After substitutions $a=2x, \ \ b=2y, \ \ c=2z$, conditions becomes $xyz=1$ , we need to prove $\sum_{cyc} \frac{xy+1}{x+1}\ge 3$ $\iff \sum_{cyc} \frac{\frac{1}{z}+1}{x+1}\ge 3 \ \ \ \iff \ \ \ \sum_{cyc} \frac{1+z}{z(x+1)}\ge 3$ By AM-GM: $ \sum_{cyc} \frac{1+z}{z(x+1)}\ge 3\sqrt[3]{\prod_{cyc} \frac{1+z}{z(x+1)} }=3$

$\frac{ab + 4}{a + 2}+\frac{bc + 4}{b + 2}+\frac{ca + 4}{c + 2}=\frac{8+4c}{c(a+2)}+\frac{8+4a}{a(b+2)}+\frac{8+4b}{b(c+2)} \geq 3\sqrt[3]{\frac{4^3(a+2)(b+2)(c+2)}{abc(a+2)(b+2)(c+2)}}=6$

parmenides51 wrote: Let $a, b, c$ be positive real numbers such that $abc = 8$. Prove that $\frac{ab + 4}{a + 2}+\frac{bc + 4}{b + 2}+\frac{ca + 4}{c + 2}\ge 6$. https://artofproblemsolving.com/community/c6h1467776p8505040

In these types of problems, the best and the most intuitive way should be that of subsitution.

Let $a=\frac{2x}{y},b=\frac{2y}{z},c=\frac{2z}{x}$,then $x^3y^3+y^3z^3+z^3x^3 \geq x^3y^2z+x^2yz^3+xy^3z^2$ By AM-GM inequality, $\frac{2}{3}x^3y^3+\frac{1}{3}z^3x^3 \geq x^3y^2z$ $\frac{2}{3}y^3z^3+\frac{1}{3}x^3y^3 \geq xy^3z^2$ $\frac{2}{3}z^3x^3+\frac{1}{3}y^3z^3 \geq x^2yz^3$ Thus we are done.

Takeya.O wrote: Let $a=\frac{2x}{y},b=\frac{2y}{z},c=\frac{2z}{x}$,then $x^3y^3+y^3z^3+z^3x^3 \geq x^3y^2z+x^2yz^3+xy^3z^2$ By AM-GM inequality, $\frac{2}{3}x^3y^3+\frac{1}{3}z^3x^3 \geq x^3y^2z$ $\frac{2}{3}y^3z^3+\frac{1}{3}x^3y^3 \geq xy^3z^2$ $\frac{2}{3}z^3x^3+\frac{1}{3}y^3z^3 \geq x^2yz^3$ Thus we are done Could you pls elaborate it

\[\Leftrightarrow \sum_{cyc} \frac{\frac{4}c+2}{a+2}\]\[\sum_{cyc} \frac{2(c+2)}{a+2} \geq 3\sqrt[3]{\prod_{cyc} \frac{2(c+2)}{a+2}} = 3\sqrt[3]{\frac{8}{abc}}=3\]

aksshatkhanna wrote: Could you pls elaborate it I'll only do the first inequation you should be able to do others yourself. \[\frac{x^3y^3 + x^3y^3+z^3x^3}2 \geq \sqrt[3]{(x^3y^3)^2\cdot z^3x^3} = (xy)^2\cdot zx = x^3y^2z\]

parmenides51 wrote: Let $a, b, c$ be positive real numbers such that $abc = 8$. Prove that $\frac{ab + 4}{a + 2}+\frac{bc + 4}{b + 2}+\frac{ca + 4}{c + 2}\ge 6$. The inequality is equivalent to $\sum_{\text {cyc}} ab^2\ge 2\sum_{\text {cyc}}ab$. Now, $abc=8\implies a^{\frac {1}{3}}b^{\frac {1}{3}}c^{\frac {1}{3}}=2$. So, the above reduces to proving $\sum_{\text {cyc}} ab^2\ge \sum_{\text {cyc}}a^{\frac {4}{3}}b^{\frac {4}{3}}c^{\frac {1}{3}}$. This follows directly from A.M.-G.M. as $ab^2+ab^2+ca^2\ge 3a^{\frac {4}{3}}b^{\frac {4}{3}}c^{\frac {1}{3}} $ and so on.

