shobber wrote: Let $a$, $b$, $c$ be positive real numbers. Prove that \[ \biggl(1+\frac{a}{b}\biggr) \biggl(1+\frac{b}{c}\biggr) \biggl(1+\frac{c}{a}\biggr) \ge 2 \biggl(1+\frac{a+b+c}{\sqrt[3]{abc}}\biggr). \] I am pretty sure this has been discussed. Here is one more solution (maybe duplicated).

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shobber wrote: Let $a$, $b$, $c$ be positive real numbers. Prove that \[ \biggl(1+\frac{a}{b}\biggr) \biggl(1+\frac{b}{c}\biggr) \biggl(1+\frac{c}{a}\biggr) \ge 2 \biggl(1+\frac{a+b+c}{\sqrt[3]{abc}}\biggr). \] I solved it this way: \[ LHS=1+\frac{a}{b}+\frac{b}{c}+\frac{c}{a}+\frac{a}{b}\frac{b}{c}+\frac{b}{c}\frac{c}{a}+\frac{c}{a}\frac{a}{b}+1 \\=2+\frac{a}{b}+\frac{b}{c}+\frac{c}{a}+\frac{b}{a}+\frac{c}{b}+\frac{a}{c} \\=2+\frac{2}{3}\left(\frac{a}{b}+\frac{a}{c}+\frac{b}{a}+\frac{b}{c}+\frac{c}{a}+\frac{c}{b}\right)+2\frac{\frac{a}{b}+\frac{b}{c}+\frac{c}{a}+\frac{b}{a}+\frac{c}{b}+\frac{a}{c}}{6} \\\geq 2+\frac{2}{3}\left(\frac{a}{b}+\frac{a}{c}+\frac{b}{a}+\frac{b}{c}+\frac{c}{a}+\frac{c}{b}\right)+2\sqrt[6]{\frac{a}{b}\frac{b}{c}\frac{c}{a}\frac{b}{a}\frac{c}{b}\frac{a}{c}} \\=2+\frac{2}{3}\left(\frac{a}{b}+\frac{a}{c}+\frac{b}{a}+\frac{b}{c}+\frac{c}{a}+\frac{c}{b}\right)+2 \\=2\left(1+\frac{\frac{a}{b}+\frac{a}{c}+1}{3}+\frac{\frac{b}{a}+\frac{b}{c}+1}{3}+\frac{\frac{c}{a}+\frac{c}{b}+1}{3}\right) \\\geq 2\left(1+\sqrt[3]{\frac{a}{b}\cdot\frac{a}{c}\cdot 1}+\sqrt[3]{\frac{b}{a}\cdot\frac{c}{a}\cdot 1}+\sqrt[3]{\frac{c}{a}\cdot\frac{c}{b}\cdot 1}\right)=RHS \]

The proof of this problem has a good application of two Useful theorems for which you dont have write a proof in Olympiads Muirhead and Rearrangement...

shobber wrote: Let $a$, $b$, $c$ be positive real numbers. Prove that \[ \biggl(1+\frac{a}{b}\biggr) \biggl(1+\frac{b}{c}\biggr) \biggl(1+\frac{c}{a}\biggr) \ge 2 \biggl(1+\frac{a+b+c}{\sqrt[3]{abc}}\biggr). \] Another solution: Since the inequality is homogeneous in a, b, c, we can WLOG assume that abc = 1. Then, $\left(1+\frac{a}{b}\right)\left(1+\frac{b}{c}\right)\left(1+\frac{c}{a}\right)=\frac{a+b}{b}\cdot\frac{b+c}{c}\cdot\frac{c+a}{a}$ $=\frac{\left(b+c\right)\left(c+a\right)\left(a+b\right)}{abc}=\left(b+c\right)\left(c+a\right)\left(a+b\right)$ and $\frac{a+b+c}{\sqrt[3]{abc}}=\frac{a+b+c}{\sqrt[3]{1}}=a+b+c$, so the inequality in question becomes $\left(b+c\right)\left(c+a\right)\left(a+b\right)\geq 2\left(1+a+b+c\right)$. Now, by AM-GM, we have $a+b+c\geq 3\sqrt[3]{abc}=3\sqrt[3]{1}=3$, so that $4\left(a+b+c-1\right)=2\left(a+b+c\right)+2\left(a+b+c\right)-4$ $\geq 2\left(a+b+c\right)+2\cdot 3-4=2\left(1+a+b+c\right)$. Thus, according to http://www.mathlinks.ro/Forum/viewtopic.php?t=6347 , we have $\left(b+c\right)\left(c+a\right)\left(a+b\right)\geq 4\left(a+b+c-1\right)\geq 2\left(1+a+b+c\right)$, and the proof is complete. Darij

