Inequalites

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Mario16

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Mario16

#1 h Feb 1, 2021, 10:08 PM • 1 Y

Y by Mango247

If a+b+c=3 ;a,b,c>=0 prove that 1/(5+a^2)+1/(5+b^2)+1/(5+c^2)<=1/2

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IAmTheHazard

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#2 h Feb 1, 2021, 10:46 PM

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You literally posted this 2 hours ago.

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Mario16

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Mario16

#3 h Feb 1, 2021, 10:50 PM

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Yes but i forgot to write Something

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IAmTheHazard

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IAmTheHazard

#4 h Feb 1, 2021, 10:52 PM

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It seems to be the exact same to me.

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Wildabandon

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Wildabandon

#5 h Feb 2, 2021, 12:22 AM

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Mario16 wrote:

Yes but i forgot to write Something

You can edit your post.

If $a,b,c\ge 0$ and $a+b+c=3$ , prove that
$\frac{1}{5+a^2} + \frac{1}{5+b^2} + \frac{1}{5+c^2}\le \frac{1}{2}$

This post has been edited 1 time. Last edited by Wildabandon, Feb 2, 2021, 12:23 AM

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Quantum_fluctuations

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Quantum_fluctuations

#6 h Feb 2, 2021, 12:23 AM • 1 Y

Y by Mango247

This is where you should go.
https://artofproblemsolving.com/community/c6h2387664

This post has been edited 1 time. Last edited by Quantum_fluctuations, Feb 2, 2021, 12:23 AM

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KP9

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KP9

#7 h Feb 2, 2021, 8:38 AM

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Another solution :

Easy to prove : $\frac{1}{5+a^2} \leq \frac{1}{5}(1-\frac{1}{18}a^3 - \frac{1}{9}a)$

So we have : $\sum \frac{1}{5+a^2} \leq \frac{3}{5} - \frac{1}{90}(a^3+b^3+c^3) - \frac{1}{45}(a+b+c) = \frac{3}{5} -\frac{3}{45} - \frac{1}{90}(a^3+b^3+c^3) = \frac{8}{15} - \frac{a^3+b^3+c^3}{90}$ (1)

We also have : $(1+1+1)(1+1+1)(a^3+b^3+c^3)\geq (a+b+c)^3$

$\Rightarrow a^3 + b^3 + c^3 \geq 3$ (2)

(1) and (2) $\Rightarrow \sum \frac{1}{5+a^2} \leq \frac{8}{15} - \frac{3}{90} = \frac{1}{2}$ ( Q.E.D)

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Quantum_fluctuations

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Quantum_fluctuations

#8 h Feb 2, 2021, 8:43 AM

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KP9 wrote:

Another solution :

Easy to prove : $\frac{1}{5+a^2} \leq \frac{1}{5}(1-\frac{1}{18}a^3 - \frac{1}{9}a)$

How did you find that?

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KP9

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KP9

#10 h Feb 2, 2021, 9:00 AM • 1 Y

Y by Mango247

oh , iam sorry , i have a problem when i prove it

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logrange

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logrange

#11 h Feb 2, 2021, 9:00 AM • 1 Y

Y by Mango247

I have another similar problem in which sum is cyclic in one variable only. In these kind of problems we can use tangent line, but I want to know that whether it can be used in problems which have expression ≤ constant. I have used it only in cases like expression ≥ constant. If you can do this by tangent line then please post the solution of this by tangent line also.
Prove that cyclic sum $\frac{a}{2a^2+a+1}\leq \frac{3}{4}$
Given a+b+c=3 (Sorry, I missed that earlier)

This post has been edited 3 times. Last edited by logrange, Feb 2, 2021, 5:39 PM

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logrange

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logrange

#12 h Feb 2, 2021, 5:40 PM

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Bump bump bump

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logrange

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logrange

#13 h Feb 3, 2021, 5:07 AM

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Anyone?
Note - I want a solution without n-1 EV (Calculus)

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Wildabandon

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Wildabandon

#14 h Feb 3, 2021, 5:21 AM

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I'm thinking Jensen but still using the calculus LOL

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logrange

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logrange

#15 h Feb 3, 2021, 5:25 AM

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@below
Thanks

This post has been edited 1 time. Last edited by logrange, Feb 3, 2021, 11:13 AM

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starchan

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starchan

#16 h Feb 3, 2021, 5:36 AM

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I think Chebyshev's kills this one..

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sqing

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sqing

#17 h Mar 17, 2025, 2:31 PM

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Wildabandon wrote:

If $a,b,c\ge 0$ and $a+b+c=3$ , prove that
$\frac{1}{5+a^2} + \frac{1}{5+b^2} + \frac{1}{5+c^2}\le \frac{1}{2}$

https://artofproblemsolving.com/community/c4h2391785p19635446
https://artofproblemsolving.com/community/c6h2387664p20486022
Let $a,b,c$ be non-negative numbers such that $ab+bc+ca+abc=4.$ Prove that
$\frac{1}{a+2}+\frac{1}{b+2}+\frac{1}{c+2}= 1$ $\frac{1}{2}\leq \frac{1}{a^2+4}+\frac{1}{b^2+4}+\frac{1}{c^2+4}\leq \frac{3}{5}$ https://artofproblemsolving.com/community/c6h1510436p8962718
Let $a,b,c>0$ and $a^2+b^2+c^2+ab+bc+ca =6.$ Prove that
$\frac{1}{a^2+5}+\frac{1}{b^2+5}+\frac{1}{c^2+5}\leq \frac{1}{2}$ ( Vasile Cîrtoaje)
$a^2b+b^2c+c^2a\leq \frac{368}{3}-\frac{176\sqrt{33}}{9}$ https://artofproblemsolving.com/community/c6h382474p2119615

This post has been edited 2 times. Last edited by sqing, Mar 17, 2025, 2:55 PM

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sqing

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sqing

#18 h Mar 17, 2025, 2:50 PM

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Let $a,b,c>0$ and $a^2+b^2+c^2+ab+bc+ca =6.$ Prove that
$ab+bc+ca- abc\leq 2$ $\frac{1}{a^2+1}+\frac{1}{b^2+1}+\frac{1}{c^2+1}\geq \frac{3}{2}$ $\frac{1}{a^2+8}+\frac{1}{b^2+8}+\frac{1}{c^2+8}\leq \frac{1}{3}$

This post has been edited 2 times. Last edited by sqing, Mar 17, 2025, 3:30 PM

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sqing

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sqing

#19 h Mar 17, 2025, 3:21 PM

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Let $a,b\geq 0$ and $a+b+a^2+ab+b^2 =5.$ Prove that
$\frac{1}{a^2+1}+ \frac{1}{b^2+1} \geq1$ $\frac{1}{a^2+2}+ \frac{1}{b^2+2} \geq \frac{2}{3}$ $\frac{1}{a^2+\frac{53}{20}}+ \frac{1}{b^2+\frac{53}{20}} \geq \frac{40}{73}$ $\frac{1}{a^2+\frac{1327}{500}}+ \frac{1}{b^2+ \frac{1327}{500}} \geq \frac{1000}{1827}$ $\frac{1}{a^2+3}+ \frac{1}{b^2+3} \geq \frac{185+3\sqrt{21}}{402}$

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