$\sum_{cyc} \frac{a+b+c+2}{c+2}\ge 5 \iff \sum_{cyc} \frac{1}{c+2} \ge 1$. By C-S, $(a+2+b+2+c+2)(\sum_{cyc} \frac{1}{c+2})\ge 9$, so we are done. Equality holds iff $a=b=c=1$.

$$\frac{a+b}{c+2} + \frac{b+c}{a+2} + \frac{c+a}{b+2} \geq 2$$$$\frac{a}{c+2} + \frac{b}{a+2} + \frac{c}{b+2}+\frac{b}{c+2} + \frac{c}{a+2} + \frac{a}{b+2} \geq 2 (?)$$By $$\frac{2(a+b+c)^2}{ab+bc+ca+6} \geq 2$$Note that $ab+bc+ca \le 3$ Proof: $$(a+b+c)^2 \geq 3(ab+bc+ca)$$And its true So we have $$\frac{2(a+b+c)^2}{9} \geq 2$$And its true.

I will provide 2 solutions to this problem: Solution 1: Jensen's inequality Notice that $a+b=3-c, b+c=3-a, c+a=3-b$, thus $\sum_{cyc}\frac{a+b}{c+2}=\sum_{cyc}\frac{3-c}{c+2}$ Furthermore let $f(x)=\frac{3-x}{x+2}\Longrightarrow f'(x)=-\frac{5}{(x+2)^2}\Longrightarrow f''(x)=\frac{10}{(x+2)^3}$, thus the function is convex over the positive reals. Now, $\sum_{cyc}\frac{3-c}{c+2}=\sum_{cyc}f(a)\overset{\text{Jensen}}{\ge}3f(\frac{a+b+c}{3})=3f(1)=2$ $\blacksquare$ Solution 2: Cauchy-Schwartz $\sum_{cyc}\frac{a+b}{c+2}\overset{\text{C-S}}{\ge}\frac{36}{4\sum_{cyc}a+2\sum_{cyc}ab}=\frac{18}{6+\sum_{cyc}ab}$ Thus the inequality boils down to $\frac{18}{6+\sum_{cyc}ab}\ge2\Longleftrightarrow 3\ge\sum_{cyc}ab\Longleftrightarrow\bigg(\sum_{cyc}a\bigg)^2\ge3\sum_{cyc}ab$, which is clearly true $\blacksquare$

leannan-capall wrote: Suppose that $a, b, c$ are positive real numbers and $a + b + c = 3$. Prove that $$\frac{a+b}{c+2} + \frac{b+c}{a+2} + \frac{c+a}{b+2} \geq 2$$ and determine when equality holds. $a+b+c=p=3, ab+ac+bc=q, abc=r$ Then we write the expression in the common denominator, so inequality becomes:$4p^2-4q+pq-3r+8p\ge 2r+4q+8p+16$ By $p=3$, we must to show that $20\ge 5r+5q$ We know that $p^2\ge 3q$ , so $3 \ge q$ and $p^3\ge 27r$ ,so $1\ge r$ by these we may say that $20\ge 5r+5q$.So we are done.

Suppose that $a, b, c$ are positive real numbers and $a + b + c = 3$. Prove that $$\frac{a+b}{c+k} + \frac{b+c}{a+k} + \frac{c+a}{b+k} \geq \frac{6}{k+1}$$Where $k>0.$ $$\frac{a+b}{c+1} + \frac{b+c}{a+1} + \frac{c+a}{b+1} \geq 3$$$$\frac{a+b}{5c+1} + \frac{b+c}{5a+1} + \frac{c+a}{5b+1} \geq 1$$

sqing wrote: Suppose that $a, b, c$ are positive real numbers and $a + b + c = 3$. Prove that $$\frac{a+b}{c+k} + \frac{b+c}{a+k} + \frac{c+a}{b+k} \geq \frac{6}{k+1}$$Where $k>0.$ $$\iff$$$$\frac{1}{a+k} + \frac{1}{b+k} +\frac{1}{c+k} \geq \frac{3}{k+1}$$

