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#1 h Jan 5, 2016, 4:11 AM • 2 Y

Y by Adventure10, Mango247

Given $a,b,c>0$ & $(a+b+c)(\frac{1}{a}+\frac{1}{b}+\frac{1}{c})=10$ . Prove that
$(a^2+b^2+c^2)(\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2})\ge \frac{27}{2}$

This post has been edited 1 time. Last edited by mrmath2006, Jan 5, 2016, 4:11 AM

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arqady

#2 h Jan 5, 2016, 5:22 AM • 1 Y

Y by Adventure10

mrmath2006 wrote:

Given $a,b,c>0$ & $(a+b+c)(\frac{1}{a}+\frac{1}{b}+\frac{1}{c})=10$ . Prove that
$(a^2+b^2+c^2)(\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2})\ge \frac{27}{2}$

Let $a+b+c=3u$ , $ab+ac+bc=3v^2$ , $abc=w^3$ and $u^2=tv^2$ .
Hence, $t\geq1$ and the condition gives $w^3=\frac{9uv^2}{10}$ .
Since $(a-b)^2(a-c)^2(b-c)^2\geq0$ gives $3u^2v^4-4v^6-4u^3w^3+6uv^2w^3-w^6\geq0$ ,
we obtain $360u^4-759u^2v^2+400v^4\leq0$ , which is $\frac{25}{24}\leq t\leq\frac{16}{15}$ .
Now $(a^2+b^2+c^2)(\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2})=\frac{(9u^2-6v^2)(9v^4-6uw^3)}{w^6}=\frac{20(3u^2-2v^2)(5v^2-3u^2)}{3u^2v^2}=\frac{20(3t-2)(5-3t)}{3t}\geq\frac{27}{2}$ .

This post has been edited 1 time. Last edited by arqady, Jan 5, 2016, 5:23 AM

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2032 posts

#3 h Jan 5, 2016, 5:27 AM • 3 Y

Y by rightways, Adventure10, Mango247

mrmath2006 wrote:

Given $a,b,c>0$ & $(a+b+c)(\frac{1}{a}+\frac{1}{b}+\frac{1}{c})=10$ . Prove that
$(a^2+b^2+c^2)(\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2})\ge \frac{27}{2}$

We have
$(a^2+b^2+c^2)(\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2})-\frac{27}{2}=\frac{(a-b)^2(b-c)^2(c-a)^2}{2a^2b^2c^2}\ge{0}$

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3302 posts

#4 h Jan 5, 2016, 5:44 AM • 6 Y

Y by bobthesmartypants, huricane, rightways, AlanZ307, Adventure10, MathIQ.

mrmath2006 wrote:

Given $a,b,c>0$ & $(a+b+c)(\frac{1}{a}+\frac{1}{b}+\frac{1}{c})=10$ . Prove that
$P=(a^2+b^2+c^2)(\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2})\ge \frac{27}{2}$

Let $x=\frac{a}{b}+\frac{b}{c}+\frac{c}{a},\,y=\frac{b}{a}+\frac{c}{b}+\frac{a}{c},$ we have
$x+y=(a+b+c)\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)-3=10-3=7.$ By Cauchy-Schwarz Inequality, we have
$P=x^2+y^2-2(x+y)+3=x^2+y^2-11 \geqslant \frac{1}{2}(x+y)^2-11=\frac{27}{2}.$ The proof is complete.

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#5 h Jan 5, 2016, 5:51 AM • 1 Y

Y by Adventure10

If $a,\,b,\,c$ are positive real numbers such that $(a+b+c)\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)=10.$ Find maximum of
$P=(a^2+b^2+c^2)\left(\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2}\right).$

This post has been edited 1 time. Last edited by Nguyenhuyen_AG, Jan 5, 2016, 5:51 AM

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arqady

#6 h Jan 5, 2016, 1:26 PM • 1 Y

Y by Adventure10

Nguyenhuyen_AG wrote:

If $a,\,b,\,c$ are positive real numbers such that $(a+b+c)\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)=10.$ Find maximum of
$P=(a^2+b^2+c^2)\left(\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2}\right).$

$\max P=\max_{\left[\frac{25}{24},\frac{16}{15}\right]}\frac{20(3t-2)(5-3t)}{3t}=140-40\sqrt{10}$ .

