We must find the greatest $ k$ such that \[ f_k(a,b,c)=\frac{a}{b}+\frac{b}{c}+\frac{c}{a} - 3 - k\left(\frac{a}{b+c}+\frac{b}{c+a}+\frac{c}{a+b}-\frac{3}{2}\right) \geq 0\] I adopt the (slightly modified) notation that $ O(x)$ is any function for which there exist real constants $ C,D$ such that $ |O(x)|\leq Cx$ for all $ x\geq D$. In particular, any monomial $ ax^r$ with $ a,r$ real and $ r\leq 1$ (including zero) is $ O(x)$, and $ O(x)+O(x)=O(x)$ (adding their constants). I claim that $ k=1$ is the solution. Firstly, I show that $ k\leq 1$. Consider letting $ a=x^2,b=1,c=\frac{1}{x}$. Then we obtain \[ f_k(x^2,1,\frac{1}{x})=x^2+x+\frac{1}{x^3} - 3 - k\left(\frac{x^3}{1+x}+\frac{x}{1+x^3}+\frac{1}{x^3+x}-\frac{3}{2}\right)\] \[ = x^2 - \frac{kx^3}{1+x} + O(x) = x^2 - \frac{k(x^3+1)}{1+x} + \frac{k}{x+1} + O(x) = x^2-k(x^2-x+1) + O(x)\] This is $ (1-k)x^2+O(x)$, so for $ x$ sufficiently large and $ k>1$, this will become negative. Thus $ k\leq 1$ as required. Now, it remains to demonstrate that \[ \frac{a}{b}+\frac{b}{c}+\frac{c}{a}\geq \frac{a}{b+c}+\frac{b}{c+a}+\frac{c}{a+b}+\frac{3}{2}\]

$ k = 2$ We shall first prove that $ k\leq2$ Substitute $ a = 1, b = 1, c = n$, for some real number $ n$ Then the original inequality is equivalent to $ n + \frac {1}{n} - 2\geq k(\frac {n}{2} + \frac {2}{n + 1} - \frac {3}{2}$) Since the LHS = $ (2 + \frac {2}{n})(\frac {n}{2} + \frac {2}{n + 1} - \frac {3}{2})$ So if $ LHS\geq RHS$, $ 2 + \frac {2}{n}\geq k$, as n tends to infinity, $ k\leq2$. Now we shall prove that $ k = 2$ is suitable. Set $ k = 2$ in the original inequality. Then it is equivalent to prove $ \frac {a}{2b} + \frac {b}{2c} + \frac {c}{2a} \geq \frac {a}{b + c} + \frac {b}{a + c} + \frac {c}{a + b}$ After expansion, it is equivalent to $ a^3b^3+b^3c^3+c^3a^3+a^2b^4+b^2c^4+c^2a^4 \geq 3a^2b^2c^2 + a^4bc +b^4ca + c^4ab$ Since $ a^3b^3+b^3c^3+c^3a^3 \geq 3a^2b^2c^2$ by AM-GM inequality and $ a^2b^4+b^2c^4+c^2a^4 \geq a^4bc +b^4ca + c^4ab$ by Muirhead inequality, so the original inequality $ \frac {a}{2b} + \frac {b}{2c} + \frac {c}{2a} \geq \frac {a}{b + c} + \frac {b}{a + c} + \frac {c}{a + b}$ is true. Thus $ k = 2$.

