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Source: special and difficult inequality

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2861 posts

Vasc

#1 h Oct 25, 2009, 2:44 AM • 2 Y

Y by Adventure10, Mango247

If $a,b,c$ are positive real numbers such that $a + b + c = 3$ , then

$\frac {24}{a^2b + b^2c + c^2a} + \frac 1{abc} \ge 9.$

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1630 posts

#2 h Oct 25, 2009, 5:23 AM • 2 Y

Y by Adventure10, Mango247

Vasc wrote:

If $a,b,c$ are positive real numbers such that $a + b + c = 3$ , then

$\frac {24}{a^2b + b^2c + c^2a} + \frac 1{abc} \ge 9.$

Generally, the inequality
$\frac {k}{a^2b + b^2c + c^2a} + \frac 1{abc} \ge {\frac {k + 3}{3},}$

which contain parameter k ,holds for

${a + b + c = 3,{a,b,c}>{0}}$

if and only if $- 12\le{k}\le{24}.$

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#3 h Oct 25, 2009, 7:04 AM • 2 Y

Y by Adventure10, Mango247

fjwxcsl wrote:

Vasc wrote:

If $a,b,c$ are positive real numbers such that $a + b + c = 3$ , then

$\frac {24}{a^2b + b^2c + c^2a} + \frac 1{abc} \ge 9.$

Generally, the inequality
$\frac {k}{a^2b + b^2c + c^2a} + \frac 1{abc} \ge {\frac {k + 3}{3},}$

which contain parameter k ,holds for

${a + b + c = 3,{a,b,c} > {0}}$

if and only if $- 12\le{k}\le{24}.$

Analogously,the inequality
$\frac {k}{a^2b + b^2c + c^2a} - \frac 1{a+b+c} \le {\frac {k -1}{3},}$

which contain parameter k ,holds for

${abc=1,{a,b,c} > {0}}$

if and only if ${k}\ge{1}.$

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2861 posts

Vasc

#4 h Oct 25, 2009, 8:21 AM • 2 Y

Y by Adventure10, Mango247

fjwxcsl wrote:

Vasc wrote:

If $a,b,c$ are positive real numbers such that $a + b + c = 3$ , then

$\frac {24}{a^2b + b^2c + c^2a} + \frac 1{abc} \ge 9.$

Generally, the inequality
$\frac {k}{a^2b + b^2c + c^2a} + \frac 1{abc} \ge {\frac {k + 3}{3},}$

which contain parameter k ,holds for

${a + b + c = 3,{a,b,c} > {0}}$

if and only if $- 12\le{k}\le{24}.$

Indeed,

$\frac 1{abc} + 3 \ge \frac {12}{a^2b + b^2c + c^2a}$

is nice, too, fjwxcsl.

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2948 posts

#5 h Oct 25, 2009, 2:38 PM • 2 Y

Y by Adventure10, Mango247

Vasc wrote:

If $a,b,c$ are positive real numbers such that $a + b + c = 3$ , then

$\frac {24}{a^2b + b^2c + c^2a} + \frac 1{abc} \ge 9.$

It is a nice and difficult one, Vasc. My ugly proof used the fact that
$2(a^2b + b^2c + c^2a) = \sum ab(a + b) + (a - b)(b - c)(a - c) \le \sum ab(a + b) + \sqrt {(a - b)^2(b - c)^2(c - a)^2}$
to transform the inequality into $pqr$ form, and then by considering a function of $q,$ I deduced the result. For $a \ge b \ge c,$ equality holds when $a = b = c = 1,$ or when $a,b,c$ are three positive roots of the equations
$x^3 - 3x^2 + \left( 1 + \sqrt {3}\right)x - \frac {3 + 2\sqrt {3}}{9} = 0.$
This proof is too complicated and I dont want to post it. I think you have found a new method to treat this inequality. May I know your solution, Vasc? Thank you very much.

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663 posts

#6 h Oct 26, 2009, 5:06 AM • 1 Y

Y by Adventure10

can_hang2007 wrote:

Vasc wrote:

If $a,b,c$ are positive real numbers such that $a + b + c = 3$ , then

$\frac {24}{a^2b + b^2c + c^2a} + \frac 1{abc} \ge 9.$

It is a nice and difficult one, Vasc. My ugly proof used the fact that
$2(a^2b + b^2c + c^2a) = \sum ab(a + b) + (a - b)(b - c)(a - c) \le \sum ab(a + b) + \sqrt {(a - b)^2(b - c)^2(c - a)^2}$
to transform the inequality into $pqr$ form, and then by considering a function of $q,$ I deduced the result. For $a \ge b \ge c,$ equality holds when $a = b = c = 1,$ or when $a,b,c$ are three positive roots of the equations
$x^3 - 3x^2 + \left( 1 + \sqrt {3}\right)x - \frac {3 + 2\sqrt {3}}{9} = 0.$
This proof is too complicated and I dont want to post it. I think you have found a new method to treat this inequality. May I know your solution, Vasc? Thank you very much.

