Inequality stroke

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#1 h Mar 16, 2025, 12:32 PM

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Let $a,b,c$ be real positive numbers such that: $a+b+c=abc-2$ . Prove that $\sum \frac{1}{\sqrt{ab}} \leq \frac{3}{2}$

This post has been edited 1 time. Last edited by giangtruong13, Mar 16, 2025, 12:32 PM

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#2 h Mar 16, 2025, 2:26 PM • 1 Y

Y by no_room_for_error

$c=\frac{a+b+2}{ab-1}$ , so

giangtruong13 wrote:

Let $a,b,c$ be real positive numbers such that: $a+b+c=abc-2$ . Prove that $\sum \frac{1}{\sqrt{ab}} \leq \frac{3}{2}$

$\xLeftrightarrow{\hspace{25pt}}\frac {\sqrt{ab-1}(\sqrt a+\sqrt b)}{\sqrt{a+b+2}}\le \frac 32 \sqrt {ab} -1$ . Note $ab\ge 1$ . Squaring,
$\begin{align*} \xLeftrightarrow{\hspace{25pt}}&\ \frac {ab-1}{a+b+2} (a+b+2\sqrt{ab})\le \frac 94 ab -3\sqrt{ab}+1\\ \xLeftrightarrow{\hspace{25pt}}&\ (ab-1)\cdot\frac{2(\sqrt{ab}-1)}{a+b+2}\le \frac 54 ab-3\sqrt{ab}+2. \end{align*}$ But $a+b\ge 2\sqrt{ab}\xRightarrow{\hspace{25pt}}\frac{2(\sqrt{ab}-1)}{a+b+2}\le\frac{2(\sqrt{ab}-1)}{2\sqrt{ab}+2}$ . So it suffices to prove $\frac 54 ab-3\sqrt{ab}+2-(ab-1)\cdot\frac{\sqrt{ab}-1}{\sqrt{ab}+1}=\frac {ab\sqrt{ab}-3ab+4}{4(\sqrt{ab}+1)}=\frac {(\sqrt{ab}-2)^2}{4}\ge0,$ which is true.

This post has been edited 2 times. Last edited by MS_asdfgzxcvb, Mar 16, 2025, 2:27 PM

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no_room_for_error

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no_room_for_error

#3 h Mar 16, 2025, 2:43 PM • 1 Y

Y by MS_asdfgzxcvb

There is one-line proof by AM-GM.

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arqady

#4 h Mar 16, 2025, 4:10 PM

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giangtruong13 wrote:

Let $a,b,c$ be real positive numbers such that: $a+b+c=abc-2$ . Prove that $\sum \frac{1}{\sqrt{ab}} \leq \frac{3}{2}$

Let $a=\frac{y+z}{x}$ and $b=\frac{x+z}{y},$ where $x,$ $y$ and $z$ are positives.
Thus, the condition gives $c=\frac{x+y}{z}$ and by AM-GM we obtain:
$\sum_{cyc}\frac{1}{\sqrt{ab}}=\sum_{cyc}\sqrt{\frac{x}{x+z}\cdot\frac{y}{y+z}}\leq\frac{1}{2}\sum_{cyc}\left(\frac{x}{x+z}+\frac{y}{y+z}\right)=\frac{3}{2}.$

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