(An problem due to Michael Rozenberg) Given three positive numbers $a,\,b,\,c$ so that $abc= 1$. Prove
$$\left ( a- 1+ \frac{1}{b} \right )\left ( b- 1+ \frac{1}{c} \right )\left ( c- 1+ \frac{1}{a} \right )\leqq \frac{2}{a+ b+ c- 1}$$
I have a solution for this, and I hope to see a nicer one(s), thanks for your interests a real lot !
Let $a= \dfrac{m}{n},\,b= \dfrac{n}{p},\,c= \dfrac{p}{m}\,(\!m,\,n,\,p> 0\!)$. Hence we need to prove that
$$2\,m^{2}n^{2}p^{2}\geqq (m^{2}n+ n^{2}p+ p^{2}m- mnp)(m+ n- p)(n+ p- m)(p+ m- n)$$
We see that it's enough to prove the last inequality for triangle. (the same idea with the author!)/Thus
Let $m= x+ y,\,n= y+ z,\,p= z+ x\,(\!x,\,y,\,z> 0\!)$. With Murihead's inequality, we easily get to see
$$\sum\limits_{cyc} \left (\!(\!x^{4}y^{2}\!- 2\,x^{4}yz\!+ x^{4}z^{2}\!)\!+ 2(\!x^{3}y^{3}\!+ x^{3}yz^{2}\!- 2\,x^{3}y^{2}z\!)\!\right )\!+ 2(\!x^{3}y^{2}z\!+ y^{3}z^{2}x\!+ z^{3}x^{2}y\!- 3\,xyz\!)\geqq 0$$
Best Answer
We can continue with your idea without full expanding!
Indeed, after your substitution we need to prove that: $$\frac{(x+y)(x+z)(y+z)}{4xyz}\geq\sum_{cyc}\frac{x+y}{y+z}-1.$$ Now, since this inequality is cyclic, we can assume that $z=\max\{x,y,z\}$ and we need to prove that $$\frac{(x+y)(x+z)(y+z)}{4xyz}-2\geq\frac{x+y}{y+z}+\frac{y+z}{z+x}+\frac{z+x}{x+y}-3$$ or $$\frac{\sum\limits_{cyc}(x^2y+x^2z-2xyz)}{4xyz}\geq\frac{y+z}{z+x}+\frac{z+x}{y+z}-2+\frac{z+x}{x+y}-\frac{z+x}{y+z}-\left(1-\frac{x+y}{y+z}\right)$$ or $$\frac{2z(x-y)^2+(x+y)(z-x)(z-y)}{4xyz}\geq\frac{(x-y)^2}{(x+z)(y+z)}+\frac{(z-x)(z-y)}{(x+y)(y+z)}.$$ Id est, it's enough to prove that $$\frac{1}{2xy}\geq\frac{1}{(x+z)(y+z)}$$ or $$z^2+xz+yz-xy\geq0,$$ which is obvious, and $$\frac{x+y}{4xyz}\geq\frac{1}{(x+y)(y+z)}$$ or $$(x+y)^2(y+z)\geq4xyz,$$ which is true by AM-GM: $$(x+y)^2(y+z)\geq\left(2\sqrt{xy}\right)^2z=4xyz.$$ Done!