Inequality a+b+c=3

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Inequality a+b+c=3

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#1 h Dec 23, 2020, 10:23 PM • 3 Y

Y by SD18, RedFireTruck, Aurn0b

Let a, b, c be positive real numbers such that a+b+c=3.
Prove that 1/(5+a^2)+1/(5+b^2)+1/(5+c^2) is less than 1/2
PS Sorry for my bad writing, I don't know how to use Latex

This post has been edited 2 times. Last edited by Mathlover2003, Dec 24, 2020, 12:21 AM
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#2 h Dec 24, 2020, 4:26 AM • 2 Y

Y by teomihai, RedFireTruck

Mathlover2003 wrote:

Let $a$ , $b$ , $c$ be positive real numbers such that $a+b+c=3$ .
Prove that $\frac{1}{5+a^2}+\frac{1}{5+b^2}+\frac{1}{5+c^2} \leq \frac{1}{2}$

Failed attempt

This post has been edited 2 times. Last edited by Quantum_fluctuations, Dec 24, 2020, 4:27 AM

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#3 h Dec 24, 2020, 7:41 AM • 1 Y

Y by RedFireTruck

Quantum_fluctuations wrote:

Mathlover2003 wrote:

Let $a$ , $b$ , $c$ be positive real numbers such that $a+b+c=3$ .
Prove that $\frac{1}{5+a^2}+\frac{1}{5+b^2}+\frac{1}{5+c^2} \leq \frac{1}{2}$

WLOG,we can assume $a\ge b \ge c$ We can easily see these:
$a\ge 1$ Thus,
$a^2+5 \ge 6$
$\frac{1}{a^2+5} \leq \frac{1}{6}$
$\frac{1}{2}\ge \frac{3}{5+a^2}$
Henceforth,
$\frac{1}{5+a^2}+\frac{1}{5+b^2}+\frac{1}{5+c^2} \leq \frac{3}{5+a^2} \leq \frac{1}{2}$

This post has been edited 4 times. Last edited by LeonhardEuler0, Dec 24, 2020, 7:47 AM
Reason: Typo

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#4 h Dec 24, 2020, 7:58 AM • 1 Y

Y by RedFireTruck

I've got a lot of failed attempts too...

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#5 h Dec 24, 2020, 8:13 AM • 2 Y

Y by RedFireTruck, Mango247

Mathlover2003 wrote:

I've got a lot of failed attempts too...

I think my solution is so easy and so clear.

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#6 h Dec 24, 2020, 9:03 AM • 1 Y

Y by RedFireTruck

LeonhardEuler0 wrote:

Mathlover2003 wrote:

I've got a lot of failed attempts too...

I think my solution is so easy and so clear.

It's wrong. 1/(5+b^2) is greater that 1/(5+a^2).

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#7 h Dec 24, 2020, 9:25 AM • 1 Y

Y by RedFireTruck

Mathlover2003 wrote:

LeonhardEuler0 wrote:

Mathlover2003 wrote:

I've got a lot of failed attempts too...

I think my solution is so easy and so clear.

It's wrong. 1/(5+b^2) is greater that 1/(5+a^2).

Oh, i am sorry

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#10 h Dec 24, 2020, 9:45 AM • 4 Y

Y by RedFireTruck, Mango247, Mango247, Mango247

Clear the denominators and homogenize. Then use A.M.-G.M. and schur.

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#12 h Dec 24, 2020, 10:04 AM • 1 Y

Y by RedFireTruck

Quantum_fluctuations wrote:

Clear the denominators and homogenize. Then use A.M.-G.M. and schur.

What does schur mean, i dont know some english contractions.

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#13 h Dec 24, 2020, 10:10 AM • 1 Y

Y by RedFireTruck

LeonhardEuler0 wrote:

Quantum_fluctuations wrote:

Clear the denominators and homogenize. Then use A.M.-G.M. and schur.

What does schur mean, i dont know some english contractions.

See schur's inequality.

https://en.wikipedia.org/wiki/Schur%27s_inequality#:~:text=with%20equality%20if%20and%20only,numbers%20x%2C%20y%20and%20z.

This post has been edited 1 time. Last edited by Quantum_fluctuations, Dec 24, 2020, 10:10 AM

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#14 h Dec 24, 2020, 10:19 AM • 1 Y

Y by RedFireTruck

Quantum_fluctuations wrote:

LeonhardEuler0 wrote:

Quantum_fluctuations wrote:

Clear the denominators and homogenize. Then use A.M.-G.M. and schur.

What does schur mean, i dont know some english contractions.