SOLUTION: Without loss of generality, suppose that $ c = max(a,b,c)$. Observe that $ \frac {a}{b} + \frac {b}{c} + \frac {c}{a}-3 = \frac {(a - b)^2}{ab} + \frac {(c - a)(c - b )}{ac} .$ $ \frac {a}{b + c} + \frac {b}{c + a} + \frac {c}{b + a} - \frac {3}{2} = \frac {(a - b)^2}{(a + c)(b + c)} + \frac {(a + b + 2c)(c - a)(c - b)}{2(a + b)(b + c)(c + a)} .$ Thus we have the equivalent inequality: $ (\frac {1}{ab} - \frac {k}{(c - a)(c - b)})(a - b)^2 - (\frac {1}{ac} - \frac {k(a + b + 2c)}{2(a + b)(b + c)(c + a)})(c - a)(c - b) .$ $ (*)$ We will show that $ 1$ is the greatest value of the positive real number $ k$ such that this inequality holds for any positive real numbers $ a, b, c$. Indeed if $ k = 1$, Because $ c = max(a,b,c)$, we have $ \frac {1}{ab} - \frac {1}{(a + c)(b + c)} \ge 0 ;$ $ \frac {1}{ac} - \frac {a+b+2c}{2(a + b)(b + c)(c + a)} \ge 0 ;$ $ (c - a)(c - b) \ge 0 .$ Therefore, the inequality $ (*)$ is true. Finally, we must show that if $ k = m + 1 > 1$ (Let $ m$ be a positive real), then there exists a triple $ (a,b,c)$ such that the inequality $ (*)$ is not true. Consider the function $ f(c) = (\frac {1}{ab} - \frac {k}{(c - a)(c - b)})(a - b)^2 - (\frac {1}{ac} - \frac {k(a + b + 2c)}{2(a + b)(b + c)(c + a)})(c - a)(c - b)$ $ = (\frac {1}{ab} - \frac {m + 1}{(c - a)(c - b)})(a - b)^2 - (\frac {2(b - ma)c^2 + (a + b)((1 - m)a + 2b)c + 2ab(a + b)}{2(a + b)(b + c)(c + a)})(c - a)(c - b).$ We choose $ a$ and $ b$ such that $ b < ma$. Hence $ \displaystyle \lim_{c\rightarrow + \infty}{(\frac {1}{ab} - \frac {k}{(c - a)(c - b)})(a - b)^2}) = \frac {(a - b)^2}{ab}$ $ \displaystyle \lim_{c\rightarrow + \infty}{(\frac {1}{ab} - \frac {m + 1}{(c - a)(c - b)})(a - b)^2 - (\frac {2(b - ma)c^2 + (a + b)((1 - m)a + 2b)c + 2ab(a + b)}{2(a + b)(b + c)(c + a)})(c - a)(c - b)) = - \infty.}$ Thus $ \displaystyle \lim_{c\rightarrow + \infty}f(c) = - \infty$. Now we choose $ c$ such that it is great enough to $ f(c) < 0$. Consequently, there exists a triple $ (a,b,c)$ such that the inequality $ (*)$ is not true. In conclude, we can finally write the full answer to the problem: $ k = 1$.

Here is my solution Let $ a=t^3,b=t,c=1$ $ (t>0)$ then the inequality becomes \[ \frac{(t-1)^2(t^3+3t^2+2t+1)}{t^3} \ge k\cdot \frac{(t-1)^2(2t^6+2t^5+3t^4+2t^3+2t^2+t+2)}{2t(t^3+1)(t^2+1)}\] \[ \Leftrightarrow k \le \frac{2(t^3+1)(t^2+1)(t^3+3t^2+2t+1)}{t^2(2t^6+2t^5+3t^4+2t^3+2t^2+t+2)}\] Now, let $ t\rightarrow +\infty$, we get $ k \le 1$. We will show that $ 1$ is the value which we find, that is \[ \frac{a}{b}+\frac{b}{c}+\frac{c}{a} -3 \ge \frac{a}{b+c}+\frac{b}{c+a}+\frac{c}{a+b} -\frac{3}{2}\] \[ \Leftrightarrow \frac{ca}{b(b+c)}+\frac{ab}{c(c+a)}+\frac{bc}{a(a+b)} \ge \frac{3}{2}\] By the Cauchy Schwarz Inequality, we get \[ \frac{ca}{b(b+c)}+\frac{ab}{c(c+a)}+\frac{bc}{a(a+b)} \ge \frac{(ab+bc+ca)^2}{abc(b+c)+abc(c+a)+abc(a+b)} =\frac{(ab+bc+ca)^2}{2abc(a+b+c)}\] Moreover, it is easy to show that \[ (ab+bc+ca)^2 \ge 3abc(a+b+c)\] And so, we get the conclusion \[ k_{\max}=1\] We have done.

