Prove Inequality x,y with $x+y=2$

[maxkor]

Let x,y>0 and x+y=2. Prove $\sqrt{x+\sqrt[3]{y^2+7}}+\sqrt{y+\sqrt[3]{x^2+7}}\geq2\sqrt3$

 

[jvicens]

Hello!

To prove this inequality, we can use the Cauchy-Schwarz inequality, which states that for any real numbers $a_1, a_2, ..., a_n$ and $b_1, b_2, ..., b_n$,

$(a_1^2+a_2^2+...+a_n^2)(b_1^2+b_2^2+...+b_n^2)\geq(a_1b_1+a_2b_2+...+a_nb_n)^2$

In our case, we can let $a_1=\sqrt{x}$, $a_2=\sqrt[3]{y^2+7}$, $b_1=1$, and $b_2=1$. Then, the Cauchy-Schwarz inequality becomes:

$(x+1)(\sqrt[3]{y^2+7}+1)\geq(\sqrt{x}+\sqrt[3]{y^2+7})^2$

Similarly, we can also let $a_1=\sqrt{y}$, $a_2=\sqrt[3]{x^2+7}$, $b_1=1$, and $b_2=1$. The Cauchy-Schwarz inequality then becomes:

$(y+1)(\sqrt[3]{x^2+7}+1)\geq(\sqrt{y}+\sqrt[3]{x^2+7})^2$

Now, we can add these two inequalities together and use the fact that $x+y=2$ to get:

$(x+1)(\sqrt[3]{y^2+7}+1)+(y+1)(\sqrt[3]{x^2+7}+1)\geq(\sqrt{x}+\sqrt[3]{y^2+7})^2+(\sqrt{y}+\sqrt[3]{x^2+7})^2$

Simplifying, we get:

$2(x+y+2+\sqrt[3]{(x^2+7)(y^2+7)})\geq2(\sqrt{x}+\sqrt{y}+\sqrt[3]{x^2+7}+\sqrt[3]{y^2+7})^2$

Using the fact that $x+y=2$, we can further simplify to:

$2(4+\