348a^5-1560a^4+2920a^3-1560a^2-348a+475-(80a^4-80a^3+18a^2-80a+80)\sqrt{40a^4-40a^3+9a^2-40a+40}$$ written as $t(a)-w(a)\sqrt{40a^4-40a^3+9a^2-40a+40}.$ Check that $t(a)=475a^6-348a^5-1560a^4+2920a^3-1560a^2-348a+475\ge 475a^6-348a^5-1560a^4+2866a^3-1560a^2-348a+475\ge 0$.

The last polynomial is $(a-1)^2(475a^4+602a^3-831a^2+602a+475)$ and it is nonnegative for $a\le 1$ and $a\ge 1$. Also $w(a)\ge 0$ as $a^4-a^3-a+1\ge 0$.

Take finally $t^2-w^2(40a^4-40a^3+9a^2-40a+40)$. This equals $$-243(a-1)^2(a+1)^2(a^2-5a+1)^2(125a^4-550a^3+847a^2-550a+125). $$ Verify only that $a^4+1-4.4(a^3+a)+\frac{847}{125}a^2=q(a)\le 0$ when $0.53 \le a \le 1.88$. Approximating its roots or the ones of its derivative or also consider $q(a)\le a^4+1-4.4(a^3+a)+6.8a^2=0.2(a-1)^2(5a^2-12a+5)$. (Calculation are assisted by GeoGebra, online).

The answer of Iosif helps one try to change variables as the solution verify $x^2+y^2=1$. This one is a bit long but oddly it has many powerful factorisation arguments.

So set $\sqrt{x^2+y^2}=a$ and $y^2=a^2-x^2$ with $-a\le x\le a$. The inequality is then: $$\sqrt{a^2+1-2x}(9xa^2-6a^2-3x^2-9x^2a+9xa-6a^3+6ax^2)\ge (x^2-a^2)(16x^2-16x+7)+a(9x+9xa^2-3a^2+3x^2-18x^2)-9x^2+18x^3-9x^4-7a^4-7x^4+14a^2x^2.$$ We check first that both sides are negative or zero: the left side factor is $l(x)=-3(a+1)(a-x)(2a-x)\le 0$. The right side is equal to: $f(x)=2x^3-x^2(2+2a^2+15a)+x(16a^2+9a+9a^3)-7a^2-3a^3-7a^4$, $f(a)=0$, $f(-a)\le 0$ then notice that $f'(x)\ge 0$, for $-a\le x\le a$, also we can write $f(x)=(x-a)(2x^2-x(2+2a^2+13a)+7a+3a^2+7a^3)$.

Simplifying by $(a-x)$ and squaring both sides we get the polynomial inequality $$(x+a)(4x^3+x^2(10(a-1)^2)+x(-5a^4-20a^3+47a^2-20a-5)+13a^5-30a^4+35a^3-30a^2+13a)=(x+a)p(x)\ge 0.$$ The inequality is true for $x=\pm a$ (by factorisation), for the third-degree polynomial $p$, the derivative has one negative and one positive root, so we need only to check $p(x)\ge 0$ when $0\le x\le a$,  (for example, $-5a^4-20a^3+47a^2-20a-5 \le -5a^4-20a^3+50a^2-20a-5$ and this is $-5(x-1)^2(x^2+6x+1)\le 0$).

We take $p'(a)$, if $p'(a)\le 0$, $p$ is decreasing on $[0; a]$ and we are done, if $p'(a)>0$ we find the positive root $x_0$ of $p'(x)=0$ for $0\le x_0\le a$. Simplifying $p'(a)$ gives $p'(a)=19a^2-5a^4-5$. We only check $\dfrac{19-\sqrt{261}}{10}\le a^2\le \dfrac{19+\sqrt{261}}{10} $ or for simplicity $0.53\le a\le 1.88;$ we have $x_0=\dfrac{\sqrt{40a^4-40a^3+9a^2-40a+40}-5(a-1)^2}{6}.$ Replacing, $p(x_0)$ has the same sign as $$475a^6-348a^5-1560a^4+2920a^3-1560a^2-348a+475-(80a^4-80a^3+18a^2-80a+80)\sqrt{40a^4-40a^3+9a^2-40a+40}$$ written as $t(a)-w(a)\sqrt{40a^4-40a^3+9a^2-40a+40}.$ Check that $t(a)=475a^6-348a^5-1560a^4+2920a^3-1560a^2-348a+475\ge 475a^6-348a^5-1560a^4+2866a^3-1560a^2-348a+475\ge 0$.

