sum(ab/4a^2+b^2) <= 3/5

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sum(ab/4a^2+b^2) <= 3/5

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Source: I remember I read it somewhere

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truongphatt2668

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truongphatt2668

#1 h Mar 31, 2025, 1:23 PM

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Let $a,b,c>0$ . Prove that:
$\dfrac{ab}{a^2+4b^2} + \dfrac{bc}{b^2+4c^2} + \dfrac{ca}{c^2+4a^2} \le \dfrac{3}{5}$

This post has been edited 1 time. Last edited by truongphatt2668, Mar 31, 2025, 1:47 PM

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truongphatt2668

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truongphatt2668

#2 h Mar 31, 2025, 1:46 PM

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In my proof, I use this:
$\dfrac{x}{x^2+4} \le \dfrac{1}{5} + \dfrac{1}{8} \ln (x), \forall x > 0$ combining this:
$\dfrac{ab}{a^2+4b^2} = \dfrac{1}{\dfrac{a}{b} + 4\dfrac{b}{a}}$ Hope to see a solution use only AM-GM

This post has been edited 2 times. Last edited by truongphatt2668, Mar 31, 2025, 1:48 PM

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arqady

#3 h Apr 1, 2025, 9:46 PM • 1 Y

Y by truongphatt2668

truongphatt2668 wrote:

In my proof, I use this:
$\dfrac{x}{x^2+4} \le \dfrac{1}{5} + \dfrac{1}{8} \ln (x), \forall x > 0$

Try $x=\frac{1}{10}$

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#6 h Wednesday at 6:30 AM • 1 Y

Y by ehuseyinyigit

truongphatt2668 wrote:

Let $a,b,c>0$ . Prove that:
$\dfrac{ab}{a^2+4b^2} + \dfrac{bc}{b^2+4c^2} + \dfrac{ca}{c^2+4a^2} \le \dfrac{3}{5}$

See here: https://nguyenhuyenag.wordpress.com/2025/04/02/inequality-41/

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truongphatt2668

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truongphatt2668

#7 h Wednesday at 11:24 AM • 1 Y

Y by ehuseyinyigit

Nguyenhuyen_AG wrote:

truongphatt2668 wrote:

Let $a,b,c>0$ . Prove that:
$\dfrac{ab}{a^2+4b^2} + \dfrac{bc}{b^2+4c^2} + \dfrac{ca}{c^2+4a^2} \le \dfrac{3}{5}$

See here: https://nguyenhuyenag.wordpress.com/2025/04/02/inequality-41/

Can you give your motivation since your assessment is quite strange

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