sqing wrote: Suppose that $a, b, c$ are positive real numbers and $a + b + c = 3$. Prove that $$\frac{a+b}{c+k} + \frac{b+c}{a+k} + \frac{c+a}{b+k} \geq \frac{6}{k+1}$$Where $k>0.$ $$\frac{a+b}{c+1} + \frac{b+c}{a+1} + \frac{c+a}{b+1} \geq 3$$$$\frac{a+b}{5c+1} + \frac{b+c}{5a+1} + \frac{c+a}{5b+1} \geq 1$$ $$\frac{a+b}{c+k} + \frac{b+c}{a+k} + \frac{c+a}{b+k} \geq \frac{6}{k+1}$$$$\frac{a}{c+k} + \frac{b}{a+k} + \frac{c}{b+k}+\frac{b}{c+k} + \frac{c}{a+k} + \frac{a}{b+k} \geq \frac{6}{k+1} (?)$$By $$\frac{2(a+b+c)^2}{ab+bc+ca+3k} \geq \frac{6}{k+1} $$Note that $ab+bc+ca \le 3$ Proof: $$(a+b+c)^2 \geq 3(ab+bc+ca)$$And its true

sqing wrote: Suppose that $a, b, c$ are positive real numbers and $a + b + c = 3$. Prove that $$\frac{a+b}{c+k} + \frac{b+c}{a+k} + \frac{c+a}{b+k} \geq \frac{6}{k+1}$$Where $k>0.$ $\frac{a+b}{c+k}=\frac{3-c}{c+k}$ so we can rewrite our inequality as $$f(a)+f(b)+f(c)\geq\cfrac{6}{k+1}$$where $f(x)=\frac{3-x}{x+k}$ We have the inequality $$f(x)\geq \cfrac{2}{1+k}-\cfrac{k+3}{(k+1)^2}(x-1) \iff \cfrac{(k+3)(x-1)^2}{(k+1)^2(k+x)}\geq 0$$clearly true for all reals. Now we have that $f(a)+f(b)+f(c)\geq \cfrac{6}{k+1}+\cfrac{k+3}{(k+1)^2}(a+b+c-3)=\cfrac{6}{k+1}$ $\square$

Suppose that $a, b, c$ are positive real numbers and $a + b + c = 3$. Prove that $$\frac{6}{k}\geq\frac{a+b}{c+k} + \frac{b+c}{a+k} + \frac{c+a}{b+k} \geq \frac{6}{k+1}$$Where $k>0.$

Suppose that $a, b, c\geq 0$ and $a + b + c = 3$. Prove that $$\frac{6}{k}\geq\frac{a+b}{c+k} + \frac{b+c}{a+k} + \frac{c+a}{b+k} \geq \frac{6}{k+1}$$Where $k>0.$

sqing wrote:

Suppose that $a, b, c\geq 0$ and $a + b + c = 3$. Prove that $$\frac{6}{k}\geq\frac{a+b}{c+k} + \frac{b+c}{a+k} + \frac{c+a}{b+k} \geq \frac{6}{k+1}$$Where $k>0.$ $$\frac{a+b}{c+k} + \frac{b+c}{a+k} + \frac{c+a}{b+k} \leq \frac{a+b}{k} + \frac{b+c}{k} + \frac{c+a}{k} =\frac{6}{k}$$

$\sum \dfrac{a+b}{c+2}\ge 2\Rightarrow \sum\dfrac{a+b}{c+2}\sum c+2\ge 18$ By Cauchy $\sum\dfrac{a+b}{c+2}\sum c+2\ge (\sum\sqrt{a+b})^{2}$ By Holder $(\sum\sqrt{a+b})^{2}\ge 3\sum a+b=18$

SQTHUSH wrote: $\sum \dfrac{a+b}{c+2}\ge 2\Rightarrow \sum\dfrac{a+b}{c+2}\sum c+2\ge 18$ By Cauchy $\sum\dfrac{a+b}{c+2}\sum c+2\ge (\sum\sqrt{a+b})^{2}$ By Holder $(\sum\sqrt{a+b})^{2}\ge 3\sum a+b=18$ The last part is wrong: It is $(x+y+z)^2\geq 3(x^2+y^2+z^2)$ where $x=\sqrt{a+b}, y=\sqrt{b+c}, z=\sqrt{c+a}$. By expanding this we get $xy+yz+zx\geq x^2+y^2+z^2$, false from am-gm.

$$\frac{a+b}{c+2}+\frac{b+c}{a+2}+\frac{c+a}{b+2}=\frac{(a+b)^2}{(c+2)(a+b)}+\frac{(b+c)^2}{(a+2)(b+c)}+\frac{(c+a)^2}{(b+2)(c+a)}\geq \frac{((a+b)+(b+c)+(c+a))^2}{(c+2)(a+b)+(a+2)(b+c)+(b+2)(c+a)}=\frac{36}{12+2(ab+bc+ca)}\geq 2$$