If $abc=1$ where $a,b,c>0$, then show that $(a-1+b^{-1})(b-1+c^{-1})(c-1+a^{-1}) \leq 1$.

Question

I tried writing everything in terms of $a$ and $c$, but got stuck at $(a-1+ac)(a+1-ac)(1-a+ac) \leq a^2c$ where I thought of trying to show for all $x,y,z>0$ that $(x-y+z)(x+y-z)(-x+y+z) \leq xyz $ and substitute $x=1, y=a$ and $z=ac$. However, I'm not sure if that inequality is even correct and I doubt that this approach will work, so I would like some hints or a piece of advice. I would be grateful for that.

Answer 1

This is a well-known inequality from some years of IMO I think. After your substitution, use the following:

$$(x-y+z)(x+y-z)\leq\dfrac{(x-y+z+x+y-z)^2}{4}=x^2.$$

Answer 2

Let $a=\frac{x}{y}$ and $b=\frac{y}{z},$ where $x$, $y$ and $z$ are positives.

Thus, $c=\frac{z}{x}$ and we need to prove that: $$\prod_{cyc}\left(\frac{x}{y}-1+\frac{z}{y}\right)\leq1$$ $$\prod_{cyc}(x+z-y)\leq xyz$$ or $$\sum_{cyc}(x^3-x^2y-x^2z+xyz)\geq0.$$ Now, let $x\geq y\geq z$.

Thus, $$\sum_{cyc}(x^3-x^2y-x^2z+xyz)=\sum_{cyc}x(x-y)(x-z)\geq$$ $$\geq x(x-y)(x-z)+y(y-x)(y-z)=(x-y)^2(x+y-z)\geq0$$ and we are done!