Annoying problem. We want to prove \[ S_1:=\frac{a}{a+b}+\frac{3c}{a+b}+\frac{c}{b+c}+\frac{3a}{b+c}+\frac{4b}{c+a}\ge 6. \]Note that by Cauchy-Schwarz (aka T2's lemma), \[ S_1S_2\ge 16(a+b+c)^2 \]where \[ S_2 = a(a+b)+3c(a+b)+c(b+c)+3a(b+c)+4b(c+a). \]So, it suffices to prove \[ \frac{16(a+b+c)^2}{S_2}\ge 6\Leftrightarrow 10a^2+16b^2+10c^2\ge 16ab + 4ac+16bc, \]after some algebra. The least inequality, however, is equivalent to \[ 8(a-b)^2 + 8(b-c)^2 + 2(c-a)^2\ge 0, \]which is clear.

vicentev wrote: Prove that if \( a \), \( b \), and \( c \) are positive real numbers, then the following inequality holds: \[ \frac{a + 3c}{a + b} + \frac{c + 3a}{b + c} + \frac{4b}{c + a} \geq 6. \] It can be written equivalently as: $$\frac{3a+2b+3c}{a+b}+\frac{3a+2b+3c}{b+c}+\frac{2a+4b+2c}{c+a}\geq 12$$ Using AM-GM $$\frac{3a+2b+3c}{a+b}+\frac{3a+2b+3c}{b+c}+\frac{2a+4b+2c}{c+a}\geq 3\sqrt[3]{\frac{(3a+2b+3c)^2(2a+4b+2c)}{(a+b)(b+c)(c+a)}}$$ It's enough to prove: $$(3a+2b+3c)^2(2a+4b+2c) \geq 64(a+b)(b+c)(c+a)$$ and this follows from AM-GM: $$3a+2b+3c=(a+b)+(b+c)+2(c+a)\geq 4\sqrt[4]{(a+b)(b+c)(c+a)^2}$$ $$2a+4b+2c=2(a+b)+2(b+c)\geq 4\sqrt{(a+b)(b+c)}$$

vicentev wrote: Prove that if \( a \), \( b \), and \( c \) are positive real numbers, then the following inequality holds: \[ \frac{a + 3c}{a + b} + \frac{c + 3a}{b + c} + \frac{4b}{c + a} \geq 6. \] Let $x=b+c,y=c+a,z=a+b$. So $$a=\frac{y+z-x}{2},b=\frac{z+x-y}{2},c=\frac{x+y-z}{2}.$$We have \begin{align*} &\frac{a + 3c}{a + b} + \frac{c + 3a}{b + c} + \frac{4b}{c + a} \ge 6\\ \iff&\frac{x+2y-z}{z}+\frac{-x+2y+z}{x}+\frac{2x-2y+2z}{y}\ge 6\\ \iff&\left(\frac{x}{z}+\frac{z}{x}\right)+2\left(\frac{y}{z}+\frac{z}{y}\right)+2\left(\frac{x}{y}+\frac{y}{x}\right)\ge10. \text{ (easy)} \end{align*}

Given positive numbers $a, b, c .$ Prove that$$ \frac{a+5 c}{a+b} +\frac{c+5 a}{b+c}+\frac{6 b}{c+a}\geq 9 $$ vicentev wrote: Prove that if \( a \), \( b \), and \( c \) are positive real numbers, then the following inequality holds: \[ \frac{a + 3c}{a + b} + \frac{c + 3a}{b + c} + \frac{4b}{c + a} \geq 6. \]

If we add 4 to both sides of the inequality, the problem is equivalent to proving that \[ \left(\frac{a + 3c}{a + b} + 1\right) + \left(\frac{c + 3a}{b + c} + 1\right) + \left(\frac{4b}{c + a} + 2\right) \geq 10. \]Now note that \[ \frac{a + 3c}{a + b} + 1 = 2\left(\frac{a + c}{a + b}\right) + \frac{b + c}{a + b}, \]\[ \frac{c + 3a}{b + c} + 1 = 2\left(\frac{a + c}{b + c}\right) + \frac{a + b}{b + c}, \]\[ \frac{4b}{c + a} + 2 = 2\left(\frac{b + c}{a + c}\right) + 2\left(\frac{a + b}{a + c}\right). \]Reorganizing, the left-hand side of our inequality becomes \[ \left(2\left(\frac{a + c}{a + b}\right) + 2\left(\frac{a + b}{a + c}\right)\right) + \left(2\left(\frac{a + c}{b + c}\right) + 2\left(\frac{b + c}{a + c}\right)\right) + \left(\frac{b + c}{a + b} + \frac{a + b}{b + c}\right). \]Using AM-GM inequality in each parenthesis, we have that \[ 2\left(\frac{a + c}{a + b}\right) + 2\left(\frac{a + b}{a + c}\right) \geq 4, \]\[ 2\left(\frac{a + c}{b + c}\right) + 2\left(\frac{b+c}{a + c}\right) \geq 4, \]\[ \frac{b + c}{a + b} + \frac{a + b}{b + c} \geq 2. \]Adding these inequalities, we obtain the desired result. By the equality case of AM-GM, it is easy to see that equality holds if and only if \(a = b = c\).