sqing wrote: Let $x,y$ be positive real numbers .Find the minimum of $x+y+\frac{|x-1|}{y}+\frac{|y-1|}{x}$. For $x=y=1$ we get $x+y+\frac{|x-1|}{y}+\frac{|y-1|}{x}=2$. If $(x-1)(y-1)\geq0$ then the inequality is obviously true. Let $x>1$ and $y<1$. Hence, $x+y+\frac{|x-1|}{y}+\frac{|y-1|}{x}-2=(x-1)\left(1+\frac{1}{y}-\frac{1}{x}+\frac{y}{x}\right)\geq0$. Id est, the answer is $2$.

If $(x-1)(y-1)\ge 0$ ,then it also possible that $x,y\le 1$ and then the inequality is not obvious.

alexheinis wrote: If $(x-1)(y-1)\ge 0$ ,then it also possible that $x,y\le 1$ and then the inequality is not obvious. If $x,y \le 1$, then we have \[x+y+\frac{|x-1|}{y}+\frac{|y-1|}{x}-2=x+y+\frac{1-x}{y}+\frac{1-y}{x}-2=\frac{x+y}{xy} \cdot (1-x)(1-y) \ge 0\]so this case is also not hard.

Nice, I didn't see that.

We will consider $4$ cases so that we can get rid of absolute value sign. Case 1: $ \quad \quad x , y \geq 1$ $$x+y+ \frac{|x-1|}{y} +\frac{|y-1|}{x} = x+y+\frac{x-1}{y}+\frac{y-1}{x}$$ We have, $$x \geq 1 \quad \quad y \geq 1 $$ Therefore, $$x+y+\frac{x-1}{y}+\frac{y-1}{x} \geq 1+1+0+0 =2$$ Case 2: $\quad \quad 1 \geq x, y \geq 0$ $$x+y+ \frac{|x-1|}{y} +\frac{|y-1|}{x} \geq x+y+\frac{1-x}{y}+\frac{1-y}{x} $$ Observe that, $$x+y+\frac{1-x}{y}+\frac{1-y}{x} \geq 2$$ since this reduces to, $$\Leftrightarrow xy(x+y)+ x-x^2 +y-y^2 \geq 2xy $$ $$ \Leftrightarrow xy(x+y) + (x+y) -(x+y)^2 \geq 0$$ $$ \Leftrightarrow (x+y)(xy +1 -x-y) \geq 0$$ $$\Leftrightarrow (x+y)(1-x)(1-y) \geq 0$$ which is obviously true since, $ \quad \quad 1 \geq x , y \geq 0$ Case 3: $ \quad \quad x \geq 1 \geq y \geq 0$ $$x+y+ \frac{|x-1|}{y} +\frac{|y-1|}{x} \geq x+y+\frac{x-1}{y}+\frac{1-y}{x}$$ Observe that, we have $$x+y+\frac{x-1}{y}+\frac{1-y}{x} \geq 2$$ since this is equivalent to $$\Leftrightarrow x^2y+xy^2+x^2-x+y-y^2 \geq 2xy $$ $$\Leftrightarrow x^2y-xy+xy^2-xy+x^2 -x+y-y^2 \geq 0$$ $$\Leftrightarrow xy(x-1) +xy(y-1) + x(x-1)+ y(1-y) \geq 0$$ $$\Leftrightarrow (xy+x)(x-1) +(1-y)(y-xy) \geq 0$$ $$ \Leftrightarrow (xy+x)(x-1) +(y-y^2)(1-x) \geq 0$$ $$ \Leftrightarrow (x-1)(xy+x-y+y^2) \geq 0$$ which is obviously true since $x\geq 1$ and $x \geq y$. Case 4 : $ \quad \quad y \geq 1 \geq x$ Due to symmetry this case is similar to the previous case. Hence, the minimum value of the expression is $\boxed{2}$.

sqing wrote: Let $x,y$ be positive real numbers .Find the minimum of $x+y+\frac{|x-1|}{y}+\frac{|y-1|}{x}$.

Attachments:

@above Wow... Very short. I missed it.

Let $x,y>0$. Prove that $$x+y+\frac{|x-1|}{y}+\frac{|2y-1|}{x}\ge \frac{3}{2}$$$$x+y+\frac{|2x-1|}{y}+\frac{|3y-1|}{x}\ge \frac{5}{6}$$$$x+2y+\frac{|3x-1|}{y}+\frac{|4y-1|}{x}\ge \frac{5}{6}$$

sqing wrote: Let $x,y$ be positive real numbers .Find the minimum of $x+y+\frac{|x-1|}{y}+\frac{|y-1|}{x}$.

Attachments:

Let $n \ge 2,x_i>0,x_1=x_{n+1.} $ We have that $ \sum\limits_{i=1}^nx_i+\sum\limits_{i=1}^n\frac{\mid{x_i-1}\mid}{x_{i+1}}\ge n. $

Let $x,y>0$. Prove that$$x+y+ |x^k-1|+\frac{|y^k-1|}{x} \ge 2$$Where $ k=1,2,3,4,5.$