Question
If $x,y>0,$ then prove $$\frac{1}{\sqrt[3]{xy}}+3\ge 2\sqrt{3}\left(\frac{1}{\sqrt{x+2}}+\frac{1}{\sqrt{y+2}}\right)$$ By Cauchy-Schwarz,$$\frac{1}{\sqrt{x+2}}+\frac{1}{\sqrt{y+2}}\le \frac{\sqrt{2(x+y+4)}}{\sqrt{xy+2(x+y)+4}}$$We will prove $$\frac{1}{\sqrt[3]{xy}}+3\ge 2\sqrt{6}.\sqrt{\frac{x+y+4}{(x+2)(y+2)}}=2\sqrt{6}.\sqrt{\frac{1}{x+2}+\frac{1}{y+2}},$$which is wrong when $a=\dfrac{1}{5}; b>29.$
I am looking for a good proof. Thanks.
Alternative proof:
Using $\sqrt[3]{xy} \le \frac{\sqrt{xy} + \sqrt{xy} + 1}{3}$ (by AM-GM), it suffices to prove that $$\frac{3}{2\sqrt{xy} + 1}+3\ge 2\sqrt{3}\left(\frac{1}{\sqrt{x+2}}+\frac{1}{\sqrt{y+2}}\right). \tag{1}$$
Letting $x = a^2, y = b^2$, it suffices to prove that $$\frac{3}{2ab + 1}+3\ge 2\sqrt{3}\left(\frac{1}{\sqrt{a^2+2}}+\frac{1}{\sqrt{b^2+2}}\right). \tag{2}$$
Squaring both sides, it suffices to prove that $$\left(\frac{3}{2ab + 1}+3\right)^2\ge 12\left(\frac{1}{a^2 + 2}+\frac{1}{b^2 + 2} + \frac{2}{\sqrt{(a^2+2)(b^2 + 2)}}\right). \tag{3}$$
By Cauchy-Bunyakovsky-Schwarz inequality, we have $$\frac{2}{\sqrt{(a^2 + 2)(b^2 + 2)}} \le \frac{2}{ab + 2}. \tag{4}$$
Also, we have $$\frac{1}{a^2+2}+\frac{1}{b^2+2} = \frac{a^2 + b^2 + 4}{a^2b^2 + 2(a^2 + b^2) + 4} = \frac12 + \frac{4 - a^2b^2}{2a^2b^2 + 4(a^2 + b^2) + 8}. \tag{5}$$
From (3)-(5), it suffices to prove that $$\left(\frac{3}{2ab + 1}+3\right)^2\ge 12\left(\frac12 + \frac{4 - a^2b^2}{2a^2b^2 + 4(a^2 + b^2) + 8} + \frac{2}{ab + 2}\right). \tag{6}$$
If $a^2b^2 > 4$, it suffices to prove that $$\left(\frac{3}{2ab + 1}+3\right)^2\ge 12\left(\frac12 + \frac{2}{ab + 2}\right) \tag{7}$$ which is written as ${\frac {6(2\,{u}^{3}-4\,{u}^{2}+5\,u+6)}{ \left( 2\,u+1 \right) ^{2} \left( u+2 \right) }} \ge 0$ where $u = ab$. True.
If $a^2b^2 \le 4$, using $a^2 + b^2 \ge 2ab$, it suffices to prove that $$\left(\frac{3}{2ab + 1}+3\right)^2\ge 12\left(\frac12 + \frac{4 - a^2b^2}{2a^2b^2 + 4\cdot 2ab + 8} + \frac{2}{ab + 2}\right)\tag{8}$$ which is written as ${\frac { 12\left( 3\,u+2 \right) \left( u-1 \right) ^{2}}{ \left( u +2 \right) \left( 2\,u+1 \right) ^{2}}} \ge 0$ where $u = ab$. True.
We are done.
Proof.
*Notice: The main idea is same as above Michael Rozenberg's proof.
Denoting $a^2=\sqrt[3]{x};b^2=\sqrt[3]{y}$, the OP turns out$$\frac{1}{a^2b^2}+3\ge 2\sqrt{3}\left(\frac{1}{\sqrt{a^6+2}}+\frac{1}{\sqrt{b^6+2}}\right).$$ By using Cauchy-Schwarz $\sqrt{3(a^6+2)}\ge a^3+2;\sqrt{3(b^6+2)}\ge b^3+2$ and we will prove $$\frac{1}{a^2b^2}+3\ge 6\left(\frac{1}{a^3+2}+\frac{1}{b^3+2}\right).$$ After expanding, we obtain $$3(ab)^5+(ab)^3+2(a^3+b^3+2)\ge 12(ab)^2.$$ Apply AM-GM to show that \begin{align*} &3(a^5b^5+a+b)\ge 9(ab)^2,\\& a^3b^3+a^3+b^3\ge 3(ab)^2,\\& a^3+1+1\ge 3a,b^3+1+1\ge 3b. \end{align*} Sum up cyclically above inequalities, we have $Q.E.D.$
Equality holds at $a=b=1$ or $x=y=1.$
Proof.