See schur's inequality.

https://en.wikipedia.org/wiki/Schur%27s_inequality#:~:text=with%20equality%20if%20and%20only,numbers%20x%2C%20y%20and%20z.

Thanks,sir.

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#15 h Dec 24, 2020, 10:26 AM • 1 Y

Y by RedFireTruck

Quantum_fluctuations wrote:

Clear the denominators and homogenize. Then use A.M.-G.M. and schur.

You're talking like it's an easy job. It we homogenize, we're going to need to compute an expression of degree 6. It is awful

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SD18

#16 h Dec 24, 2020, 10:33 AM • 1 Y

Y by RedFireTruck

Take $a+b+c=3u$ , $ab+bc+ca=3v^2$ , $abc=w^3$ this will ease the job.

This post has been edited 1 time. Last edited by SD18, Feb 27, 2021, 4:08 AM

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#17 h Dec 25, 2020, 4:10 AM • 2 Y

Y by RedFireTruck, Mango247

The desired inequality is equivalent to,

$\frac{1}{2} \geq \frac{1}{5+a^2} + \frac{1}{5+b^2} +\frac{1}{5+c^2}$

$\iff (5+a^2)(5+b^2)(5+c^2) \enspace \geq \enspace 2(5+a^2)(5+b^2) \enspace + \enspace 2(5+b^2)(5+c^2) \enspace + \enspace 2(5+c^2)(5+a^2)$
$\iff a^2b^2c^2 \enspace + \enspace 5\sum_{ \text{cyc}} a^2b^2\enspace + \enspace 25\sum_{ \text{cyc}} a^2\enspace +\enspace 125 \geq 2\sum_{ \text{cyc}} a^2b^2 \enspace + \enspace 20 \sum_{ \text{cyc}} a^2\enspace + \enspace 150$
$\iff a^2b^2c^2 \enspace + \enspace 3\sum_{ \text{cyc}} a^2b^2\enspace + \enspace 5\sum_{ \text{cyc}} a^2 \enspace \geq \enspace 25$
Now, we homogenize this inequality using the condition $a+b+c=3$

$\iff \enspace a^2b^2c^2 \enspace + \enspace \frac{1}{3} \left( \sum_{ \text{cyc}} a^2b^2 \right) \left( \sum_{ \text{cyc}} a \right)^2\enspace +\enspace \frac{5}{81} \left(\sum_{ \text{cyc}} a^2\right)\left( \sum_{ \text{cyc}} a \right)^4 \enspace \geq \enspace \frac{25}{729} \left( \sum_{ \text{cyc}} a \right)^6$
$\iff 729a^2b^2c^2 \enspace + \enspace 243 \left( \sum_{ \text{cyc}} a^2b^2 \right) \left( \sum_{ \text{cyc}} a \right)^2 \enspace + \enspace 45\left(\sum_{ \text{cyc}} a^2\right)\left( \sum_{ \text{cyc}} a\right)^4 \enspace \geq \enspace 25 \left( \sum_{ \text{cyc}} a \right)^6$
$\iff \enspace 729a^2b^2c^2 \enspace + \enspace 243\left( \sum_{ \text{cyc}} a^4b^2 \enspace + \enspace \sum_{ \text{cyc}} a^4c^2 \enspace + \enspace \sum_{ \text{cyc}} a^2b^2c^2 \enspace + \enspace \sum_{ \text{cyc}}a^3b^3 \enspace + \enspace \sum_{ \text{cyc}} a^3b^2c \enspace + \enspace \sum_{ \text{cyc}} a^3bc^2 \right)\enspace +$
$+ \enspace 45 \left( \sum_{ \text{cyc}} a^6 \enspace + \enspace 7 \sum_{ \text{cyc}} a^4b^2 \enspace +\enspace 7 \sum_{ \text{cyc}} a^4c^2 \enspace +\enspace 4 \sum_{ \text{cyc}} a^5b \enspace + \enspace 4 \sum_{ \text{cyc}} a^5c \enspace + \enspace 8 \sum_{ \text{cyc}} a^3b^3\enspace + \enspace 6 \sum_{ \text{cyc}} a^2b^2c^2\enspace + \enspace 12 \sum_{ \text{cyc}} a^4bc\enspace + \enspace 16 \sum_{ \text{cyc}} a^3b^2c \enspace +\enspace 16 \sum_{ \text{cyc}} a^3bc^2 \right) \enspace$
$\geq 25\left(\sum_{ \text{cyc}} a^6\enspace +\enspace 15 \sum_{ \text{cyc}} a^4b^2\enspace +\enspace 15\sum_{ \text{cyc}} a^4c^2\enspace +\enspace 6 \sum_{ \text{cyc}} a^5b\enspace +\enspace 6 \sum_{ \text{cyc}} a^5c\enspace +\enspace 20 \sum_{ \text{cyc}} a^3b^3\enspace +\enspace 30\sum_{ \text{cyc}} a^2b^2c^2\enspace +\enspace 60\sum_{ \text{cyc}} a^3b^2c\enspace +\enspace 60\sum_{ \text{cyc}} a^3bc^2\enspace +\enspace 30 \sum_{ \text{cyc}} a^4bc \right)$
which can be written in chinese dumass notation as,