If $x, y, z$ are positive numbers, then the following inequality \[\left(1+\frac{x^{3}}{y^{3}}\right)\left(1+\frac{y^{3}}{z^{3}}\right)\left(1+\frac{z^{3}}{x^{3}}\right)\geq8+k\left(\frac{x^{3}+y^{3}+z^{3}}{xyz}-3\right)\] holds if and only if $k\leq\frac{3\sqrt{665}}{8}\cos{\left(\frac{\pi}{3}-\frac{1}{3}\arccos{\frac{13117}{665\sqrt{665}}}\right)}-\frac{15}{16}=5.70288090452726654\cdots$. With equality if $k=\frac{3\sqrt{665}}{8}\cos{\left(\frac{\pi}{3}-\frac{1}{3}\arccos{\frac{13117}{665\sqrt{665}}}\right)}-\frac{15}{16}$ and $x=2\sqrt{2}\cos{\left(\frac{\pi}{3}-\frac{1}{3}\arccos{\frac{1}{4\sqrt{2}}}\right)}-1=1.36146876618582657\cdots,y=z=1$.

Ji Chen wrote: If $x, y, z$ are positive numbers, then the following inequality \[\left(1+\frac{x^{3}}{y^{3}}\right)\left(1+\frac{y^{3}}{z^{3}}\right)\left(1+\frac{z^{3}}{x^{3}}\right)\geq8+k\left(\frac{x^{3}+y^{3}+z^{3}}{xyz}-3\right) \] holds if and only if $k\leq\frac{3\sqrt{665}}{8}\cos{\left(\frac{\pi}{3}-\frac{1}{3}\arccos{\frac{13117}{665\sqrt{665}}}\right)}-\frac{15}{16}=5.70288090452726654\cdots$. With equality if $k=\frac{3\sqrt{665}}{8}\cos{\left(\frac{\pi}{3}-\frac{1}{3}\arccos{\frac{13117}{665\sqrt{665}}}\right)}-\frac{15}{16}$ and $x=2\sqrt{2}\cos{\left(\frac{\pi}{3}-\frac{1}{3}\arccos{\frac{1}{4\sqrt{2}}}\right)}-1=1.36146876618582657\cdots,y=z=1$. I am curious how do you calculate the range of k?

Notice $LHS = (a+b+c)(\frac{1}{a}+\frac{1}{b}+\frac{1}{c})-1$. So our inequality is equivalent to proving: $(a+b+c)(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}) \geq 3+\frac{2(a+b+c)}{ \sqrt [3]{abc}}$ Let: $\frac{a+b+c}{3}= A$, $\frac{3}{\frac{1}{a}+\frac{1}{b}+\frac{1}{c}}= H$, $\sqrt[3]{abc}=G$ So our inequality is equivalent to: $3AG \geq GH+2AH$ Since $A \geq G \geq H$ Thus $3AG = AG+2 AG \geq GH+2AH$ as desired.

darij grinberg wrote: Another solution: Since the inequality is homogeneous in a, b, c, we can WLOG assume that abc = 1. Darij Hello !! I doubt about this because if we put $x=a/b$... we get $xyz=1$ but we need to proove another inequalitie, see? Can someone explain please? and what does homogenous mean?? Thanks !

FOURRIER wrote: Hello !! I doubt about this because if we put $x=a/b$... we get $xyz=1$ but we need to proove another inequalitie, see? Can someone explain please? and what does homogenous mean?? Thanks ! hi FOURRIER homogeneous means of the same degree. let $abc=x^{3}$ where $x$ is some real number this gives $\frac{a}{x}\frac{b}{x}\frac{c}{x}=1$ now choose $u=\frac{a}{x}$ ,similarly choose $v,w$ now the given inequality is equal to $\prod_{cyc}(1+\frac{u}{v})$ $\geq$ $2(1+\frac{u+v+w}{\sqrt[3]{uvw}})$ where $uvw=1$ now replace WALOG $u$ by $a$ and so this was only possible since the inequation was of uniform degree

Ok I got the method that you used. -Does the same degree means:we have all the variables like this : $a^{1}, b^{1}, c^{1}$ and $a^{1/3}, b^{1/3}, c^{1/3}$? -So, whenever when its homogenous we can assume abc=1? Thanks again

FOURRIER wrote: Ok I got the method that you used. -Does the same degree means:we have all the variables like this : $a^{1}, b^{1}, c^{1}$ and $a^{1/3}, b^{1/3}, c^{1/3}$? Thanks again yes FOURRIER wrote: So, whenever when its homogenous we can assume abc=1? yes but the choice should be tricky and according to the need of the problem .....

