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Let $a,b,c\in\mathbb{R}^*_+$, $abc=1$.

How can i show that $\left(a-1+\frac{1}b\right)\left(b-1+\frac{1}c\right)\left(c-1+\frac{1}a\right)\leq1$ ?

I got $\left(ab-b+1\right)\left(bc-c+1\right)\left(ca-a+1\right)\leq abc=1$, but i can't go any further ...

If i expand it :

$2-ab-\frac{b}a-c-ac-\frac{a}c-b+2a+\frac{2}a-bc-\frac{c}b-a+2b+\frac{2}b+2c+\frac{2}c-4$

Which gives using $c=\frac{1}{ab}$ :

$5ab-\frac{b}a-\frac{1}{ab}-a^2b+\frac{1}a+b+\frac{1}b-2$

Hippalectryon
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  • I would either expand these parentheses (perhaps pulling out $(a-1),\ (b-1),\ (c-1)$ might be of use), or simply write $c=1/ab$ and see what it gives. – Berci Jun 01 '14 at 21:43
  • it is same for the problem http://math.stackexchange.com/questions/806673/proving-lefta-1-frac1b-right-leftb-1-frac1c-right-leftc-1-frac1a-right?rq=1 – chenbai Jun 02 '14 at 04:16

1 Answers1

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Make the substitution $a=x/y$, $b=y/z$, $c=z/x$ where $x$, $y$ and $z$ are positive reals. The inequality becomes $$ \left(\frac xy - 1 + \frac zy\right) \left( \frac yz - 1 + \frac xz \right) \left( \frac zx - 1 + \frac yx \right) \stackrel{?}{\leq} 1 $$ or $$ (x-y+z)(x+y-z)(-x+y+z) \stackrel{?}{\leq} xyz. $$ At most one of the factors on the left hand side can be nonpositive and in such a case the inequality is obvious. If all factors are positive, the result follows by multiplying $$ (x-y+z)(x+y-z) \leq \left(\frac{(x-y+z)+(x+y-z)}{2}\right)^2 = x^2 $$ and similar inequalities. The last inequality follows by AM-GM on $x-y+z$ and $x+y-z$.

user133281
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