$\begin{tabular}{rccccccccccccc} & & & & & & 20\\\noalign{\smallskip\smallskip} & & & & & 30 & & 30\\\noalign{\smallskip\smallskip} & & & & 183 & & -210 & & 183\\\noalign{\smallskip\smallskip} & & & 346 & & -294 & & -294 & & 346\\\noalign{\smallskip\smallskip} & & 183 & & -294 & & 18 & & -294 & & 183\\\noalign{\smallskip\smallskip} & 30 & & -210 & & -294 & & -294 & & -210 & & 30\\\noalign{\smallskip\smallskip} 20 & & 30 & & 183 & & 346 & & 183 & & 30 & & 20\\\noalign{\smallskip\smallskip} \end{tabular}$
$\geq 0 \quad \quad \quad \quad \quad(1)$
Now, using weighted A.M.-G.M. inequality, we have

$20\sum_{\text{cyc}} a^6\enspace+\enspace30 \sum_{\text{cyc}} a^5b \enspace+ \enspace30\sum_{\text{cyc}} a^5c\enspace+\enspace 346\sum_{\text{cyc}} a^3b^3\enspace+\enspace18\sum_{\text{cyc}} a^2b^3c\enspace+ \enspace18\sum_{\text{cyc}} a^2bc^3\enspace \geq\enspace 1386a^2b^2c^2 \quad \quad \quad (2)$
and since squares are nonnegative, we have

$105\sum_{\text{cyc}} a^4(b-c)^2 \enspace + \enspace 78 \sum_{\text{cyc}} a^2(b-c)^4 \enspace \geq \enspace 0 \quad \quad \quad \quad \quad(3)$

Finally, adding the inequalities $(2)$ and $(3)$ we obtain $(1)$ , which proves the original inequality. $\quad \quad \quad \quad \quad \quad \quad \blacksquare$

This post has been edited 11 times. Last edited by Quantum_fluctuations, Feb 6, 2021, 7:58 AM

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#18 h Dec 25, 2020, 4:32 AM • 1 Y

Y by RedFireTruck

Can this be solved using Titu's Lemma? :maybe:

:maybe:

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#19 h Dec 25, 2020, 4:43 AM • 2 Y

Y by Mathlover2003, RedFireTruck

Quantum_fluctuations wrote:

LeonhardEuler0 wrote:

Quantum_fluctuations wrote:

Clear the denominators and homogenize. Then use A.M.-G.M. and schur.

What does schur mean, i dont know some english contractions.

See schur's inequality.

https://en.wikipedia.org/wiki/Schur%27s_inequality#:~:text=with%20equality%20if%20and%20only,numbers%20x%2C%20y%20and%20z.

Beautiful!
Or with Cebisev: we have to prove:

$\frac{1}{5+a^2}+\frac{1}{5+b^2}+\frac{1}{5+c^2} \leq \frac{1}{2}$ or

$\frac{a^2-1}{5+a^2}+\frac{b^2-1}{5+b^2}+\frac{c^2-1}{5+c^2} \geq{0}$ or

$(a-1)\frac{a+1}{5+a^2}+(b-1)\frac{b+1}{5+b^2}+(c-1)\frac{c+1}{5+c^2} \geq{0}$
Now we may assume $a\geq{b}\geq{c}$ result $a-1\geq{b-1}\geq{c-1}$ and

$\frac{a+1}{5+a^2}\geq{\frac{b+1}{5+b^2}}\geq{\frac{c+1}{5+c^2}}$ result with CEBISEV:

$(a-1)\frac{a+1}{5+a^2}+(b-1)\frac{b+1}{5+b^2}+(c-1)\frac{c-1}{5+c^2} \geq{0}$

This post has been edited 7 times. Last edited by teomihai, Sep 15, 2022, 4:55 AM

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#20 h Dec 25, 2020, 4:50 AM • 2 Y

Y by teomihai, RedFireTruck

Redacted
I messed up the derivative

Thanks @2 below

This post has been edited 1 time. Last edited by BOBTHEGR8, Dec 25, 2020, 6:55 AM