pardesi wrote: FOURRIER wrote: Ok I got the method that you used. -Does the same degree means:we have all the variables like this : $a^{1}, b^{1}, c^{1}$ and $a^{1/3}, b^{1/3}, c^{1/3}$? Thanks again yes But according to this, this inequalitie must be homogenous http://www.mathlinks.ro/viewtopic.php?t=153230 Yet his owner says that its not pardesi wrote: FOURRIER wrote: So, whenever when its homogenous we can assume abc=1? yes but the choice should be tricky and according to the need of the problem ..... For sure

FOURRIER wrote: But according to this, this inequalitie must be homogenous http://www.mathlinks.ro/viewtopic.php?t=153230 Yet his owner says that its not Where and why

pardesi wrote: FOURRIER wrote: But according to this, this inequalitie must be homogenous http://www.mathlinks.ro/viewtopic.php?t=153230 Yet his owner says that its not Where and why homogenous in $x,y$ because we have : $x^{2},y^{2}$ and $x,y$ and $(xy)^{1}$ AS: $a^{1},b^{1},c^{1}$and $(abc)^{1/3}$ Well, I think if it was like this : $x^{1},y^{1}$ and $(xy)^{1}$ OR $x^{2},y^{2}$ and $(xy)^{1}$ it would be homogenous Is that it??

FOURRIER wrote: $x^{2},y^{2}$ and $x,y$ and $(xy)^{1}$ AS: $a^{1},b^{1},c^{1}$and $(abc)^{1/3}$ Well, I think if it was like this : $x^{1},y^{1}$ and $(xy)^{1}$ OR $x^{2},y^{2}$ and $(xy)^{1}$ it would be homogenous Is that it?? perfect

shobber wrote: Let $a$, $b$, $c$ be positive real numbers. Prove that \[\biggl(1+\frac{a}{b}\biggr) \biggl(1+\frac{b}{c}\biggr) \biggl(1+\frac{c}{a}\biggr) \ge 2 \biggl(1+\frac{a+b+c}{\sqrt[3]{abc}}\biggr). \] using $\frac{a}{b}+\frac{b}{c}+\frac{c}{a}\ge \frac{a+b+c}{\sqrt[3]{abc}}$ it's obvious.

another solution: wlog $a\ge b\ge c$. then we have $\frac{a+b}{c}+\frac{b+c}{a}+\frac{c+a}{b}\ge 2\frac{a+b+c}{\sqrt[3]{abc}}$ and we are done!

BanishedTraitor wrote: Notice $ LHS = (a+b+c)(\frac{1}{a}+\frac{1}{b}+\frac{1}{c})-1$. So our inequality is equivalent to proving: $ (a+b+c)(\frac{1}{a}+\frac{1}{b}+\frac{1}{c})\geq 3+\frac{2(a+b+c)}{\sqrt [3]{abc}}$ Let: $ \frac{a+b+c}{3}= A$, $ \frac{3}{\frac{1}{a}+\frac{1}{b}+\frac{1}{c}}= H$, $ \sqrt [3]{abc}= G$ So our inequality is equivalent to: $ 3AG\geq GH+2AH$ Since $ A\geq G\geq H$ Thus $ 3AG = AG+2 AG\geq GH+2AH$ as desired. This is just: VERY NICE SOLUTION

I see no-one has mentioned the fact that Muirhead kills this, after expanding it becomes $\sum_{sym}a^1c^0b^{-1}\ge\sum_{sym}a^{\frac{2}{3}}b^{\frac{-1}{3}}c^{\frac{-1}{3}}$.

shobber wrote: Let $a$, $b$, $c$ be positive real numbers. Prove that \[ \biggl(1+\frac{a}{b}\biggr) \biggl(1+\frac{b}{c}\biggr) \biggl(1+\frac{c}{a}\biggr) \ge 2 \biggl(1+\frac{a+b+c}{\sqrt[3]{abc}}\biggr). \] It is equivalent to $$x/y+y/z+z/x\geq \frac{x+y+z}{\sqrt[3]{xyz}},$$it is trivial by AM-GM, so we're done.

Let $a$, $b$, $c$ be positive real numbers. Prove that $$\biggl(1+\frac{a}{b}\biggr) \biggl(1+\frac{b}{c}\biggr) \biggl(1+\frac{c}{a}\biggr) \ge 5+\frac{a+b+c}{\sqrt[3]{abc}} $$$$\biggl(1+\frac{a}{b}\biggr) \biggl(1+\frac{b}{c}\biggr) \biggl(1+\frac{c}{a}\biggr) \ge \frac{8}{k+3} \biggl(k+\frac{a+b+c}{\sqrt[3]{abc}}\biggr)$$Where $k\in N^+.$

sqing wrote: Let $a$, $b$, $c$ be positive real numbers. Prove that $$\biggl(1+\frac{a}{b}\biggr) \biggl(1+\frac{b}{c}\biggr) \biggl(1+\frac{c}{a}\biggr) \ge 5+\frac{a+b+c}{\sqrt[3]{abc}} $$$$\biggl(1+\frac{a}{b}\biggr) \biggl(1+\frac{b}{c}\biggr) \biggl(1+\frac{c}{a}\biggr) \ge \frac{8}{k+3} \biggl(k+\frac{a+b+c}{\sqrt[3]{abc}}\biggr)$$Where $k\in N^+.$ here