I proved a+b+c>=6 how can I get to the next step?

So what I did was write each numerator "in terms of its denominator" .. Let the expression on the LHS of the inequality be S, we can get $S = \sum_{cyc}^{} \frac{(a+2)b - 2(b-2)}{a+2} $ , simplifying and grouping gives $S= a+b+c -2 \sum_{cyc}^{} \frac{b-2}{a+2} . $ You have already proven that a+b+c is greater than zero, and the remaining art in the above can easily be zero, so we have figured out the equality case. The rest idk tbh!!

$\frac{ab + 4}{a + 2} = \frac{ab + \frac{abc}{2}}{a + 2} = ab\frac{1 + \frac{c}{2}}{a + 2} = \frac{ab }{2}\frac{c + 2}{a + 2}$ Similarly $\frac{bc + 4}{b + 2} = \frac{bc }{2}\frac{a + 2}{b + 2}$ and $\frac{ca + 4}{b + 2} = \frac{ca }{2}\frac{b + 2}{c + 2}$ Thus $$\frac{ab + 4}{a + 2} + \frac{bc + 4}{b + 2} + \frac{ca + 4}{c + 2} = \frac{ab }{2}\frac{c + 2}{a + 2} + \frac{bc }{2}\frac{a + 2}{b + 2} + \frac{ca }{2}\frac{b + 2}{c + 2} \geq 3\sqrt[3]{\frac{ab }{2}\frac{c + 2}{a + 2}\frac{bc }{2}\frac{a + 2}{b + 2}\frac{ca }{2}\frac{b + 2}{c + 2}} = 3\sqrt[3]{\frac{a^{2}b^{2}c^{2}}{2^3}\frac{(c+2)(a+2)(b+2)}{(c+2)(a+2)(b+2)}} = \frac{3}{2}\sqrt[3]{a^{2}b^{2}c^{2}} = \frac{3}{2} 8^{\frac{2}{3}} = \frac{3}{2} 2^{2} = \frac{3}{2}4 = 6 $$$Q.E.D.$

Mathplusplus wrote: $\frac{ab + 4}{a + 2} = \frac{ab + \frac{abc}{2}}{a + 2} = ab\frac{1 + \frac{c}{2}}{a + 2} = \frac{ab }{2}\frac{c + 2}{a + 2}$ Similarly $\frac{bc + 4}{b + 2} = \frac{bc }{2}\frac{a + 2}{b + 2}$ and $\frac{ca + 4}{b + 2} = \frac{ca }{2}\frac{b + 2}{c + 2}$ Thus $$\frac{ab + 4}{a + 2} + \frac{bc + 4}{b + 2} + \frac{ca + 4}{c + 2} = \frac{ab }{2}\frac{c + 2}{a + 2} + \frac{bc }{2}\frac{a + 2}{b + 2} + \frac{ca }{2}\frac{b + 2}{c + 2} \geq 3\sqrt[3]{\frac{ab }{2}\frac{c + 2}{a + 2}\frac{bc }{2}\frac{a + 2}{b + 2}\frac{ca }{2}\frac{b + 2}{c + 2}} = 3\sqrt[3]{\frac{a^{2}b^{2}c^{2}}{2^3}\frac{(c+2)(a+2)(b+2)}{(c+2)(a+2)(b+2)}} = \frac{3}{2}\sqrt[3]{a^{2}b^{2}c^{2}} = \frac{3}{2} 8^{\frac{2}{3}} = \frac{3}{2} 2^{2} = \frac{3}{2}4 = 6 $$$Q.E.D.$ See also #3