Denoting $a=\sqrt[3]{x};b=\sqrt[3]{y}$, the OP becomes $$\frac{1}{ab}+3\ge 2\sqrt{3}\left(\frac{1}{\sqrt{a^3+2}}+\frac{1}{\sqrt{b^3+2}}\right).$$ By using Cauchy-Schwarz $$ 4\left(\frac{\dfrac{1}{\sqrt{a}}}{\sqrt{\dfrac{a^3+2}{3a}}}+\frac{\dfrac{1}{\sqrt{b}}}{\sqrt{\dfrac{b^3+2}{3b}}}\right)^2\le 12\left(\frac{1}{a}+\frac{1}{b}\right)\left(\frac{a}{a^3+2}+\frac{b}{b^3+2}\right).$$ Also, by AM-GM $$\left(\frac{1}{ab}+3\right)^2=\frac{1}{a^2b^2}+\frac{6}{ab}+9\ge 8\left(\frac{1}{ab}+1\right).$$ Hence, it is enough to prove $$2\left(\frac{1}{ab}+1\right)\ge 3\left(\frac{1}{a}+\frac{1}{b}\right)\left(\frac{a}{a^3+2}+\frac{b}{b^3+2}\right),$$ or $$\frac{2}{3}.\frac{ab+1}{a+b}\ge\frac{a}{a^3+2}+\frac{b}{b^3+2}.$$ Notice that we can rewrite it as$$\frac{(a+2)(a-1)^2}{a^3+2}+\frac{(b+2)(b-1)^2}{b^3+2}\ge \frac{-2(a-1)(b-1)}{a+b}.\tag{*}$$ Now, we split $(*)$ into two cases.
In conlusion, $(*)$ is true and we end proof here. Equality holds at $a=b=1$ or $x=y=1.$
Let $x=a^6$ and $y=b^6$, where $a>0$ and $b>0$.
Thus, we need to prove that: $$\frac{1}{a^2b^2}+3\geq2\sqrt3\left(\frac{1}{\sqrt{a^6+2}}+\frac{1}{\sqrt{b^6+2}}\right).$$ Now, by C-S $$\sqrt{a^6+2}=\frac{1}{\sqrt3}\sqrt{(1+2)(a^6+2)}\geq\frac{1}{\sqrt{3}}(a^3+2),$$ which says that it's enough to prove: $$\frac{1}{a^2b^2}+3\geq6\left(\frac{1}{a^3+2}+\frac{1}{b^3+2}\right)$$ or $$3a^5b^5+a^3b^3-12a^2b^2+2a^3+2b^3+4\geq0,$$ which is true by AM-GM: $$3a^5b^5+a^3b^3-12a^2b^2+2a^3+2b^3+4\geq$$ $$\geq12\sqrt[12]{\left(a^5b^5\right)^3\cdot a^3b^3\cdot\left(\frac{a^3+b^3}{2}\right)^4\cdot1^4}-12a^2b^2\geq$$ $$\geq12\sqrt[12]{\left(a^5b^5\right)^3\cdot a^3b^3\cdot\left(\sqrt{a^3b^3}\right)^4\cdot1^4}-12a^2b^2=0.$$
My second proof:
We can use the isolated fudging.
It suffices to prove that $$\left(\frac{1}{\sqrt[3]{xy}}+3\right)\cdot \frac{y^{1/6}}{x^{1/6} + y^{1/6}}\ge 2\sqrt{3}\cdot \frac{1}{\sqrt{x+2}}. \tag{1}$$ (Note: Take cyclic sum on (1), we get the desired inequality.)
Letting $x = a^6, y = b^6$, (1) is written as $$\frac{3a^2b^2 + 1}{a^2b(a + b)} \ge \frac{2\sqrt 3}{\sqrt{a^6 + 2}} = \frac{6}{\sqrt{3(a^6 + 2)}}. \tag{2}$$
Using $\sqrt{3(a^6 + 1 + 1)} \ge a^3 + 1 + 1$ (using $3(u^2 + v^2 + w^2)\ge (u + v + w)^2$), it suffices to prove that $$\frac{3a^2b^2 + 1}{a^2b(a + b)} \ge \frac{6}{a^3 + 2}$$ or $$\frac{3a^5b^2 - 6a^3 b + a^3 + 2}{a^2b(a+b)(a^3 + 2)} \ge 0$$ which is true using $a^5b^2 + a^5b^2 + a^5b^2 + a^3 + 1 + 1 \ge 6\sqrt[6]{(a^5b^2)^3 a^3} = 6a^3 b$.
We are done.