The last polynomial is $(a-1)^2(475a^4+602a^3-831a^2+602a+475)$ and it is nonnegative for $a\le 1$ and $a\ge 1$. Also $w(a)\ge 0$ as $a^4-a^3-a+1\ge 0$.

Take finally $t^2-w^2(40a^4-40a^3+9a^2-40a+40)$. This equals $$-243(a-1)^2(a+1)^2(a^2-5a+1)^2(125a^4-550a^3+847a^2-550a+125). $$ Verify only that $a^4+1-4.4(a^3+a)+\frac{847}{125}a^2=q(a)\le 0$ when $0.53 \le a \le 1.88$. Approximating its roots or the ones of its derivative or also consider $q(a)\le a^4+1-4.4(a^3+a)+6.8a^2=0.2(a-1)^2(5a^2-12a+5)$. (Calculation are assisted by GeoGebra, online).

The answer of Iosif helps one try to change variables as the solution verify $x^2+y^2=1$. This one is a bit long but oddly it has many powerful factorisation arguments.

So set $\sqrt{x^2+y^2}=a$ and $y^2=a^2-x^2$ with $-a\le x\le a$. The inequality is then: $$\sqrt{a^2+1-2x}(9xa^2-6a^2-3x^2-9x^2a+9xa-6a^3+6ax^2)\ge (x^2-a^2)(16x^2-16x+7)+a(9x+9xa^2-3a^2+3x^2-18x^2)-9x^2+18x^3-9x^4-7a^4-7x^4+14a^2x^2.$$ We check first that both sides are negative or zero: the left side factor is $l(x)=-3(a+1)(a-x)(2a-x)\le 0$. The right side is equal to: $f(x)=2x^3-x^2(2+2a^2+15a)+x(16a^2+9a+9a^3)-7a^2-3a^3-7a^4$, $f(a)=0$, $f(-a)\le 0$ then notice that $f'(x)\ge 0$, for $-a\le x\le a$, also we can write $f(x)=(x-a)(2x^2-x(2+2a^2+13a)+7a+3a^2+7a^3)$.

Simplifying by $(a-x)$ and squaring both sides we get the polynomial inequality $$(x+a)(4x^3+x^2(10(a-1)^2)+x(-5a^4-20a^3+47a^2-20a-5)+13a^5-30a^4+35a^3-30a^2+13a)=(x+a)p(x)\ge 0.$$ The inequality is true for $x=\pm a$ (by factorisation), for the third degree polynomial $p$, the derivative has one negative and one positive root so we need only to check $p(x)\ge 0$ when $0\le x\le a$, (for example $-5a^4-20a^3+47a^2-20a-5\le -5a^4-20a^3+50a^2-20a-5=-5(x-1)^2(x^2+6x+1)\le 0$).

We take $p'(a)$, if $p'(a)\le 0$, $p$ is decreasing on $[0;a]$ and we are done, if $p'(a)>0$ we find the positive root $x_0$ of $p'(x)=0$ for $0\le x_0\le a$. Simplifying $p'(a)$ gives $p'(a)=19a^2-5a^4-5$. We only check $\dfrac{19-\sqrt{261}}{10}\le a^2\le \dfrac{19+\sqrt{261}}{10} $ or for simplicity $0.53\le a\le 1.88;$ we have $x_0=\dfrac{\sqrt{40a^4-40a^3+9a^2-40a+40}-5(a-1)^2}{6}.$ Replacing, $p(x_0)$ has the same sign as $$475a^6-348a^5-1560a^4+2920a^3-1560a^2-348a+475-(80a^4-80a^3+18a^2-80a+80)\sqrt{40a^4-40a^3+9a^2-40a+40}$$ written as $t(a)-w(a)\sqrt{40a^4-40a^3+9a^2-40a+40}.$ Check that $t(a)=475a^6-348a^5-1560a^4+2920a^3-1560a^2-348a+475\ge 475a^6-348a^5-1560a^4+2866a^3-1560a^2-348a+475\ge 0$.

The last polynomial is $(a-1)^2(475a^4+602a^3-831a^2+602a+475)$ and it is nonnegative for $a\le 1$ and $a\ge 1$. Also $w(a)\ge 0$ as $a^4-a^3-a+1\ge 0$.