Rearrangement can be used to make it symmetric (it becomes $$\sum{ \frac{ab+4}{c+2} }=\sum{ \frac{ \frac{8}{c} +4}{c+2}}=\sum{\frac{4}{c}}$$) Then it can be finished off with AM-GM

RagvaloD wrote: $\frac{ab + 4}{a + 2}+\frac{bc + 4}{b + 2}+\frac{ca + 4}{c + 2}=\frac{8+4c}{c(a+2)}+\frac{8+4a}{a(b+2)}+\frac{8+4b}{b(c+2)} \geq 3\sqrt[3]{\frac{4^3(a+2)(b+2)(c+2)}{abc(a+2)(b+2)(c+2)}}=6$ Nice

We use the substituition $a=\frac{2x}{y},b=\frac{2y}{z},c=\frac{2z}{x}$ So we have to prove $\sum \frac{xy+yz}{xz+yz} \geq 3$ Or equivalently $\sum \frac{xy}{zx+yz} \geq \sum \frac{xz}{xz+yz} $ which follows from rearrangement as well as Murhead(After full manipulation) Q.E.D

Si a,b,c>0 Dq: (ab+4)/(a+2)+(bc+4)/(b+2)+(ac+4)/(c+2)>=6 sea a=2x/y, b=2y/z, c=2z/x, reemplazando en lo que piden (4x/z+4)/(2x/y+2)+(4y/x+4)/(2y/z+2)+(4z/y+4)/(2z/x+2)>=6 y(x+z)/z(x+y)+y(x+z)/z(x+y)+y(x+z)/z(x+y)>=3 por MA y MG [y(x+z)/z(x+y)+y(x+z)/z(x+y)+y(x+z)/z(x+y)]/3>=[y(x+z)/z(x+y)*y(x+z)/z(x+y)*y(x+z)/z(x+y)]^1/3 [y(x+z)/z(x+y)+y(x+z)/z(x+y)+y(x+z)/z(x+y)]>=3 l.q.q.d

the given inequality is equivalent to $\frac{2}{ac+2}+\frac{2}{ab+2}+\frac{2}{bc+2}\geq 1$ Multiplying both sides with $\left ( ab+2 \right )\left ( bc+2 \right )\left ( ca+2 \right )$ we get $4ab+4bc+4ca+16\geq64$ which is true from AM-GM

JBMO 2016 A1 wrote: Let $a, b, c$ be positive real numbers such that $abc = 8$. Prove that $\frac{ab + 4}{a + 2}+\frac{bc + 4}{b + 2}+\frac{ca + 4}{c + 2}\ge 6$. \begin{align*} \frac{ab+4}{a+2}+\frac{bc+4}{b+2}+\frac{ca+4}{c+2}\geq 6 &\iff \frac{8+4c}{c(a+2)}+\frac{8+4a}{a(b+2)}+\frac{8+4b}{b(c+2)}\geq 6 &\iff \frac{2+c}{c(a+2)}+\frac{2+a}{a(b+2)}+\frac{2+b}{b(c+2)}\geq \frac{3}{2}\end{align*} Now $\frac{2+c}{c(a+2)}+\frac{2+a}{a(b+2)}+\frac{2+b}{b(c+2)}\overset{AM-GM}{\geq} 3\sqrt[3]{\frac{1}{abc}}=\frac{3}{2}$. Hence Proved. $\blacksquare$

Let $a = \frac{2x}{y}, b = \frac{2y}{z}, c = \frac{2z}{x}$. Then, we know that $$\frac{ab + 4}{a + 2} = \frac{\frac{4x}{z} + 4}{\frac{2x}{y} +2} = \frac{4xy + 4yz}{2xz+2yz} = 2\frac{y(x+z)}{z(x+y)}$$Similarly, we have that: $$\frac{bc + 4}{b+2} = 2\frac{z (x+y)}{x (y+z)}$$$$\frac{ac + 4}{c+2} = 2\frac{x (y+z)}{y (x+z)}$$It follows that by the AM-GM inequality, $$\frac{ab + 4}{a + 2}+\frac{bc + 4}{b + 2}+\frac{ca + 4}{c + 2}\ge 6$$as desired.

