∑1/(a+b²)≥27/4 . (a+b+c = 1)

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∑1/(a+b²)≥27/4 . (a+b+c = 1)

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Source: ∑1/(a(1+b))≥27/4

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sqing

#1 h May 27, 2012, 10:45 PM • 1 Y

Y by Adventure10

For $a,b,c>0,a+b+c=1,$ prove that

${{\frac{1}{a+b^2}+\frac{1}{b+c^2}+\frac{1}{c+a^2} \ge\frac{27}{4}}}$

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#2 h May 28, 2012, 3:55 AM • 3 Y

Y by sqing, Adventure10, Mango247

Are you sure that is the correct inequality?

I think it is:

${{\frac{1}{a-b^2}+\frac{1}{b-c^2}+\frac{1}{c-a^2} \ge\frac{27}{4}}}$

Solution

This post has been edited 1 time. Last edited by Amir Hossein, May 29, 2012, 5:11 AM
Reason: Hided the solution.

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sqing

#3 h May 28, 2012, 4:46 AM • 3 Y

Y by Ramonjms, Euler149, Adventure10

Error is the mother of success.

For $a,b,c>0,a+b+c=1,$ prove that

${{\frac{1}{a-a^2}+\frac{1}{b-b^2}+\frac{1}{c-c^2} \ge\frac{27}{2}}}$

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#4 h May 28, 2012, 5:07 AM • 2 Y

Y by Adventure10, Mango247

As $a,b,c>0,a<a+b+c=1, b<1, c<1,$ it implies $a^2<a, b^2<b, c^2<c.$
Now i can do what i did earlier.

This post has been edited 1 time. Last edited by Amir Hossein, May 29, 2012, 5:12 AM
Reason: Do not quote the whole post immediate before you.

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sqing

#5 h May 28, 2012, 6:05 AM • 2 Y

Y by Adventure10, Mango247

For $a,b,c>0,a+b+c=1,$ We have
${{\frac{1}{1+a+a^2}+\frac{1}{1+b+b^2}+\frac{1}{1+c+c^2} \ge\frac{27}{13}}}$

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arqady

#6 h May 28, 2012, 7:17 AM • 3 Y

Y by sqing, Adventure10, Mango247

It seems that the following inequality is true.
Let $a$ , $b$ and $c$ are positives such that $a+b+c=1$ . Prove that:
$\frac{1}{5a+4b^2}+\frac{1}{5b+4c^2}+\frac{1}{5c+4a^2} \ge\frac{27}{19}$

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#7 h May 28, 2012, 11:23 AM • 3 Y

Y by sqing, Adventure10, Mango247

sqing wrote:

For $a,b,c>0,a+b+c=1,$ We have
${{\frac{1}{1+a+a^2}+\frac{1}{1+b+b^2}+\frac{1}{1+c+c^2} \ge\frac{27}{13}}}$

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#8 h May 29, 2012, 2:36 AM • 3 Y

Y by sqing, Adventure10, Mango247

sqing wrote:

For $a,b,c>0,a+b+c=1,$ prove that

${{\frac{1}{a+b^2}+\frac{1}{b+c^2}+\frac{1}{c+a^2} \ge\frac{27}{4}}}$

I remembered JiChen's very old one but quite hard one:
$\frac{1}{a+b^2}+\frac{1}{b+c^2}+\frac{1}{c+a^2}\geq \frac{13}{2-2abc}>\frac{13}{2}$
It seems that there is no solution yet(without full expanding).

:

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#9 h May 29, 2012, 10:13 AM • 2 Y

Y by Adventure10, Mango247

arqady wrote:

It seems that the following inequality is true.
Let $a$ , $b$ and $c$ are positives such that $a+b+c=1$ . Prove that:
$\frac{1}{5a+4b^2}+\frac{1}{5b+4c^2}+\frac{1}{5c+4a^2} \ge\frac{27}{19}$

Maybe use Cauchy-Schwarz inequality in this way:
$\sum{\frac{1}{5a+4b^2}}\geq \frac{49(a+b+c)^2}{\sum{(a+2b+4c)^2(5a+4b^2)}}$
But finnally we have to prove that
$364\sum{a^4}+1537\sum{bc^3}\geq 570\sum{a^2b^2}+839\sum{a^3b}+492abc\sum{a}$
The number seems to big

Arqady, maybe you have another solution?

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arqady

#10 h May 29, 2012, 10:37 AM • 3 Y

Y by sqing, Adventure10, Mango247

Tourish wrote:

Arqady, maybe you have another solution?

The idea is the same! What else we have here?

But your last inequality is wrong. Try $a=2$ , $b=1$ and $c\rightarrow0^+$ .

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#11 h May 30, 2012, 2:38 PM • 1 Y

Y by Adventure10

arqady wrote:

But your last inequality is wrong. Try $a=2$ , $b=1$ and $c\rightarrow0^+$ .

But when $a=2,b=1$ , it becomes
$8505c+270-3834c^2-141c^3+364c^4\geq 0$
what's wrong with it

:

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arqady

#12 h May 30, 2012, 2:48 PM • 2 Y

Y by Adventure10, Mango247

Sorry Tourish. My mistake.

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sqing

#13 h Oct 16, 2015, 8:27 AM • 2 Y

Y by Adventure10, Mango247

Let $a$ , $b$ and $c$ are non-negative numbers such that $a+b+c=1$ . Prove that $\frac{5}{2}\leq\frac{1+3bc}{1+a^2}+\frac{1+3ca}{1+b^2}+\frac{1+3ab}{1+c^2}\leq \frac{18}{5}.$

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arqady

#14 h Oct 29, 2015, 2:55 PM • 2 Y

Y by Adventure10, Mango247

sqing wrote:

Let $a$ , $b$ and $c$ are non-negative numbers such that $a+b+c=1$ . Prove that $\frac{1+3bc}{1+a^2}+\frac{1+3ca}{1+b^2}+\frac{1+3ab}{1+c^2}\leq \frac{18}{5}.$

It's obviously true by SOS.

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Wangzu

#15 h Oct 29, 2015, 4:24 PM • 3 Y

Y by Whycannot, Adventure10, Mango247

sqing wrote:

Let $a$ , $b$ and $c$ are non-negative numbers such that $a+b+c=1$ . Prove that $\frac{5}{2}\leq\frac{1+3bc}{1+a^2}+\frac{1+3ca}{1+b^2}+\frac{1+3ab}{1+c^2}\leq \frac{18}{5}.$

I haven't tried S.O.S before but $uvw$ is very well

. RHS equivalent to $f(w^3) \leq 0$ , $f(w^3)$ is a convex function of $w^3$ .

LHS equivalent to $f(w^3) \geq 0$ , $f(w^3)$ is a decreasing function of $w^3$

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