Since $abc = 1$, there exist positive real numbers $x$, $y$ and $z$ such that $a = x/y$, $b = y/z$ and $c = z/x$. The inequality then reduces to \[ \left( \frac{x-y+z}y \right) \left( \frac{y-z+x}z \right) \left( \frac{z-x+y}x \right) \leq 1. \] Substitute $p= b + c - a$, $q = c + a - b$ and $r = a + b - c$. The inequality then reduces to \[ 8pqr \le (p+q)(q+r)(r+p) \] which is obvious from the AM-GM since at most one of $p,q,r$ can be negaitve..

There's no need to quote my post for giving a solution Peter

I'm sorry, it's an old habit. Maybe you can configure the resource page so that it ignores quotes? (just like it ignores attachements)

From problem every one can take as $abc\geq (a+b-c)(a-b+c)(-a+b+c)$ so we have to prove this WLOG we can assume $a\geq b\geq c$ then $a+c>b$ and $a+b>c$ 1) $b+c>a$ an others then it ia sides of triangle so take $a=x+y, b=y+z, c=z+x$ so from it follows. (we will look next step) 2) $a\geq b+c$ . Since $a,b,c$ are positive so $abc>0\geq (a+b-c)(a-b+c)(-a+b+c)$ Abdurashid

Peter VDD wrote: \[8pqr \le (p+q)(q+r)(r+p)\] which is obvious from the AM-GM since at most one of $p,q,r$ can be negaitve.. I don't understand this step. Please, can anyone explain me?

José wrote: Peter VDD wrote: \[8pqr \le (p+q)(q+r)(r+p)\] which is obvious from the AM-GM since at most one of $p,q,r$ can be negaitve.. I don't understand this step. Please, can anyone explain me? The numbers $(p+q),(q+r),(r+p)$ are positive by definition. Now obviously at most one of the numbers $p,q,r$ can be negative. If one of the numbers is negative, then it is trivial since the left side would be negative and the right side positive. Hence we can assume that they are are all nonnegative and then it is an application of AmGm.

I feel ignorant. How do I pass from AM-GM inequality to the inequality that Peter gave? I understood everything in his proof but this (the passage)

you can expand it and apply AM-GM with 8 variables or you can apply $p+q\ge2\sqrt{pq}$ and multiply each factor

Now yes, thanks!

Peter VDD wrote: you can expand it and apply AM-GM with 8 variables or you can apply $p+q\ge2\sqrt{pq}$ and multiply each factor Nice options...

I think this is easier______CHN student

oh wow thats nice also hehe i was stuck with that inequality for a while...

In fact, it's not entirely correct. Why would $x+y-z\ge0$ for example? If not, you can't appy AM-GM. There is a solution in that sense, but not as trivial as the one posted there.

I certainly know.But I'm a chinese who is not good at English,so I pass over a lot of words,sorry.

write it in chinese and pass it through an online translator

themonster wrote:

Sorry that's incorrect; also, combined with the following inequality, its absurd!! (Just flip $y$ and $z$ to see)... Also, there is no way to prove this without any casework or a stronger inequality

That's it.

$ \left( a - 1 + \frac 1b \right) \left( b - 1 + \frac 1c \right) \left( c - 1 + \frac 1a \right) \leq 1$ inequality is equal as: 1. $ ( ab - b + 1)( cb - c + 1) ( ac - a + 1) \leq 1$ 2. $ ( a - 1 + ac)( b - 1 + ab) ( bc - 1 + c) \leq 1$ 3. $ ( 1 - bc + c)( 1 - ac + a) ( 1 - ab + b) \leq 1$ with abc=1 $ x = ( ab - b + 1) = ( 1 - bc + c); y = ( cb - c + 1) = ( 1 - ac + a); z = ( ac - a + 1) = ( 1 - ab + b)$ if x,y,z<0 or one of them is less then 0 and others greater than 0 then equality is obvius if x<0 and y<0 than $ cb + 1 \leq c$ $ a + 2 = a(cb + 1) + 1\leq ac + 1 \leq a$ which isn't possible so we have contradiction on the same way you can prove that it isn't possible that x<0 and z<0 ; y<=0 and z<=0 so now we only must prove that inequality worths for x>0, y>0 and z>0. $ xyz \leq {(\frac {x + y + z}{3})}^3$ $ ( ab - b + 1)( cb - c + 1)( ac - a + 1) \leq {(\frac {3 + ab + bc + ac + - a - b - c}{3})}^3$ $ ( 1 - bc + c)( 1 - ac + a) ( 1 - ab + b) \leq {(\frac {3 - ab - bc - ac + a + b + c}{3})}^3$ and now we see that one of 2 rigth sides of upper inequalities must be less than 1 and that implicates that the other one must be less than 1 as well and we are done.

Replace $(a,b,c) = \left(\tfrac{x}{y}, \tfrac{y}{z}, \tfrac{z}{x}\right)$. Then: \begin{align*} \left( a - 1 + \frac 1b \right) \left( b - 1 + \frac 1c \right) \left( c - 1 + \frac 1a \right) \leq 1 \quad \iff \quad (x+z-y)(y+z-x)(x+y-z) \leq xyz. \end{align*}Now, replace $(x,y,z) = (p+q,q+r,r+s)$. Then: \begin{align*} (x+z-y)(y+z-x)(x+y-z) \leq xyz \quad \iff \quad 8pqr \leq (p+q)(q+r)(r+s). \end{align*}But, by AM-GM, note that $p+q \geq 2\sqrt{pq}$. Multiplying cyclically yields the desired result.

