parmenides51 wrote: Positive real numbers $a, b, c$ satisfy $abc = 1$. Prove that $$\frac{a}{1+b}+\frac{b}{1+c}+\frac{c}{1+a} \ge \frac32$$ here $$2(a^2 + b^2 + c^2) + ab + bc + ca \geq (a+b+c)^2 \geq 3(a+b+c)$$$$2(a^2 + b^2 + c^2) + ab + bc + ca \geq 3(a+b+c)\iff \frac{(a+b+c)^2}{a+b+c+ab+bc+ca} \geq \frac{3}{2} $$$$\frac{a}{1+b}+\frac{b}{1+c}+\frac{c}{1+a} \geq \frac{(a+b+c)^2}{a+b+c+ab+bc+ca} \geq \frac{3}{2} $$here

Let $a, b, c$ be positive real numbers. Prove that $$\frac{a}{1+b}+\frac{b}{1+c}+\frac{c}{1+a} \ge \frac{9abc}{4(1+abc)}$$

Let $a, b, c$ be positive real numbers. Prove that$$\frac{a}{1+b}+\frac{b}{1+c}+\frac{c}{1+a} \ge \frac{a+b+c+3abc}{2(1+abc)}$$

@below thanks for detect my mistake.

Albert123 wrote: This solution is okay? By AM-AG: $\frac{a}{1+b}+\frac{b}{1+c}+\frac{c}{1+a} \ge 3\sqrt[3]{\frac{abc}{(1+b)(1+c)(1+a)}}\ge 3\sqrt[3]{\frac{abc}{8abc}}\ge \frac{3}{2}$.$\blacksquare$ Why does $(a+1)(b+1)(c+1)\le8abc$?

Just let a=x/y,b=y/z,c=z/x Prove that xz/y(y+z)+...=(xz)^2/xyz(y+z)+... >=3/2 This step just use C-S

jasperE3 wrote: Albert123 wrote: This solution is okay? By AM-AG: $\frac{a}{1+b}+\frac{b}{1+c}+\frac{c}{1+a} \ge 3\sqrt[3]{\frac{abc}{(1+b)(1+c)(1+a)}}\ge 3\sqrt[3]{\frac{abc}{8abc}}\ge \frac{3}{2}$.$\blacksquare$ Why does $(a+1)(b+1)(c+1)\le8abc$? Good point

Old Idea We can write $a=\frac{x}{y}$ , $a=\frac{y}{z}$ , $a=\frac{z}{x}$ . Then we need to prove : $$\sum \frac{a}{1+b} = \sum \frac{xy}{y(y+z)} \geq \frac{3}{2}$$ Also we know $\sum \frac{xy}{y(y+z)} = \sum \frac{(xz)^2}{y+z} \geq \frac{(xy+yz+zx)^2}{2(x+y+z)} \geq \frac{3}{2}$ Its mean we need to prove : $$(\sum xy)^2 \geq 3(\sum x)$$ And it is true ( This famous inequality ) : $(\sum xy)^2 \geq 3xyz(\sum x)$ $\blacksquare$

By Cauchy Schwarz inequality \[\sum{\frac{a}{b+1}=\sum{\frac{a^2}{ab+a}}} \geq \frac{(a+b+c)^2}{ab+bc+ca+a+b+c}\]Let's prove that \[\frac{(a+b+c)^2}{ab+bc+ca+a+b+c} \geq \frac{3}{2}\]\[2.\sum{a^2}+4.\sum{ab} \geq 3.\sum{ab}+3.\sum{a}\]\[2.\sum{a^2}+\sum{ab} \geq 3.\sum{a}\]\[2.\sum{a^2}+\sum{\frac{1}{a}} \geq 3.\sum{a}\]By AM-GM \[a^2+a^2+\frac{1}{a} \geq 3a\]\[b^2+b^2+\frac{1}{b} \geq 3b\]\[c^2+c^2+\frac{1}{c} \geq 3c\]By adding these three inequalities, we get the desired conclusion.