Given $0 < x \le 1 \le y$. Calculate the minimum value of $\frac{x}{y + 1} + \frac{y}{x + 1} + \frac{x^2 + 3y^2 + 2}{6xy(xy + 1)}$.

Question

Given $0 < x \le 1 \le y$. Calculate the minimum value of $$\large \dfrac{x}{y + 1} + \dfrac{y}{x + 1} + \dfrac{x^2 + 3y^2 + 2}{6xy \cdot (xy + 1)}$$

We have that $$\frac{x^2 + 3y^2 + 2}{6xy \cdot (xy + 1)} \ge \left[ \begin{align} \frac{2 \cdot (x + 3y + 2)^2}{12xy \cdot (6xy + 6)}\\ \frac{2 \cdot (2xy + 4y)}{12xy \cdot (xy + 1)} \end{align} \right. = \frac{8 \cdot y(\sqrt{x} + 2)^2}{9 \cdot (3xy + 1)^2}$$

But that's all I have.

Answer 1

Show that $$\frac{x}{y+1}+\frac{y}{x+1}+\frac{x^2+3y^2+2}{6xy(xy+1)}\geq \frac{3}{2}$$ The equal sign holds if $x=y=1$

Answer 2

For $x=y=1$ we'll get a value $\frac{3}{2}$.

We'll prove that it's a minimal value.

Indeed, let $y=1+a$.

Thus, $a\geq0$ and we need to prove that: $$6x^2a^4+(3+9x+21x^2-9x^3)a^3+3(4+9x+3x^2-10x^3+2x^4)a^2+$$ $$+(1-x)(17+37x+14x^2-12x^3)a+2(x+1)^2(1-x)(5-3x)\geq0,$$ which is obvious.