Inequality with abc = 1

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Katy729
Posts: 47
Joined: Sat May 06, 2017 2:30 am

Inequality with abc = 1

Unread post by Katy729 » Mon May 15, 2017 2:03 am

Let $ x, y, z$ be positive real numbers so that $ xyz = 1$. Prove that

\[ \left( x - 1 + \frac 1y \right) \left( y - 1 + \frac 1z \right) \left( z - 1 + \frac 1x \right) \leq 1.
\]

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asif e elahi
Posts: 183
Joined: Mon Aug 05, 2013 12:36 pm
Location: Sylhet,Bangladesh

Re: Inequality with abc = 1

Unread post by asif e elahi » Mon May 15, 2017 11:06 am

Hint:
Prove that
\begin{eqnarray}
\left( x - 1 + \dfrac 1y \right) \left( y - 1 + \dfrac 1z \right)\leq \dfrac{x}{z}
\end{eqnarray}

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Atonu Roy Chowdhury
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Location: Chittagong, Bangladesh

Re: Inequality with abc = 1

Unread post by Atonu Roy Chowdhury » Sat May 20, 2017 12:07 am

My solution. Same as Asif Bhai's.
$(x-1+\frac{1}{y})(y-1+\frac{1}{z})=xy-x+\frac{x}{z}-y+1-\frac{1}{z}+1-\frac{1}{y}+\frac{1}{yz}=\frac{x}{z}+2-(y+\frac{1}{y})\le \frac{x}{z}$ because $y+\frac{1}{y} \ge 2$
Similarly we get, $(y-1+\frac{1}{z})(z-1+\frac{1}{x}) \le \frac{y}{x}$ and $(z-1+\frac{1}{x})(x-1+\frac{1}{y}) \le \frac{z}{y}$ . Multiplying these three, we get the result.
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Katy729
Posts: 47
Joined: Sat May 06, 2017 2:30 am

Re: Inequality with abc = 1

Unread post by Katy729 » Sun May 21, 2017 2:42 am

Good solution Atonu! :)

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