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shobber wrote: Let $a$, $b$, $c$ be positive real numbers. Prove that \[ \biggl(1+\frac{a}{b}\biggr) \biggl(1+\frac{b}{c}\biggr) \biggl(1+\frac{c}{a}\biggr) \ge 2 \biggl(1+\frac{a+b+c}{\sqrt[3]{abc}}\biggr). \]

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Let $a$, $b$, $c$ be positive real numbers. Prove that $$\biggl(1+\frac{ka}{b}\biggr) \biggl(1+\frac{kb}{c}\biggr) \biggl(1+\frac{kc}{a}\biggr) \geq \frac{(k+1)^3}{4}\biggl(1+\frac{a+b+c}{\sqrt[3]{abc}}\biggr)$$Where $k\in N^+.$

k=2,3

Can I have a hint? The inequality is homogeneous, so we can let $abc = 1$. Our inequality then simplifies to $(a+b)(b+c)(c+a) \geq 2(1+a+b+c)$, or $a^2(b+c) + b^2(a+c) + c^2(a+b) + 2abc \geq 2 + 2(a+b+c) \implies a^2(b+c)+b^2(a+c)+c^2(a+b) \geq 2(a+b+c)$. I'm struggling to prove this. What should I use?

Bumping this. (Please don't yell at me because I bumped less than 30 minutes before 24 hours passed.)

Hint

franzliszt wrote:

Hint

Remark

Notice that the inequality is homogenous so WLOG assume $abc=1$. Our inequality then simplifies to $$(a+b)(b+c)(c+a) \geq 2(1+a+b+c).$$Expanding and using $abc=1$ gives $$a^2b+a^2c+b^2a+b^2c+c^2a+c^2b \geq 2(a+b+c).$$Now rehomogenize the inequality by multiplying the RHS by $1=a^{\frac23}b^{\frac23}c^{\frac23}$. It suffices to show $$a^2b+a^2c+b^2a+b^2c+c^2a+c^2b \geq a^{\frac53}b^{\frac23}c^{\frac23}+a^{\frac53}b^{\frac23}c^{\frac23}+a^{\frac23}b^{\frac53}c^{\frac23}+a^{\frac23}b^{\frac53}c^{\frac23}+a^{\frac23}b^{\frac23}c^{\frac53}+a^{\frac23}b^{\frac23}c^{\frac53}$$which is true by Muirhead since $(2,1,0)\succ\left(\frac53,\frac23,\frac23\right)$.

Really nice solution, I did not think of that at all!

Not sure if this method has been posted before, but I'll post it anyways. Observe that $$\prod(1+\frac xy)=\sum_{cyc}(\frac xy+\frac yx)+2\ge 8\quad (1);\quad\frac xx+\frac xy+\frac xz\ge\frac{3x}{\sqrt[3]{xyz}}\stackrel{(1)}{\implies}\frac{\prod(1+\frac xy)}2\ge\frac{4+\prod(1+\frac xy)}3=1+\frac{3+\sum(\frac xy+\frac yx)}3\ge1+\frac{x+y+z}{\sqrt[3]{xyz}},$$as desired. $\blacksquare$ edit: oops its was a,b,c but wtv in the problem i found it in said x,y,z

The original problem is a special case of the following generalization where $$\lambda=2,k=p=1$$.

Generalization 1 Let $a,b,c$ positive reals. Then prove that $$\left(\lambda +\dfrac{a^{k}}{b^{p}}\right)\left(\lambda +\dfrac{b^{k}}{c^{p}}\right)\left(\lambda +\dfrac{c^{k}}{a^{p}}\right)\geq \left(\dfrac{3\sum_{cyc}{\left(a^{k}\sqrt{b^{k-p}}\right)}-\sum_{cyc}{a^{p}}}{\sqrt{\left(abc\right)^{p}}}+\lambda^{2}+\dfrac{\left(abc\right)^{k-p}}{\lambda }\right)$$

sqing wrote: Let $a,b,c$ be positive real numbers. Proof that $$\frac{a}{b}+\frac{b}{c}+\frac{c}{a} \geq \frac{a+b+c}{\sqrt[3]{abc}}$$ Let $a,b,c$ be positive real numbers. Prove that $$\frac{b+c}{a} + \frac{c+a}{b} + \frac{a+b}{c} \geq \frac{2(a+b+c)}{\sqrt[3]{abc}}$$

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