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#1 h Apr 2, 2025, 6:11 PM

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Let $x,y,z>0$ and $xy+yz+zx=3$ : Prove that :
$\sum \ \frac{x}{y+z}\ge\sum \frac{1}{\sqrt{x+3}}$

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#4 h Apr 3, 2025, 9:03 AM

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...........

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arqady

#5 h Apr 3, 2025, 10:37 AM

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Cobedangiu wrote:

Let $x,y,z>0$ and $xy+yz+zx=3$ : Prove that :
$\sum \ \frac{x}{y+z}\ge\sum \frac{1}{\sqrt{x+3}}$

Use Jensen and $uvw$ .

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Math-Problem-Solving

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Math-Problem-Solving

#6 h Apr 22, 2025, 3:32 PM

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arqady wrote:

Cobedangiu wrote:

Let $x,y,z>0$ and $xy+yz+zx=3$ : Prove that :
$\sum \ \frac{x}{y+z}\ge\sum \frac{1}{\sqrt{x+3}}$

Use Jensen and $uvw$ .

What does this $uvw$ stand for?

This post has been edited 1 time. Last edited by Math-Problem-Solving, Apr 22, 2025, 3:33 PM
Reason: Err

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#7 h Apr 22, 2025, 3:39 PM

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$x+y+z=u$ , $xy+yz+zx=v$ , $xyz=w$

This post has been edited 1 time. Last edited by giangtruong13, Apr 22, 2025, 3:39 PM

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#8 h Apr 22, 2025, 3:54 PM

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giangtruong13 wrote:

$x+y+z=u$ , $xy+yz+zx=v$ , $xyz=w$

That's $pqr$ ,not $uvw$ .

$uvw$ should be $a+b+c=3u , ab+bc+ca=3v^2 , abc=w^3.$

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#9 h Apr 22, 2025, 4:01 PM

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pqr LMAO
I prefer pqr

This post has been edited 1 time. Last edited by Jackson0423, Apr 22, 2025, 4:01 PM
Reason: d

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#10 h Apr 22, 2025, 7:48 PM

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Math-Problem-Solving wrote:

What does this $uvw$ stand for?

$x+y+z=3u$ , $xy+xz+yz=3v^2$ and $xyz=w^3$ .

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#11 h Apr 23, 2025, 1:52 PM

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Jackson0423 wrote:

$pqr$ LMAO
I prefer $pqr$

I also prefer $pqr$ ,but $uvw$ just seems stronger-
Why is this though?

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