Let $ f(a,b,c,k)=LHS-RHS$. Then $ 2a^{2}(1+a)b(a+b)(a^{2}+b)f(1,\frac{1}{a},\frac{b}{a^{2}},k)$ $ =(2-2b(k-1))a^{7}+(2+2b^{2}+b(k-2))a^{6}+b(1+b)(k-2)a^{5}+2b(1-3b+b^{3}-k)a^{4}+b^{2}(1+b)(k-2)a^{3}+b^{3}(k-2)a^{2}+2b^{3}(1+b-bk)a+2b^{4}$. If $ k>1$, we can choose $ b$ large enough so that $ 2-2b(k-1)<0$. Then the coefficient of $ a^{7}$ become negative, so if we choose $ a$ large enough, $ f(1,\frac{1}{a},\frac{b}{a^{2}})<0$. For $ k=1$, $ 2abc(a+b)(b+c)(c+a)f(a,b,c,1)=2\sum_{cyc}a^{3}b^{3}+2\sum_{cyc}a^{2}b^{4}-\sum_{sym}a^{3}b^{2}c-6a^{2}b^{2}c^{2}$. This is positive since $ 2\sum_{cyc}a^{3}b^{3}=\sum_{sym}a^{3}b^{3}\ge \sum_{sym}a^{3}b^{2}c$ (Muirhead Inequality) $ 2\sum_{cyc}a^{2}b^{4}\ge 6a^{2}b^{2}c^{2}$ (AM-GM) Therefore minimum $ k$ is $ 1$, equality occurs iff $ a=b=c$.

Ilthigore wrote: We must find the greatest $ k$ such that \[ f_k(a,b,c)=\frac{a}{b}+\frac{b}{c}+\frac{c}{a} - 3 - k\left(\frac{a}{b+c}+\frac{b}{c+a}+\frac{c}{a+b}-\frac{3}{2}\right) \geq 0\]I adopt the (slightly modified) notation that $ O(x)$ is any function for which there exist real constants $ C,D$ such that $ |O(x)|\leq Cx$ for all $ x\geq D$. In particular, any monomial $ ax^r$ with $ a,r$ real and $ r\leq 1$ (including zero) is $ O(x)$, and $ O(x)+O(x)=O(x)$ (adding their constants). I claim that $ k=1$ is the solution. Firstly, I show that $ k\leq 1$. Consider letting $ a=x^2,b=1,c=\frac{1}{x}$. Then we obtain \[ f_k(x^2,1,\frac{1}{x})=x^2+x+\frac{1}{x^3} - 3 - k\left(\frac{x^3}{1+x}+\frac{x}{1+x^3}+\frac{1}{x^3+x}-\frac{3}{2}\right)\]\[ = x^2 - \frac{kx^3}{1+x} + O(x) = x^2 - \frac{k(x^3+1)}{1+x} + \frac{k}{x+1} + O(x) = x^2-k(x^2-x+1) + O(x)\]This is $ (1-k)x^2+O(x)$, so for $ x$ sufficiently large and $ k>1$, this will become negative. Thus $ k\leq 1$ as required. Now, it remains to demonstrate that \[ \frac{a}{b}+\frac{b}{c}+\frac{c}{a}\geq \frac{a}{b+c}+\frac{b}{c+a}+\frac{c}{a+b}+\frac{3}{2}\] $$\Leftrightarrow \frac{ac}{b(b+c)}+\frac{cb}{a(a+b)}+\frac{ba}{c(c+a)}\geq \frac{3}{2}$$By C-S inequality: $$LHS\geq \frac{(bc+ca+ab)^{2}}{2abc(a+b+c)}\geq \frac{3}{2}$$https://artofproblemsolving.com/community/c4h2054805p14768263 https://artofproblemsolving.com/community/c6h583302p3448303

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