$\displaystyle 1+\frac{3}{a+b+c}\geq \frac{6}{ab+bc+ca}$ is equivalent to $\displaystyle \frac{ab+bc+ca}{3} + \frac{ab+bc+ca}{a+b+c} \geq 2$. By AM-GM, it suffices to prove $\displaystyle \frac{ab+bc+ca}{3} \cdot \frac{ab+bc+ca}{a+b+c} \geq 1$. That is, $\displaystyle (ab+bc+ca)^2 \geq 3(a+b+c)$. Let $x=1/a, y=1/b, z=1/c$, then $xyz=1$ and it reduces to prove $(x+y+z)^2\geq 3(xy+yz+zx)$ which is obvious.

$ 1 + \frac {3}{a + b + c}\ge{\frac {6}{ab + bc + ca}}$ is equivalent to $ (3+a + b + c)(ab + bc + ca)\ge 6(a + b + c)$. Now lets consider two cases. First $ ab + bc + ca\ge a + b + c$ and then $ ab + bc + ca\le a + b + c$. If $ ab + bc + ca\ge a + b + c$ then using AM-GM $ (3+a + b + c)(ab + bc + ca)\ge (3+3)(ab + bc + ca))\ge 6(ab + bc + ca)\ge 6(a + b + c)$. If $ ab + bc + ca\le a + b + c$ then let $ x= a + b + c$. $ (3+a + b + c)(ab + bc + ca)\ge (3+a + b + c)(a + b + c) = (3+x)x$. Now we net that $ (3+x)x\ge 6x$ but this equivalent to $ x(x-3)\ge 0$ and this is true becouse $ a+b+c\ge 3$ by AM-GM.

Maverick wrote: Let $ a,b,c$ be three positive real numbers such that $ abc = 1$. Prove that: \[ 1 + \frac {3}{a + b + c}\ge{\frac {6}{ab + bc + ca}} . \] We set \[ a=\frac{1}{x} ; b=\frac{1}{y} , c=\frac{1}{z}\] and observe xyz=1. The inequality is equivalent to: \[ 1+\frac{3}{xy+yz+xz}\geq \frac{6}{x+y+z}\] Clearly \[ 1+\frac{3}{xy+yz+xz}\geq 1+\frac{9}{(x+y+z)^{2}}\geq \frac{6}{x+y+z}\]

Maverick wrote: Let $ a,b,c$ be three positive real numbers such that $ abc = 1$. Prove that: \[ 1 + \frac {3}{a + b + c}\ge{\frac {6}{ab + bc + ca}} . \] Let $ p = a + b + c$, $ q = ab + bc + ca$ and $ r = abc$. Then we have $ 1 + \frac {3}{p} \ge \frac {6}{q}$. By AM-GM: $ 1 + \frac {3}{p} \ge 2\sqrt {\frac {3}{p}}$. So we have to prove: $ 2\sqrt {\frac {3}{p}} \ge \frac {6}{q}$, which is equivalent to $ q^2 \ge 3pr$. This is obviously true upon expanding. Edit: An one liner: $ 1 + \frac {3}{a + b + c} - {\frac {6}{ab + bc + ca}} =$ $ \left(1 - \sqrt {\frac {3}{a + b + c} \right)^2 +}$ $ \frac {\sqrt {3}}{\sqrt {a + b + c}(ab + bc + ca)(ab + bc + ca + \sqrt {3a + 3b + 3c})} \cdot$ $ ( (ab - bc)^2 + (ab - ac)^2 + (ca - cb)^2 ) \ge 0$ (If it can fit in one line )