dudade wrote: Now, replace $(x,y,z) = (p+q,q+r,r+s)$. Then: \begin{align*} (x+z-y)(y+z-x)(x+y-z) \leq xyz \quad \iff \quad 8pqr \leq (p+q)(q+r)(r+s). \end{align*}But, by AM-GM, note that $p+q \geq 2\sqrt{pq}$. Multiplying cyclically yields the desired result. I don’t understand your logic. There is no guarantee that p,q and r is positive

sqing wrote: Valentin Vornicu wrote: Let $ a, b, c$ be positive real numbers so that $ abc =1$. Prove that \[ \left(a-1+\frac{1}{b}\right)\left(b-1+\frac{1}{c}\right)\left(c-1+\frac{1}{a}\right) \leq 1\] Let $ a, b, c$ be positive real numbers so that $ abc =1$. Prove that $$\left(a-1+\frac{1}{b}\right)\left(b-1+\frac{1}{c}\right)\left(c-1+\frac{1}{a}\right) \leq \frac{1}{a+2}+ \frac{1}{b+2}+\frac{1}{c+2}\leq 1$$(Ji chen) Does muirhead work

Since $abc = 1$, let $a = \frac{x}{y}$, $b = \frac{y}{z}$, and $\frac{z}{x}$. Thus it suffices to prove \[\prod_{\text{cyc}} \frac{x+z-y}{y} \le 1\]Expanding, it suffices to prove that \[\sum_{\text{cyc}} x^3 + 3xyz \ge \sum_{\text{sym}} x^2y\]which is true by Schur.

Gold is old heh (or whatever )

Let $\frac xy = a$ and cyclic variants, then we can rewrite as $\prod_{cyc} (x - y + z) \le xyz$. Observe at most one of the cyclic terms can be negative, since if the variable being negated is not maximized we can easily see the term is positive. If one term is negative, we instantly win by sign, otherwise we see that $x < y + z$ and cyclic variants so $x,y,z$ are sides of a triangle, set $p,q,r$ with $q + r = x$ and cyclic variants, then we are reduced to proving $8pqr \le (p +q)(q + r)(p + r)$, which is obvious by AM-GM on each term on the RHS.

Perform the substitution $a = \frac xy, b = \frac yz, c = \frac zx$ with $x, y, z >0$. Then we have \[ \left( a - 1 + \frac 1b \right) \left( b - 1 + \frac 1c \right) \left( c - 1 + \frac 1a \right) \leq 1 \iff (x+y-z)(x-y+z)(-x+y+z) \le xyz. \]Now we have two cases: Case 1: The number of terms out of $(x+y-z), (x-y+z), (-x+y+z)$ which are negative is at least $2$. WLOG $$x + y < z, x + z < y,$$then adding yields a contradiction. Case 2: There is one negative term among $(x+y-z), (x-y+z), (-x+y+z)$. Then $$(x+y-z)(x-y+z)(-x+y+z) \le xyz < 0 < xyz.$$Case 3: All of the terms $(x+y-z), (x-y+z), (-x+y+z)$ are nonnegative. Perform the substitution $x = p+q, y = q+r, z = r+p$. with $p, q, r \ge 0$. Then the inequality reduces to $$8pqr \le (p+q)(q+r)(r+p).$$which is true by AM-GM. Since all of the cases have been exhausted, we have proven the inequality. $\square$

We have $a,b,c \in \mathbb{R^{+}}$ with $abc=1$. Claim: $\left( a - 1 + \frac 1b \right) \left( b - 1 + \frac 1c \right) \left( c - 1 + \frac 1a \right) \leq 1$ Proof: On expanding the inequality and using $abc=1$, the above expression is equivalent to: $$ a+b+c+ab+bc+ca-(a^{2}b+b^{2}c+c^{2}a) \leq 3$$Now by using the substitution $\left(a=\dfrac{u}{v},b=\dfrac{v}{w},c=\dfrac{w}{u}\right)$ we can homogenize the above inequality. Thus we have to show that, $$ u^{3}+v^{3}+w^{3}+3uvw \geq uv(u+v)+vw(v+w)+wu(w+u)$$which is trivial by Schur's and hence, we are done. $\blacksquare$

Write $a=\frac{x}{y}$ and so on. Multiply the product of denominators on both sides of the inequality and expand the $LHS$ and simplify a bit. The resulting inequality is just a case of Schur's.

Let $$a=\frac{x}{y}$$$$b=\frac{y}{z}$$$$c=\frac{z}{x}.$$We wish to show that $$(\frac{x+z-y}{y})(\frac{y+x-z}{z})(\frac{z+y-x}{x})\le 1,$$or $$(-x+y+z)(x-y+z)(x+y-z)\le xyz.$$Expanding gives Schur's inequality, which is true.

Let's substitute as follows: a= y/x, b= z/y, c= x/z. (a -1 +1/b)(b -1 +1/c)(c -1 +1/a)=< 1 ⟺ (xy+yz-zx)(xy-yz+zx)(-xy+yz+zx)=<(x^2)(y^2)(z^2). The right-hand side is always positive. Only two terms on the left-hand side cannot be negative. Let xy+yz-zx=k, xy-yz+zx=m, -xy+yz+zx=n. (xy+yz-zx)(xy-yz+zx)(-xy+yz+zx)=<(x^2)(y^2)(z^2)⟺kmn=<{(k+m)(m+n)(n+k)/8}. It holds by the arithmetic-geometric mean inequality. ∴The problem has been proven.

by not opening full but grouping and using abc=1 and from titu and some cases will give easily

I have discussed this problem in my youtube channel in my inequalities tutorial playlist. Here is the link

Two years ago this problem took me $2$ hours