For $x,y \geqslant 0.$ Prove$:$ $$ \left( xy+2\,x+2\,y+1 \right) ^{2} \left( x+y+2 \right) ^{2}\geqslant \frac14
\left[xy(x+y)+2({x}^{2}+{y}^{2})-18\,xy+5(x+y)+2
\right] ^{2}+144\,xy \left( {x}^{2}+{y}^{2}+2 \right) \,\,(\ast)$$
First$,$ I checked it by [Wolfram|Alpha: Computational Intelligence](https://www.wolframalpha.com/)$,$ and know it's [true](https://tinyurl.com/y3ryfts9).

Morever$,$ it's stronger than [KaiRain's problem](https://artofproblemsolving.com/community/c6t243f6h1970813_prove_inequality_with_xy_gt0).

Equality holds when $\{x = 0, y = -2\}, \{x = -2, y = 0\}, \\\{x = 0, y = -\frac12\}, \{x = -\frac12, y = 0\},\\ \{x = 1, y = 3\} , \{x = 3, y = 1\}\,\text{and}\,\{x = 1, y = 1\},$

Now I try to prove it!

Let $S=x+y,\, P=xy.$ The problem can be writen as:

$$ ( P+2\,S+1 ) ^{2} ( S+2 ) ^{2}\geqslant \frac14 ( SP+
2{S}^{2}-22P+5S+2 ) ^{2}+144\cdot \text{P} ( {S}^{2}-2P+2
 ) 
\,\,(\text{1})$$

Or $$\frac{3}{4} ( P{S}^{2}+2{S}^{3}+20PS-71{S}^{2}+228P+52S-172
 ) ^{2}+48 ( 6{S}^{3}-13{S}^{2}+8S-28 ) 
 ( S-4 ) ^{2} \geqslant 0$$
If $ 6{S}^{3}-13{S}^{2}+8S-28 \geqslant 0$ then the inequality is true!

Which means $(\text{1})$ is true when $S\geqslant \frac{1}{18}\sqrt [3]{12997+324\,\sqrt {1609}}+{\frac {25}{18}}{\frac {1}{
\sqrt [3]{12997+324\,\sqrt {1609}}}}+{\frac {13}{18}}\approx 2.4148004292977.$

From here I don't know how to finish this proof.

I hope to see if there is a way to end my proof or another approach and also the *Sum Of Squares* of the inequality $(\ast)$ if if could be.

**Edit.** I found that the inequality $$ \left( xy+2\,x+2\,y+1 \right) ^{2} \left( x+y+2 \right) ^{2}\geqslant k
\left[xy(x+y)+2({x}^{2}+{y}^{2})-18\,xy+5(x+y)+2
\right] ^{2}+144\,xy \left( {x}^{2}+{y}^{2}+2 \right) \,\,$$
is true for all $k\leqslant k\_0 \approx 0.3074105436.$ Where $k\_0$ is a root of ${X}^{3}+{\frac {151}{150}}\,{X}^{2}+{\frac {13}{240}}\,X-{\frac {169}
{1200}}=0$

Who can prove this also$?$