first ,we use $ {abc = 1}$,so the inequality $ {\Leftrightarrow}$ $ {\frac {6}{\frac {1}{a} + \frac {1}{b} + \frac {1}{c}} < = 1 + \frac {3}{a + b + c}}$;we know that :$ {\frac {3}{\frac {1}{a} + \frac {1}{b} + \frac {1}{c}} \le \frac {a + b + c}{3}}$ so we only need to prove that :$ {\frac {2}{3}(a+b+c) \le 1 + \frac {3}{a + b + c}}$_______[#];let $ {a + b + c = x}$ by $ {a + b + c \le 3*(abc)^{\frac {1}{3}}}$ ,we know $ {x \in (0,3] }$; so [#] $ {\Longleftrightarrow}$ $ {\frac {2}{3}x^2 - x - 3 > = 0}$ ; then consider the function: $ {\phi (x) = \frac {2}{3}(x - \frac {3}{4})^2 - \frac {27}{8}}$ ; we know that : $ {\phi (x) \ge 0}$ on $ {x \in (0,3]}$. then # is true ,so we have completed!

Maverick wrote: Let $ a,b,c$ be three positive real numbers such that $ abc = 1$. Prove that: \[ 1 + \frac {3}{a + b + c}\ge{\frac {6}{ab + bc + ca}} .\] $ LHS = 1 + \frac{9}{{3abc(a + b + c)}} \ge 1 + \frac{9}{{{{(ab + bc + ca)}^2}}} \ge 2\sqrt {\frac{9}{{{{(ab + bc + ca)}^2}}}} = \frac{6}{{ab + bc + ca}}$ ^^!

Maranna SortList - 2009 - Italy: Let $ a, b, c$ be positive numbers and $ abc=1$. Prove that\[ \frac{2}{a+b+c}+\frac{4}{3}\geq1+\frac{3}{ab+bc+ca}\geq\frac{6}{a+b+c}.\]

sqing wrote: Maranna SortList - 2009 - Italy: Let $ a, b, c$ be positive numbers and $ abc=1$. Prove that\[ \frac{2}{a+b+c}+\frac{4}{3}\geq1+\frac{3}{ab+bc+ca}\geq\frac{6}{a+b+c}.\] I think Buffalo Way helps us here (but looking for something quicker). Any ideas arqady sir ?

Let $ a , b, c$ be positive real numbers such that $abc= 1 .$ Show that $$ ab+bc+ca+\frac{9}{10(a^2+b^2+c^2)}\geq \frac{33}{10}. $$

Let $a_1,a_2,\cdots,a_n (n\ge 2)$ be positive real numbers . Prove that $$1+\frac{n(n-1)}{2\sum_{1\le i<j\le n}a_ia_j}\geq\frac{2n}{a_1+a_2+\cdots+a_n}$$SXJX, 2(2020),Q1082 h

Let $a,b,c$ be three positive real numbers such that $a+b+c=1$. Prove that$$ 1+\frac{3}{\frac{1}{a}+\frac{1}{b}+\frac{1}{c}}\ge 4(ab+bc+ca).$$(Crux) $$ 1+\frac{3}{\frac{1}{a}+\frac{1}{b}+\frac{1}{c}}\ge1+\frac{9abc}{(a+b+c)^2}=(a+b+c)^3+9abc\ge 4(ab+bc+ca).$$

sqing wrote: Let $ a , b, c$ be positive real numbers such that $abc= 1 .$ Show that $$ ab+bc+ca+\frac{9}{10(a^2+b^2+c^2)}\geq \frac{33}{10}. $$ We use the well known inequality, that is

and use $(ab+bc+ca)^2\geq 3abc(a+b+c)\implies ab+bc+ca\geq\sqrt{3(a+b+c)}$ And so by AM-GM $$ ab+bc+ca+\frac{9}{10(a^2+b^2+c^2)}\geq \sqrt{3(a+b+c)}+\frac{9\cdot 81}{10(a+b+c)^5}=10\cdot\frac{\sqrt{3(a+b+c)}}{10}+\frac{9\cdot 81}{10(a+b+c)^5}\geq 11\sqrt[11]{\left(\frac{\sqrt{3(a+b+c)}}{10}\right)^{10}\cdot\frac{9\cdot 81}{10(a+b+c)^5}}=\frac{33}{10}.\quad\blacksquare$$

