Old but nice Inequality

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Old but nice Inequality

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Source: Jack garfunkel

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111 posts

akai

#1 h Aug 26, 2012, 2:35 PM • 4 Y

Y by HWenslawski, donotoven, Adventure10, Mango247

Let $a,b,c\ge 0$ , show that
$\frac{a^2+b^2+c^2}{ab+bc+ca}+\frac{8abc}{(a+b)(b+c)(c+a)}\ge2$

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2130 posts

Sayan

#2 h Aug 26, 2012, 3:16 PM • 5 Y

Y by MathNinja7, HWenslawski, donotoven, jokehim, Adventure10

Proof 1
Symmetric inequality of 5th degree so by $uvw$ we are left to check the cases when $a=b=1$ and $c=0$
Proof 2
WLOG assume $a\ge b\ge c$
Rewrite the inequality as
$\frac{a^2+b^2+c^2}{ab+bc+ca}-1\ge 1-\frac{8abc}{(a+b)(b+c)(c+a)} \iff \\ \sum \frac{(a-b)^2}{2(ab+bc+ca)} \ge \sum \frac{c(a-b)^2}{(a+b)(b+c)(c+a)}$
thus the inequality is equivalent to
$S_a(b-c)^2+S_b(c-a)^2+S_c(a-b)^2 \ge 0$
where $S_a=b^2(c+a)+c^2(a+b)-a^2(b+c)$
Due to our earlier assumption we have $S_c \ge 0$ and clearly $S_a+S_b \ge 0$
So
$S_a(b-c)^2+S_b(c-a)^2+S_c(a-b)^2 \\ \ge S_a(b-c)^2+S_b(c-a)^2 \ge (b-c)^2(S_a+S_b) \ge 0$
Hence the inequality is proved.

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3377 posts

mudok

#3 h Aug 26, 2012, 4:45 PM • 3 Y

Y by donotoven, Adventure10, Mango247

is it Jack Garfunkel's inequality? I tought it is Pham Kim Hung's inequality.

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2130 posts

Sayan

#4 h Aug 26, 2012, 4:57 PM • 3 Y

Y by mudok, donotoven, Adventure10

mudok wrote:

is it Jack Garfunkel's inequality? I tought it is Pham Kim Hung's inequality.

Jack Garfunkel proposed the following inequality about 25 years ago,
$A,B,C$ be angles of a triangle then,
$\sum_{cyc} \tan^2\frac{A}{2} \ge 2- 8\prod_{cyc}\sin\frac{A}2$
Its algebraic form is the above discussed inequality.

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3377 posts

mudok

#5 h Aug 26, 2012, 5:11 PM • 6 Y

Y by vankhea, 123steeve, golema11, donotoven, Adventure10, Mango247

Proof 3.
WLOG assume that $a\ge b\ge c$ , then it is easy to check $\frac{8abc}{(a+b)(b+c)(c+a)}\ge \frac{4ac}{(a+c)^2}$ . And

$\frac{a^2+b^2+c^2}{ab+bc+ca}+\frac{4ac}{(a+c)^2}\ge 2 \ \ \iff \ \ \ (a^2+c^2-ab-bc)^2\ge 0$

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younesmath2012maroc

621 posts

younesmath2012maroc

#6 h Aug 26, 2012, 7:12 PM • 3 Y

Y by donotoven, Adventure10, Mango247

mudok wrote:

Proof 3.
WLOG assume that $a\ge b\ge c$ , then it is easy to check $\frac{8abc}{(a+b)(b+c)(c+a)}\ge \frac{4ac}{(a+c)^2}$ . And

$\frac{a^2+b^2+c^2}{ab+bc+ca}+\frac{4ac}{(a+c)^2}\ge 2 \ \ \iff \ \ \ (a^2+c^2-ab-bc)^2\ge 0$

very good a beautiful proof!!!

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younesmath2012maroc

621 posts

younesmath2012maroc

#7 h Aug 26, 2012, 7:16 PM • 3 Y

Y by donotoven, Adventure10, Mango247

younesmath2012maroc wrote:

mudok wrote:

Proof 3.
WLOG assume that $a\ge b\ge c$ , then it is easy to check $\frac{8abc}{(a+b)(b+c)(c+a)}\ge \frac{4ac}{(a+c)^2}$ . And

$\frac{a^2+b^2+c^2}{ab+bc+ca}+\frac{4ac}{(a+c)^2}\ge 2 \ \ \iff \ \ \ (a^2+c^2-ab-bc)^2\ge 0$

very good a beautiful proof!!!

ho can write an other beautiful proof 4 !!!!
thank's!!!!

