Prove $( xy+2x+2y+1 ) ^{2} ( x+y+2 ) ^{2}\geqslant \frac14 [xy(x+y)+2(x^2+y^2)-18xy+5(x+y)+2 ] ^{2}+144xy ( {x}^{2}+{y}^{2}+2) $

Question

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$,$ and know it's true.

Morever$,$ it's stronger than KaiRain's problem.

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$?$

Answer 1

You made most of the work!

I like another substitutions and I'll end you work in the $uvw$'s substitutions.

Let $x+y=2u$ and $xy=v^2$.

Thus, we need to prove that $$(u^2+10u+57)v^4+2(4u^3-71u^2+26u-43)v^2+(4u+1)(u+1)^2\geq0,$$ which is true for $$(4u^3-71u^2+26u-43)^2-(u^2+10u+57)(4u+1)(u+1)^2\leq0$$ or $$64(u-2)^2(12u^3-13u^2+4u-7)\geq0.$$ Id est, it's enough to prove our inequality for $$12u^3-13u^2+4u-7\leq0,$$ which gives $$0\leq u\leq1.2...$$ and it's enough to prove that $$v^2\leq\frac{-4u^3+71u^2-26u+43-8(2-u)\sqrt{7-4u+13u^2-12u^3}}{u^2+10u+57}$$ and since by AM-GM $v^2\leq u^2$, it's enough to prove that $$u^2\leq\frac{-4u^3+71u^2-26u+43-8(2-u)\sqrt{7-4u+13u^2-12u^3}}{u^2+10u+57}$$ or $$43-26u+14u^2-14u^3-u^4\geq8(2-u)\sqrt{7-4u+13u^2-12u^3}$$ and since $$43-26u+14u^2-14u^3-u^4>0$$ for any $0<u\leq1.2...,$ it's enough to prove that $$(43-26u+14u^2-14u^3-u^4)^2\geq64(2-u)^2(7-4u+13u^2-12u^3)$$ or $$(u-1)^2(u^2+10u+57)(u^4+20u^3-30u^2+12u+1)\geq0$$ which is obvious even for any positive value of $u$.

Answer 2

Proceeding along your approach:

If $S \ge \frac{5}{2}$, it is true.

If $0 \le S < \frac{5}{2}$, since $S^2 \ge 4P$, we have \begin{align} &P{S}^{2}+2{S}^{3}+20PS-71{S}^{2}+228P+52S-172\\ =\ & (S^2+20S+228)P+2S^3-71S^2+52S-172 \\ \le\ & (S^2+20S+228)\frac{S^2}{4}+2S^3-71S^2+52S-172\\ =\ & \frac{1}{4}S^4+7S^3-14S^2+52S-172 \\ <\ & 0 \end{align} and thus it suffices to prove that $$\frac{3}{4} ( \frac{1}{4}S^4+7S^3-14S^2+52S-172 ) ^{2}+48 ( 6{S}^{3}-13{S}^{2}+8S-28 ) ( S-4 ) ^{2} \ge 0$$ that is $$\frac{3}{64}(S^4+40S^3-120S^2+96S+16)(S^2+20S+228)(S-2)^2 \ge 0$$ which is true.

We are done.