Inequalites

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#1 h Feb 1, 2021, 8:31 PM

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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

This post has been edited 1 time. Last edited by Mario16, Feb 1, 2021, 9:09 PM

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#2 h Feb 1, 2021, 8:46 PM

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Oh, I proved the opposite inequality like this
where did I go wrong?

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#3 h Feb 1, 2021, 9:00 PM

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Mario16 wrote:

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

Counterexample: $(a,b,c)=\left(\frac32,\frac32,0\right)$ .
$LHS=\frac{33}{65}>\frac12$

@above seems to be correct

This post has been edited 1 time. Last edited by jasperE3, Feb 1, 2021, 9:01 PM

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#5 h Feb 1, 2021, 9:10 PM • 1 Y

Y by KP9

Sory i forgot to write a^2

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BelieverofMaths

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BelieverofMaths

#6 h Feb 2, 2021, 7:28 PM • 1 Y

Y by Mango247

IF a+b+c = 3 ; a,b,c $\geq$ 0prove that $\sum_{cyc}\frac{1}{5+a^2}\leq\frac{1}{2}$

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#7 h Mar 12, 2021, 2:06 PM

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Please help me with This problem

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WonderInAliceLand

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WonderInAliceLand

#8 h Mar 12, 2021, 3:07 PM

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Can we use jensen inequality because $\frac{1}{5+a^2}$ concave for non negative a.
$\frac{1}{5+a^2}+\frac{1}{5+b^2}+\frac{1}{5+c^2} \leq \frac{1}{5+\left(\frac{a+b+c}{3}\right)^2}= \frac{1}{2}$

This post has been edited 2 times. Last edited by WonderInAliceLand, Mar 12, 2021, 3:08 PM

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Quantum_fluctuations

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Quantum_fluctuations

#9 h Mar 12, 2021, 4:32 PM

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WonderInAliceLand wrote:

$\frac{1}{5+a^2}$ concave for non negative a.

WHY? Who told you it's concave??

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Quantum_fluctuations

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Quantum_fluctuations

#10 h Mar 12, 2021, 4:35 PM • 1 Y

Y by HamstPan38825

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. $\quad \quad \quad \quad \quad \quad \quad \blacksquare$

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#11 h Mar 12, 2021, 4:47 PM

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Quantum_fluctuations wrote:

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. $\quad \quad \quad \quad \quad \quad \quad \blacksquare$

That's a bit overkill...

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#12 h Mar 12, 2021, 7:53 PM

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I have a solution better quiqkly
The condition can be write
Sigma [1/6-1/(a^2+5)]>=0
Sigma (a^2-1)/(a^2+5)>=0
But Sigma (a^2-1)/(a^2+5)>=Sigma (2a-2)/(a^2+5)
Now using the Cebîșev inequalites is easy

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WonderInAliceLand

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WonderInAliceLand

#13 h Mar 13, 2021, 1:21 AM

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Quantum_fluctuations wrote:

WonderInAliceLand wrote:

$\frac{1}{5+a^2}$ concave for non negative a.

WHY? Who told you it's concave??

Because it's second derivative is negative

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#14 h Mar 13, 2021, 1:45 AM

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$f(x)=\frac1{x^2+5}$
$f'(x)=-\frac{2x}{\left(x^2+5\right)^2}$
$f''(x)=\frac{6x^2-10}{\left(x^2+5\right)^3}$
So if $x\ge2$ , then $f''(x)\ge0$ , however, $x\in[0,2)\Rightarrow f''(x)<0$
$f$ is thus neither convex nor concave.

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#15 h Mar 13, 2021, 3:22 AM

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@#11
Is it just me, or do I like that solution because CDA makes me satisfied?

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