Maverick wrote: Let $a,b,c$ be three positive real numbers such that $abc=1$. Prove that: \[ 1+\frac{3}{a+b+c}\ge{\frac{6}{ab+bc+ca}} . \] Let $a,b,c$ be three positive real numbers such that $abc=1$. Prove that:\[ 1+\frac{8}{a+b+c}\ge{\frac{11}{ab+bc+ca}} . \]

Let $a,b,c$ be three positive real numbers such that $\frac{5}{2}+\frac{3}{\frac{1}{a}+\frac{1}{b}+\frac{1}{c}}=4(ab+bc+ca) $. Prove that$$ a+b+c\ge \frac{3}{2}$$Let $a,b,c$ be three positive real numbers such that $\frac{5}{3}+\frac{3}{\frac{1}{a}+\frac{1}{b}+\frac{1}{c}}=4(ab+bc+ca) $. Prove that$$ a+b+c\ge \frac{5}{4}$$here

sqing wrote: Let $a,b,c$ be three positive real numbers such that $a+b+c=1$. Prove that$$ 1+\frac{3}{\frac{1}{a}+\frac{1}{b}+\frac{1}{c}}\ge 4(ab+bc+ca).$$Let $a,b,c$ be three positive real numbers such that $1+\frac{3}{\frac{1}{a}+\frac{1}{b}+\frac{1}{c}}=4(ab+bc+ca) $. Prove that$$ a+b+c\ge 1$$Let $a,b,c$ be three positive real numbers such that $\frac{5}{2}+\frac{3}{\frac{1}{a}+\frac{1}{b}+\frac{1}{c}}=4(ab+bc+ca) $. Prove that$$ a+b+c\ge \frac{3}{2}$$Let $a,b,c$ be three positive real numbers such that $\frac{5}{3}+\frac{3}{\frac{1}{a}+\frac{1}{b}+\frac{1}{c}}=4(ab+bc+ca) $. Prove that$$ a+b+c\ge \frac{5}{4}$$here

Maverick wrote: Let $a,b,c$ be three positive real numbers such that $abc=1$. Prove that: \[ 1+\frac{3}{a+b+c}\ge{\frac{6}{ab+bc+ca}} . \] We have \[\frac{1}{abc}+\frac{3}{a+b+c} \geqslant {\frac{6}{ab+bc+ca}},\]for all $a,b,c$ are positive real numbers.

asdrojas wrote: $ 1 + \frac {3}{a + b + c}\ge{\frac {6}{ab + bc + ca}}$ is equivalent to $ (3+a + b + c)(ab + bc + ca)\ge 6(a + b + c)$. Now lets consider two cases. First $ ab + bc + ca\ge a + b + c$ and then $ ab + bc + ca\le a + b + c$. If $ ab + bc + ca\ge a + b + c$ then using AM-GM $ (3+a + b + c)(ab + bc + ca)\ge (3+3)(ab + bc + ca))\ge 6(ab + bc + ca)\ge 6(a + b + c)$. If $ ab + bc + ca\le a + b + c$ then let $ x= a + b + c$. $ (3+a + b + c)(ab + bc + ca)\ge (3+a + b + c)(a + b + c) = (3+x)x$. Now we net that $ (3+x)x\ge 6x$ but this equivalent to $ x(x-3)\ge 0$ and this is true becouse $ a+b+c\ge 3$ by AM-GM. You can prove that $ab+bc+ca\leq a+b+c$: $$ab+bc+ca\leq a+b+c \Longleftrightarrow ab+bc+ca\leq \sqrt[3]{abc}(a+b+c) \Longleftrightarrow \sum_{sym} ab\leq\sum_{sym} a^{4/3}b^{1/3}c^{1/3}$$And that is true by Muirhead since $\left[\frac{4}{3},\frac{1}{3},\frac{1}{3}\right]\succ[1,1,0]$