Take finally $t^2-w^2(40a^4-40a^3+9a^2-40a+40)$. This equals $$-243(a-1)^2(a+1)^2(a^2-5a+1)^2(125a^4-550a^3+847a^2-550a+125). $$ Verify only that $a^4+1-4.4(a^3+a)+\frac{847}{125}a^2=q(a)\le 0$ when $0.53 \le a \le 1.88$. Approximating its roots or the one of its derivative or also consider $q(a)\le a^4+1-4.4(a^3+a)+6.8a^2=0.2(a-1)^2(5a^2-12a+5)$. (Calculation are assisted by GeoGebra, online).

The answer of Iosif helps one try to change variables as the solution verify $x^2+y^2=1$. This one is a bit long but oddly it has many powerful factorisation arguments.

So set $\sqrt{x^2+y^2}=a$ and $y^2=a^2-x^2$ with $-a\le x\le a$. The inequality is then: $$\sqrt{a^2+1-2x}(9xa^2-6a^2-3x^2-9x^2a+9xa-6a^3+6ax^2)\ge (x^2-a^2)(16x^2-16x+7)+a(9x+9xa^2-3a^2+3x^2-18x^2)-9x^2+18x^3-9x^4-7a^4-7x^4+14a^2x^2.$$ We check first that both sides are negative or zero: the left side factor is $l(x)=-3(a+1)(a-x)(2a-x)\le 0$. The right side is equal to: $f(x)=2x^3-x^2(2+2a^2+15a)+x(16a^2+9a+9a^3)-7a^2-3a^3-7a^4$, $f(a)=0$, $f(-a)\le 0$ then notice that $f'(x)\ge 0$, for $-a\le x\le a$, also we can write $f(x)=(x-a)(2x^2-x(2+2a^2+13a)+7a+3a^2+7a^3)$.

Simplifying by $(a-x)$ and squaring both sides we get the polynomial inequality $$(x+a)(4x^3+x^2(10(a-1)^2)+x(-5a^4-20a^3+47a^2-20a-5)+13a^5-30a^4+35a^3-30a^2+13a)=(x+a)p(x)\ge 0.$$ The inequality is true for $x=\pm a$ (by factorisation), for the third degree polynomial $p$, the derivative has one negative and one positive root so we need only to check $p(x)\ge 0$ when $0\le x\le a$, (for example $-5a^4-20a^3+47a^2-20a-5\le -5a^4-20a^3+50a^2-20a-5=-5(x-1)^2(x^2+6x+1)\le 0$).

We take $p'(a)$, if $p'(a)\le 0$, $p$ is decreasing on $[0;a]$ and we are done, if $p'(a)>0$ we find the positive root $x_0$ of $p'(x)=0$ for $0\le x_0\le a$. Simplifying $p'(a)$ gives $p'(a)=19a^2-5a^4-5$. We only check $\dfrac{19-\sqrt{261}}{10}\le a^2\le \dfrac{19+\sqrt{261}}{10} $ or for simplicity $0.53\le a\le 1.88;$ we have $x_0=\dfrac{\sqrt{40a^4-40a^3+9a^2-40a+40}-5(a-1)^2}{6}.$ Replacing, $p(x_0)$ has the same sign as $$475a^6-348a^5-1560a^4+2920a^3-1560a^2-348a+475-(80a^4-80a^3+18a^2-80a+80)\sqrt{40a^4-40a^3+9a^2-40a+40}$$ written as $t(a)-w(a)\sqrt{40a^4-40a^3+9a^2-40a+40}.$ Check that $t(a)=475a^6-348a^5-1560a^4+2920a^3-1560a^2-348a+475\ge 475a^6-348a^5-1560a^4+2866a^3-1560a^2-348a+475\ge 0$.

The last polynomial is $(a-1)^2(475a^4+602a^3-831a^2+602a+475)$ and it is nonnegative for $a\le 1$ and $a\ge 1$. Also $w(a)\ge 0$ as $a^4-a^3-a+1\ge 0$.

Take finally $t^2-w^2(40a^4-40a^3+9a^2-40a+40)$. This equals $$-243(a-1)^2(a+1)^2(a^2-5a+1)^2(125a^4-550a^3+847a^2-550a+125). $$ Verify only that $a^4+1-4.4(a^3+a)+\frac{847}{125}a^2=q(a)\le 0$ when $0.53 \le a \le 1.88$. Approximating its roots or the ones of its derivative or also consider $q(a)\le a^4+1-4.4(a^3+a)+6.8a^2=0.2(a-1)^2(5a^2-12a+5)$. (Calculation are assisted by GeoGebra, online).