parmenides51 wrote: Show that for all real numbers $x$ and $y$ with $x > -1$ and $y > -1$ and $x + y = 1$ the inequality $$\frac{x}{y + 1} +\frac{y}{x + 1} \ge \frac23$$holds. When does equality apply? (Walther Janous) Because $$\frac{x}{y + 1} +\frac{y}{x + 1}-\frac23=\frac{2(x-y)^2}{3(x+1)(y+1)}\geq0.$$

Replace $x + 1$ by $x$ and $y + 1$ by $y$. Then we need to show that for positive reals $x, y$, $\frac{x - 1}{y} + \frac{y - 1}{x} \ge \frac{2}{3}$. Homogenizing, we need to show that $\frac{2x - y}{3y} + \frac{2y - x}{3x} \ge \frac{2}{3}$, which is equivalent to showing that $\frac{x}{y} + \frac{y}{x} \ge 2$, which is just AM-GM.

$\frac{x}{y+1}+\frac{y}{x+1} \geq \frac{2(x+y)}{x+y+2}$ with equality for $x=y$

$\frac{x}{y+1}+\frac{y}{x+1}\geq \frac{2}{3}$ $3(x^2+y^2)+3 \geq 2(xy)+4$ $2(x^2+y^2)\geq1$ $x^2+y^2\geq 2xy$

Let $x,y > -1$ and $x + y = 1.$ Prove that $$\frac{x}{y + 1} +\frac{2y}{x + 1} \ge \frac{4\sqrt 2}3-1$$$$\frac{x}{y + 1} +\frac{ y}{x +2}\ge \sqrt{\frac{3}2}-\frac{3}4$$

$3(x^2+x+y^2+y)\geq2(xy+x+y+1)$ $3(x^2+y^2+1)\geq2(xy+2)$ $3x^2+3y^2+3\geq 2xy+4$ $3x^2-2xy+3y^2\geq1$ Replacing $y=1-x$ gives us $$2(2x-1)^2\geq0$$Equality applies if $x=y=\frac12$