Substitute $a=2x,b=2y,c=2z$. Then the condition is $xyz=1$ and we need to prove $\sum_{cyc}\frac{xy+1}{x+y}\geq 3$. Substitute $x=\frac{p}{q},y=\frac{q}{r},z=\frac{r}{p}$. Then we need to prove $\sum_{cyc}\frac{q(p+r)}{r(p+q)}\geq 3$ which is straightforward AM-GM since the product of all summands is $1$. $\blacksquare$

Beautiful one and not so easy. The inequality can be written as $$\frac{12\sum_{cyc}a+4\sum_{cyc}ab +2\sum_{cyc}ab^2+48}{2\sum_{cyc}a+\sum_{cyc}ab+8} \ge 6 \Leftrightarrow \sum_{cyc}ab^2 \ge 2\sum_{cyc}ab $$As $abc=8$ we've $(abc)^{\frac{1}{3}}=2$ using this the inequality becomes $$\sum_{cyc}ab^2 \ge \sum_{cyc}ab \cdot (abc)^{\frac{1}{3}}$$Then using the AM-GM inequality we get $ 2ab^2+ca^2 \ge 3a^{\frac{4}{3}}b^{\frac{4}{3}}c^{\frac{1}{3}} \Leftrightarrow \sum_{cyc}(2ab^2 +ca^2)=3\sum_{cyc}ab^2 \ge 3\sum_{cyc}a^{\frac{4}{3}}b^{\frac{4}{3}}c^{\frac{1}{3}}=3\sum_{cyc}ab \cdot(abc)^{\frac{1}{3}} \Leftrightarrow \sum_{cyc}ab^2 \ge \sum_{cyc}ab \cdot(abc)^{\frac{1}{3}}$$ QED

ab=8/c bc=8/a ca=8/b 8/c+4= 4(c+2)/c 4(c+2)/(c(a+2)) + 4(a+2)/(a(b+2)) + 4(b+2)/(b(c+2)) ≥ 6 by AM-GM 4(c+2)/(c(a+2)) + 4(a+2)/(a(b+2)) + 4(b+2)/(b(c+2)) ≥ 3* 2

AM-GM bash after multiplying each term with the unknown that is missing in that term.

First i found two ways. But i think one of them is most useful. So i share this one. abc is given. We will multiple c, b, a in a order. Then we will have; 8+4c/ (c*(a+2)) + 8+4a/ (a(b+2)) + 8+4b/ (b(c+2))>=6 If we make it more simple, dividing into 4 the two sides; 2+c / c(a+2) + 2+a / (a(b+2)) + 2+b / b(c+2)>= 3/2 From AM-GM; 2+c / c(a+2) + 2+a / a(b+2) + 2+b / b(c+2) >= 3* root of 1/abc (from 3 degrees) = 3/2 So we proved)

First, let $a = 2x$, $b = 2y$, and $c = 2z$. We are now trying to prove that when $xyz = 1$, \[\sum_{cyc}\frac{xy+1}{x+1}\geq 3\] Note that $xy = \frac{1}{z}$ from the initial condition. Our inequality is now \[\sum_{cyc}\frac{\frac{1}{z}+1}{x+1}\geq 3\] We now use AM-GM and must show that \[3\sqrt[3]{\prod_{cyc}\frac{\frac{1}{z}+1}{x+1}}\geq 3\]\[\prod_{cyc}\frac{\frac{1}{z}+1}{x+1}\geq 1\] Simplifying the fraction $\frac{\frac{1}{x} + 1}{x + 1}$, we obtain $\frac{1}{x}$. Thus, our inequality is now \[\prod_{cyc}\frac{1}{x}\geq 1\] which is true from the initial condition ($xyz = 1$).