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3377 posts

mudok

#8 h Aug 27, 2012, 3:39 PM • 3 Y

Y by donotoven, Adventure10, Mango247

Proof 4

$\ \frac{a^2+b^2+c^2}{ab+bc+ca}+\frac{8abc}{(a+b)(b+c)(c+a)}\ge 2\ \ \iff \ \ a^2+b^2+c^2+\frac{8abc(ab+bc+ca)}{(a+b)(b+c)(c+a)}\ge 2(ab+bc+ca)$

$\frac{8abc(ab+bc+ca)}{(a+b)(b+c)(c+a)}\ge \frac{6abc(a+b+c)}{a^2+b^2+c^2+ab+bc+ca}\ \ \iff \ \ \sum ab(a-b)^2 \ge 0$

$\ a^2+b^2+c^2+\frac{6abc(a+b+c)}{a^2+b^2+c^2+ab+bc+ca}\ge 2(ab+bc+ca) \ \ \iff \ \ \sum a^2(a-b)(a-c)\ge 0$

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224 posts

#9 h Aug 29, 2012, 2:28 AM • 3 Y

Y by donotoven, Adventure10, Mango247

mudok wrote:

is it Jack Garfunkel's inequality? I tought it is Pham Kim Hung's inequality.

You know Pham Kim Hung's inequality. I from in Viet Nam and i don't know you know it. Where are you from.
This is a nice inequality in Pham Kim Hung's inequality:
With $x_{1},x_{2},...,x_{n}\ge0$ and $x_{1}x_{2}...x_{n}=1$ . Prove that
$x_{1}+x_{2}+...+x_{n}\ge \frac{2}{1+x_{1}}+\frac{2}{1+x_{2}}+...\frac{2}{1+x_{n}}$
We will use AM-GM's inequality

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3377 posts

mudok

#10 h Sep 1, 2012, 5:35 PM • 2 Y

Y by donotoven, Adventure10

Sayan wrote:

mudok wrote:

is it Jack Garfunkel's inequality? I tought it is Pham Kim Hung's inequality.

Jack Garfunkel proposed the following inequality about 25 years ago,
$A,B,C$ be angles of a triangle then,
$\sum_{cyc} \tan^2\frac{A}{2} \ge 2- 8\prod_{cyc}\sin\frac{A}2$
Its algebraic form is the above discussed inequality.

I am very curious about Jack Garfunkel's own solution !

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2130 posts

Sayan

#11 h Sep 2, 2012, 2:41 AM • 3 Y

Y by donotoven, Adventure10, Mango247

mudok wrote:

I am very curious about Jack Garfunkel's own solution !

Crux wrote:

An asterisk ( $\star$ ) after a problem number indicates that the problem was proposed without a solution

This was problem 825, marked with an asterisk.

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41401 posts

sqing

#12 h Feb 18, 2019, 8:27 AM • 5 Y

Y by donotoven, Adventure10, Mango247, Mango247, Mango247

Let $a,b,c\ge 0$ , show that
$\sqrt{\frac{a^2+b^2+c^2}{ab+bc+ca}}+2(2-\sqrt 2)\sqrt{\frac{abc}{(a+b)(b+c)(c+a)}} \ge \sqrt 2$

akai wrote:

Let $a,b,c\ge 0$ , show that
$\frac{a^2+b^2+c^2}{ab+bc+ca}+\frac{8abc}{(a+b)(b+c)(c+a)}\ge2$

$\implies$ $\frac{(a+b+c)^2}{ab+bc+ca}+\frac{8abc}{(a+b)(b+c)(c+a)}\ge4$ $\implies$ Let $a,b,c\ge 0, ab+bc+ca=1$ , show that
$\frac{a^2}{a^2+1}+\frac{b^2}{b^2+1}+\frac{c^2}{c^2+1}\leq\frac{1}{4}(a+b+c)^2.$ https://artofproblemsolving.com/community/c6h1254314p6478767

Attachments:

This post has been edited 2 times. Last edited by sqing, Dec 29, 2020, 12:18 AM

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41401 posts

sqing

#13 h Jul 15, 2021, 10:38 PM • 1 Y

Y by donotoven

Let $a,b,c>0$ , show that
$5\sqrt{\frac{a^2+b^2+c^2}{ab+bc+ca}}+2\sqrt{\frac{8abc}{(a+b)(b+c)(c+a)}}\geq 7$ $\frac{ab+bc+ca}{a^2+b^2+c^2}+\frac{(a+b)(b+c)(c+a)}{4abc}\geq 3$ https://artofproblemsolving.com/community/c6h603226p3581852

This post has been edited 1 time. Last edited by sqing, Apr 7, 2022, 2:07 AM

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512 posts

VMF-er

#14 h Jul 16, 2021, 4:10 AM • 1 Y

Y by donotoven

sqing wrote:

akai wrote:

Let $a,b,c\ge 0$ , show that
$\frac{a^2+b^2+c^2}{ab+bc+ca}+\frac{8abc}{(a+b)(b+c)(c+a)}\ge2$

Let $a,b,c> 0$ , show that
$\frac{ab+bc+ca}{a^2+b^2+c^2}+\frac{(a+b)(b+c)(c+a)}{4abc}\geq 3$

Stronger: $\frac{(b+c)(c+a)(a+b)}{abc}+\frac{4\sqrt{2}(bc+ca+ab)}{a^{2}+b^{2}+c^{2}}\geq 8+4\sqrt{2}.$

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