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#1 h Mar 31, 2025, 5:20 AM

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problem

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arqady

#2 h Mar 31, 2025, 9:56 AM

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Cobedangiu wrote:

$a,b,c\geq0,\sum ab>0$ Prove that: $\sum\frac{ab}{a^2+4b^2}\leq\frac{3}{5}$

Use C-S and $uvw$ . There is a nice solution.

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#3 h Mar 31, 2025, 11:07 AM • 1 Y

Y by xytunghoanh

arqady wrote:

Cobedangiu wrote:

$a,b,c\geq0,\sum ab>0$ Prove that: $\sum\frac{ab}{a^2+4b^2}\leq\frac{3}{5}$

Use C-S and $uvw$ . There is a nice solution.

Is there any other method besides $uvw$ method?

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truongphatt2668

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truongphatt2668

#5 h Apr 1, 2025, 2:43 PM • 1 Y

Y by Cobedangiu

Cobedangiu wrote:

problem

https://artofproblemsolving.com/community/u1030419h3538853p34415868

Oh its yours

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#6 h Apr 1, 2025, 3:02 PM

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Sorry for my fake solve :blush:

:blush:

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#7 h Apr 1, 2025, 3:10 PM

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Let $a,b,c>0$ , $5\ge k\ge \frac {1}{5}$ . Prove that
$\frac{ab}{a^2+kb^2}+\frac{bc}{b^2+kc^2}+\frac{ca}{c^2+ka^2}\leq \frac{3}{1+k}$

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truongphatt2668

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truongphatt2668

#8 h Apr 1, 2025, 3:38 PM

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Doing the same thing by MV

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arqady

#9 h Apr 2, 2025, 8:34 AM

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xytunghoanh wrote:

Suppose $z=\min \{x,y,z \}$
By MV, we have
$\frac{x}{x^2+4}+\frac{y}{y^2+4}\le \frac{2\sqrt{xy}}{xy+4}$

Are you sure?

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#10 h Apr 2, 2025, 1:38 PM

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arqady wrote:

xytunghoanh wrote:

Suppose $z=\min \{x,y,z \}$
By MV, we have
$\frac{x}{x^2+4}+\frac{y}{y^2+4}\le \frac{2\sqrt{xy}}{xy+4}$

Are you sure?

I think I can't suppose $z=\min \{x,y,z \}$ in here because it is $a^2\ge bc$ (not true). Then it's a fake solve :v. Thanks you arqady

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#11 h Apr 3, 2025, 8:23 AM • 1 Y

Y by MS_asdfgzxcvb

equivalent to
$12(\sum{a^2b}-2\sum{ab^2}+3abc)^2+28\sum{ab(ab+c^2-2ac+\frac25(ac-bc))^2}+\frac{43}{25}abc(\sum{a^2b}-3abc)+\frac{328}{25}abc(\sum{ab^2}-3abc)\ge0$

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