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#21 h Dec 25, 2020, 5:03 AM

Y by

BOBTHEGR8 wrote:

Mathlover2003 wrote:

Let $a$ , $b$ , $c$ be positive real numbers such that $a+b+c=3$ .
Prove that $\frac{1}{5+a^2}+\frac{1}{5+b^2}+\frac{1}{5+c^2} \leq \frac{1}{2}$

Jensen kills it

Let $f(x) = \dfrac{1}{x^2+5}$ , $f''(x) = \dfrac{-(x-1)^2-4}{(x^2+5)^3} < 0$

And so $f(a)+f(b)+f(c) \leq 3f(\dfrac{a+b+c}{3}) = 3 f(1) = \dfrac 1 2$
Hence proved , equality iff $a=b=c=1$ [/quo

This post has been edited 1 time. Last edited by teomihai, Dec 25, 2020, 8:16 AM

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#22 h Dec 25, 2020, 5:08 AM • 2 Y

Y by Mathlover2003, RedFireTruck

BOBTHEGR8 wrote:

Jensen kills it

Let $f(x) = \dfrac{1}{x^2+5}$ , $f''(x) = \dfrac{-(x-1)^2-4}{(x^2+5)^3} < 0$

And so $f(a)+f(b)+f(c) \leq 3f(\dfrac{a+b+c}{3}) = 3 f(1) = \dfrac 1 2$
Hence proved , equality iff $a=b=c=1$

Well...Jensen along with Tangent line method were the first to came in my mind.

But unfortunately, tangent is not always above the curve of $\frac{1}{5+x^2}$ and correct second derivative of $\frac{1}{5+x^2}$ is

$\dfrac{2\left(3x^2-5\right)}{\left(x^2+5\right)^3}$
which is not always negative.

This post has been edited 1 time. Last edited by Quantum_fluctuations, Dec 25, 2020, 5:08 AM

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#23 h Jan 28, 2021, 10:26 AM • 2 Y

Y by RedFireTruck, Mango247

teomihai wrote:

$\frac{a+1}{5+a^2}\geq{\frac{b+1}{5+b^2}}\geq{\frac{c+1}{5+c^2}}$ result with CEBISEV:

This is not true. So your proof fails.

Actually $n-1$ EV kills it. As noted above, $f(x)=\dfrac{1}{5+x^2}$ has single inflexion point. So by by $n-1$ EV, extremum is attained when say, $a=b$ . So the ineq becomes $\dfrac{2}{5+\frac{x^2}{4}}+\dfrac{1}{5+y^2}\le \dfrac{1}{2}$ with condition $x+y=3$
This is same as, after clearing denominators and manipulations,
$(x^2+4)(y^2+3)\ge 32$ which is true bcoz
$(x^2+4)(y^2+3)=(x^2+2+2)(2+y^2+1)\ge 2(x+y+1)^2=32$ by cauchy schwarz

Quantum_fluctuations wrote:

Well...Jensen along with Tangent line method were the first to came in my mind.

You should keep n-1 EV in mind too while dealing with such ineqs

This post has been edited 2 times. Last edited by kapilpavase, Jan 28, 2021, 12:36 PM

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#24 h Feb 19, 2021, 3:05 PM • 4 Y

Y by TerenceTao11235, Mango247, Mango247, Mango247

Mathlover2003 wrote:

Let a, b, c be positive real numbers such that a+b+c=3.
Prove that $\frac{1}{5+a^2}+\frac{1}{5+b^2}+\frac{1}{5+c^2} \leq 1/2$
PS Sorry for my bad writing, I don't know how to use Latex

Mistake in calculation

This post has been edited 1 time. Last edited by LeonhardEuler0, Feb 19, 2021, 8:01 PM

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#25 h Feb 19, 2021, 4:11 PM • 1 Y

Y by teomihai

Equivalently $f(a,b,c)=\frac{a^2}{5+a^2}+\frac{b^2}{5+b^2}+\frac{b^2}{5+b^2}\ge\frac{1}{2}$ . WLOG $a\ge b,c$ , then $a\ge 1$ and $b+c\le 2$ .
We have $f(a,b,c)-f\left(a,\frac{b+c}{2},\frac{b+c}{2}\right)=\frac{5(b-c)^2\left(10-b^2-4bc-c^2\right)}{\left(5+b^2\right)\left(5+c^2\right)\left((b+c)^2+20\right)}\ge 0$ because $10-b^2-4bc-c^2\ge 10-\frac{3}{2}(b+c)^2\ge 10-\frac{3}{2}\cdot 2^2>0$ .
It suffices to prove that $f(3-2t,t,t)\ge\frac{1}{2}$ which is $\frac{5(t-1)^2\left(t^2-t+1\right)}{\left(5+t^2\right)\left(2t^2-6t+7\right)}\ge 0.$

