Inequalities

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#1 h Apr 23, 2025, 6:36 PM

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If $x, y, z \in \mathbb{R}$ , then prove that the following inequality holds:
$\sum_{\text{cyc}} \sqrt{1 + \left(x\sqrt{1 + y^2} + y\sqrt{1 + x^2}\right)^2} \geq \sum_{\text{cyc}} xy + 2\sum_{\text{cyc}} x$

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#2 h Apr 23, 2025, 8:42 PM

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LHS is equivalent to:
$\sum_{\text{cyc}} \sqrt{1 + \left(x\sqrt{1 + y^2} + y\sqrt{1 + x^2}\right)^2} = \sum_{\text{cyc}} \sqrt{1 + x^2(1+y^2) + y^2(1+x^2) + 2xy\sqrt{(1+x^2)(1+y^2)}}$ $= \sum_{\text{cyc}} \sqrt{1 + x^2 + y^2 + 2x^2y^2 + 2xy\sqrt{(1+x^2)(1+y^2)}}$ Using $AM \geq GM$ inequality we have:
$x^2y^2 + 1 \geq 2xy$ Using $CS$ inequality we have:
$(1 + x^2)(1 + y^2) = (1 + x^2)(y^2 + 1) \geq (x + y)^2$ that gives:
$\sqrt{(1 + x^2)(1 + y^2)} \geq x + y$ Now we have:
$\sum_{\text{cyc}} \sqrt{1 + x^2 + y^2 + 2x^2y^2 + 2xy\sqrt{(1+x^2)(1+y^2)}} \geq \sum_{\text{cyc}} \sqrt{x^2 + y^2 + x^2y^2 + 2xy + 2xy(x+y)}$ $= \sum_{\text{cyc}} \sqrt{(x+y)^2 + x^2y^2 + 2xy(x+y)} = \sum_{\text{cyc}} \sqrt{(x+ y + xy)^2} = \sum_{\text{cyc}} (x+ y + xy) = \sum_{\text{cyc}} xy + 2\sum_{\text{cyc}} x$ Equality holds for $x = y = z = 1$ . $\blacksquare$

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arqady

#3 h Apr 24, 2025, 3:39 AM

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Bergo1305 wrote:

\]
Using $CS$ inequality we have:
$(1 + x^2)(1 + y^2) = (1 + x^2)(y^2 + 1) \geq (x + y)^2$ that gives:
$\sqrt{(1 + x^2)(1 + y^2)} \geq x + y$ Now we have:
$\sum_{\text{cyc}} \sqrt{1 + x^2 + y^2 + 2x^2y^2 + 2xy\sqrt{(1+x^2)(1+y^2)}} \geq \sum_{\text{cyc}} \sqrt{x^2 + y^2 + x^2y^2 + 2xy + 2xy(x+y)}$

$xy$ may be negative

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#5 h Apr 26, 2025, 6:00 PM • 2 Y

Y by arqady, kiyoras_2001

Here is the solution for real number.
$\sum_{\text{cyc}} \sqrt{1 + \left(x\sqrt{1 + y^2} + y\sqrt{1 + x^2}\right)^2} = \sum_{\text{cyc}} \sqrt{1 + x^2(1+y^2) + y^2(1+x^2) + 2xy\sqrt{(1+x^2)(1+y^2)}}$
$= \sum_{\text{cyc}} \sqrt{1 + x^2 + y^2 + 2x^2y^2 + 2xy\sqrt{(1+x^2)(1+y^2)}} = \sum_{\text{cyc}} \sqrt{(xy + \sqrt{(1+x^2)(1+y^2)})^2}$
$= \sum_{\text{cyc}} \mid xy + \sqrt{(1+x^2)(1+y^2)} \mid$
By using inequality (which is true for all real numbers)
$\sqrt{(1 + x^2)(1 + y^2)} \geq x + y$
and using inequality
$\mid x \mid \geq x$
we have:

$= \sum_{\text{cyc}} \mid xy + \sqrt{(1+x^2)(1+y^2)} \mid \geq \sum_{\text{cyc}} xy + \sum_{\text{cyc}} x + y = \sum_{\text{cyc}} xy + 2\sum_{\text{cyc}} x$
Equality holds for $x = y = z = 1.$

This post has been edited 1 time. Last edited by Bergo1305, Apr 27, 2025, 2:11 AM
Reason: Type mistake

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arqady

#6 h Apr 26, 2025, 7:32 PM

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Bergo1305 wrote:

$\sum_{\text{cyc}} \sqrt{1 + x^2 + y^2 + 2x^2y^2 + 2xy\sqrt{(1+x^2)(1+y^2)}} = \sum_{\text{cyc}} \sqrt{(xy + 2xy\sqrt{(1+x^2)(1+y^2)})^2}$

Try $x=y=z=0$ .
If it means $\geq$ , so are you sure?

This post has been edited 1 time. Last edited by arqady, Apr 26, 2025, 7:33 PM

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#7 h Apr 27, 2025, 2:12 AM • 1 Y

Y by arqady

arqady wrote:

Bergo1305 wrote:

$\sum_{\text{cyc}} \sqrt{1 + x^2 + y^2 + 2x^2y^2 + 2xy\sqrt{(1+x^2)(1+y^2)}} = \sum_{\text{cyc}} \sqrt{(xy + 2xy\sqrt{(1+x^2)(1+y^2)})^2}$

Try $x=y=z=0$ .
If it means $\geq$ , so are you sure?

I corrected, it was typing mistake. Sorry

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