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sqing

#7 h Jan 9, 2016, 7:06 AM • 2 Y

Y by Adventure10, Mango247

arqady wrote:

Nguyenhuyen_AG wrote:

If $a,\,b,\,c$ are positive real numbers such that $(a+b+c)\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)=10.$ Find maximum of
$P=(a^2+b^2+c^2)\left(\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2}\right).$

$\max P=\max_{\left[\frac{25}{24},\frac{16}{15}\right]}\frac{20(3t-2)(5-3t)}{3t}=140-40\sqrt{10}$ .

http://artofproblemsolving.com/community/c6h1133750p5282070

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sqing

#8 h Feb 13, 2016, 2:44 AM • 2 Y

Y by Adventure10, Mango247

mrmath2006 wrote:

Given $a,b,c>0$ & $(a+b+c)(\frac{1}{a}+\frac{1}{b}+\frac{1}{c})=10$ . Prove that
$(a^2+b^2+c^2)(\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2})\ge \frac{27}{2}$

http://www.artofproblemsolving.com/community/c6h528881p3010905
Let $a$ , $b$ and $c$ are positives such that $(a+b+c)\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)=11$ . Prove that:
$(a^3+b^3+c^3)\left(\frac{1}{a^3}+\frac{1}{b^3}+\frac{1}{c^3}\right)\geq41$

This post has been edited 1 time. Last edited by sqing, Feb 13, 2016, 2:45 AM

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41435 posts

sqing

#9 h Feb 13, 2016, 4:03 AM • 1 Y

Y by Adventure10

SAMC 2015 (Saudi Arabia Mathematical Competitions)：
Let $a,b,c$ be positive real numbers satisfying the condition $(a+b+c)(\frac{1}{a}+\frac{1}{b}+\frac{1}{c})=10$ . Find the greatest value and the least value of $T=(a^2+b^2+c^2)(\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2}).$

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4613 posts

#10 h Apr 16, 2016, 6:51 AM • 2 Y

Y by Adventure10, Mango247

The following inequality is also true
//cdn.artofproblemsolving.com/images/1/7/4/174028ad7a461fe816c43a02b922af27946ec494.jpg

//cdn.artofproblemsolving.com/images/1/7/4/174028ad7a461fe816c43a02b922af27946ec494.jpg

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#11 h Apr 16, 2016, 10:36 AM • 3 Y

Y by luofangxiang, Adventure10, Mango247

luofangxiang wrote:

The following inequality is also true

http://s11.sinaimg.cn/middle/0018zOAxgy70Xii0XEu5a&690

Because if $a,b,c$ are positive real numbers, then $5+\sqrt{2(a^2+b^2+c^2)\left ( \frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2} \right )-2} \geqslant (a+b+c)\left ( \frac{1}{a}+\frac{1}{b}+\frac{1}{c} \right ).$

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41435 posts

sqing

#12 h Jan 9, 2020, 12:46 PM • 2 Y

Y by Adventure10, Mango247

sqing wrote:

mrmath2006 wrote:

Given $a,b,c>0$ & $(a+b+c)(\frac{1}{a}+\frac{1}{b}+\frac{1}{c})=10$ . Prove that
$(a^2+b^2+c^2)(\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2})\ge \frac{27}{2}$

http://www.artofproblemsolving.com/community/c6h528881p3010905
Let $a$ , $b$ and $c$ are positives such that $(a+b+c)\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)=11$ . Prove that:
$(a^3+b^3+c^3)\left(\frac{1}{a^3}+\frac{1}{b^3}+\frac{1}{c^3}\right)\geq41$

Let $a$ , $b$ and $c$ are positives such that $(a+b+c)\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)=11$ . Prove that $(a^2+b^2+c^2)(\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2})\geq 19$ Proof of no_room_for_error:
Let $x=\frac{a}{b}+\frac{b}{c}+\frac{c}{a},\,y=\frac{b}{a}+\frac{c}{b}+\frac{a}{c},$ then $x+y= 8$ . Cauchy-Schwarz:
$(a^2+b^2+c^2)\left(\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2}\right)$
$=x^2+y^2-2(x+y)+3=x^2+y^2-13 \geqslant \frac{1}{2}(x+y)^2-13=19$

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41435 posts

sqing

#13 h Jul 17, 2020, 1:04 AM

Y by

sqing wrote:

Let $a$ , $b$ and $c$ are positives such that $(a+b+c)\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)=11$ . Prove that:
$(a^3+b^3+c^3)\left(\frac{1}{a^3}+\frac{1}{b^3}+\frac{1}{c^3}\right)\geq41$

Let $a$ , $b$ and $c$ are positives such that $(a+b+c)\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)=11$ . Prove that:
$(a^4+b^4+c^4)\left(\frac{1}{a^4}+\frac{1}{b^4}+\frac{1}{c^4}\right)\geq 99$

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