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#26 h Feb 19, 2021, 4:28 PM

Y by

LeonhardEuler0 wrote:

Mathlover2003 wrote:

Let a, b, c be positive real numbers such that a+b+c=3.
Prove that $\frac{1}{5+a^2}+\frac{1}{5+b^2}+\frac{1}{5+c^2} \leq 1/2$
PS Sorry for my bad writing, I don't know how to use Latex

Just apply T2's lemma,
$\frac{1}{5+a^2}+\frac{1}{5+b^2}+\frac{1}{5+c^2} \leq \frac{9}{a^2+b^2+c^2+15}$
Then use CS,
$\frac{(a+b+c)^2}{3}=3 \leq a^2+b^2+c^2$
Thus,
$\frac{9}{a^2+b^2+c^2+15} \leq \frac{9}{18}=\frac{1}{2}$

T2's lemma is incorrect.

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#27 h Feb 19, 2021, 7:51 PM • 2 Y

Y by Mango247, Mango247

If we use u,v,w and fix u,v we will have second degree polynomial which is decreasing and we want to show it is greater than 0. Therefore we look for max w^3 which is a=b. The rest is easy.
Is this solution correct?
If it is not, can you post the correct solution?

This post has been edited 1 time. Last edited by SerdarBozdag, Feb 19, 2021, 7:54 PM
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#28 h Feb 19, 2021, 8:01 PM

Y by

Mathlover2003 wrote:

Let a, b, c be positive real numbers such that a+b+c=3.
Prove that 1/(5+a^2)+1/(5+b^2)+1/(5+c^2) is less than 1/2
PS Sorry for my bad writing, I don't know how to use Latex

UVW kills it.

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#29 h Feb 19, 2021, 8:37 PM

Y by

1) a,b,c<=2 : use tangent line
2) a>2 : jensn ineqality+cauchy-schwarz

I think this problem is so tight

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#30 h Feb 19, 2021, 9:06 PM

Y by

LeonhardEuler0 wrote:

Mathlover2003 wrote:

Let a, b, c be positive real numbers such that a+b+c=3.
Prove that 1/(5+a^2)+1/(5+b^2)+1/(5+c^2) is less than 1/2
PS Sorry for my bad writing, I don't know how to use Latex

UVW kills it.

As an another solution,
Expand $0 \leq \frac{2}{5+a^2} + \frac{2}{5+b^2} +\frac{2}{5+c^2} \leq 1$
Then show that $denominator \geq numerator$ with using $a+b+c=3$ .

This post has been edited 2 times. Last edited by LeonhardEuler0, Feb 19, 2021, 9:10 PM
Reason: Typo

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sqing

#32 h Sep 15, 2022, 12:10 AM

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Let $a,b,c$ be positive real numbers such that $a+b+c=3$ . Prove that
$\frac{1}{a^2+4} +\frac{1}{b^2+4} +\frac{1}{c^2+4} \leq \frac{3}{5}$ $\frac{1}{a^2+2} +\frac{1}{b^2+2} +\frac{1}{c^2+2} \leq \frac{1}{10}\left (9+\frac{1}{abc}\right)$ $\frac{1}{a^2+3} +\frac{1}{b^2+3} +\frac{1}{c^2+3} \leq \frac{1}{20}(14+\frac{1}{abc})$ $\frac{1}{a^2+3} +\frac{1}{b^2+3} +\frac{1}{c^2+3} \leq \frac{1}{93}\left ( \frac{275}{4}+\frac{1}{abc}\right)$ h h

This post has been edited 2 times. Last edited by sqing, Sep 15, 2022, 1:03 AM

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sqing

#33 h May 6, 2023, 11:59 PM

Y by

Let $a,b,c$ be real numbers such that $a+b+c=3$ . Prove that
$\frac{1}{a^2+7} +\frac{1}{b^2+7} +\frac{1}{c^2+7}\leq \frac{3}{8}$ $\frac{1}{a^2+5} +\frac{1}{b^2+5} +\frac{1}{c^2+5}\leq \frac{1}{2}$ Let $a,b,c$ be real numbers such that $a+b+c=1$ . Prove that
$\frac{1}{a^2+3} + \frac{1}{b^2+3} + \frac{1}{c^2+3} \leq \frac{27}{28}$ h h

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