Let $x,y,z>\frac 12$ satisfy $xyz=1$. Prove that
\[\frac{1}{2x-1}+\frac{1}{2y-1}+\frac{1}{2z-1}\ge 3.\]
Inequality with xyz=1
"Everything should be made as simple as possible, but not simpler." - Albert Einstein
- asif e elahi
- Posts:185
- Joined:Mon Aug 05, 2013 12:36 pm
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Re: Inequality with xyz=1
Let
\begin{align*}
\dfrac{1}{2x-1}=a\\
\dfrac{1}{2y-1}=b\\
\dfrac{1}{2z-1}=c
\end{align*}
So
\begin{align*}
x=\dfrac{a+1}{2a}\\
y=\dfrac{b+1}{2b}\\
z=\dfrac{c+1}{2c}
\end{align*}
Now $xyz=1$ implies $(a+1)(b+1)(c+1)=8abc$
So
\begin{align*}
8abc & = \prod\limits_{cyc}^{}(a+1)\\
& \geq \prod\limits_{cyc}^{}2\sqrt[]{a}\\
& \geq 8\sqrt[]{abc}\\
& \therefore abc \geq 1
\end{align*}
Now by AM-GM, $a+b+c\geq 3\sqrt[3]{abc}\geq 3$.
So $\sum \dfrac{1}{2x-1} \geq 3$
\begin{align*}
\dfrac{1}{2x-1}=a\\
\dfrac{1}{2y-1}=b\\
\dfrac{1}{2z-1}=c
\end{align*}
So
\begin{align*}
x=\dfrac{a+1}{2a}\\
y=\dfrac{b+1}{2b}\\
z=\dfrac{c+1}{2c}
\end{align*}
Now $xyz=1$ implies $(a+1)(b+1)(c+1)=8abc$
So
\begin{align*}
8abc & = \prod\limits_{cyc}^{}(a+1)\\
& \geq \prod\limits_{cyc}^{}2\sqrt[]{a}\\
& \geq 8\sqrt[]{abc}\\
& \therefore abc \geq 1
\end{align*}
Now by AM-GM, $a+b+c\geq 3\sqrt[3]{abc}\geq 3$.
So $\sum \dfrac{1}{2x-1} \geq 3$