Dear Can,is the following step in your proof?
$abc\leq \frac {1}{9 - \frac {24}{3 + \frac {2}{9}(9 - 3q)^{1.5}}}$
where $q = ab + bc + ca$

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663 posts

#7 h Oct 26, 2009, 5:16 AM • 2 Y

Y by Adventure10, Mango247

Vasc wrote:

fjwxcsl wrote:

Vasc wrote:

If $a,b,c$ are positive real numbers such that $a + b + c = 3$ , then

$\frac {24}{a^2b + b^2c + c^2a} + \frac 1{abc} \ge 9.$

Generally, the inequality
$\frac {k}{a^2b + b^2c + c^2a} + \frac 1{abc} \ge {\frac {k + 3}{3},}$

which contain parameter k ,holds for

${a + b + c = 3,{a,b,c} > {0}}$

if and only if $- 12\le{k}\le{24}.$

Indeed,

$\frac 1{abc} + 3 \ge \frac {12}{a^2b + b^2c + c^2a}$

is nice, too, fjwxcsl.

$\frac {12}{\frac {1}{abc} + 3}\leq 3 - \frac {2}{9}[9 - 3(ab + bc + ca)]^{1.5}\leq a^2b + b^2c + c^2a$
is it?

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#8 h Oct 26, 2009, 7:57 AM • 1 Y

Y by Adventure10

Tourish wrote:

Vasc wrote:

fjwxcsl wrote:

Vasc wrote:

If $a,b,c$ are positive real numbers such that $a + b + c = 3$ , then

$\frac {24}{a^2b + b^2c + c^2a} + \frac 1{abc} \ge 9.$

Generally, the inequality
$\frac {k}{a^2b + b^2c + c^2a} + \frac 1{abc} \ge {\frac {k + 3}{3},}$

which contain parameter k ,holds for

${a + b + c = 3,{a,b,c} > {0}}$

if and only if $- 12\le{k}\le{24}.$

Indeed,

$\frac 1{abc} + 3 \ge \frac {12}{a^2b + b^2c + c^2a}$

is nice, too, fjwxcsl.

Tourish wrote:

$\frac {12}{\frac {1}{abc} + 3}\leq 3 - \frac {2}{9}[9 - 3(ab + bc + ca)]^{1.5}\leq a^2b + b^2c + c^2a$
is it?

The first inequality does not hold, the second one holds.

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2861 posts

Vasc

#9 h Sep 18, 2013, 3:56 PM • 2 Y

Y by Adventure10, Mango247

can_hang2007 wrote:

It is a nice and difficult one, Vasc. My ugly proof used the fact that
$2(a^2b + b^2c + c^2a) = \sum ab(a + b) + (a - b)(b - c)(a - c) \le \sum ab(a + b) + \sqrt {(a - b)^2(b - c)^2(c - a)^2}$ .

Indeed, using this inequality, it suffices to show that
$\frac{48}{3q-3r+\sqrt{-27r^2+54(q-2)r+9q^2-4q^3}}+\frac 1{r}\ge 9,$ where $r=abc$ and $q=ab+bc+ca$ . This is true if
$3[9r^2+15r-(9r-1)q]\ge (9r-1)\sqrt{-27r^2+54(q-2)r+9q^2-4q^3}.$ Before squaring this inequality, we need to show that $9r^2+15r-(9r-1)q\ge 0$ . From Schur's inequality
$p^3+9r\ge 4pq,$ we get
$q\le \frac{3(r+3)}{4},$ hence
$9r^2+15r-(9r-1)q\ge 9r^2+15r-\frac{3(r+3)(9r-1)}{4}=\frac{9(r-1)^2}{4}\ge 0.$
By squaring, the inequality becomes
$Aq^3+C\ge 3Bq,$ where
$A=(9r-1)^2, \ \ \ B=18r(9r-1)(3r+1), \ \ \ C=27r(27r^3+99r^2+r+1).$ Since, by the AM-GM inequality,
$Aq^3+C\ge 3\sqrt[3]{Aq^3.(C/2)^2}=3q\sqrt[3]{A(C/2)^2},$ it suffices to show that
$AC^2\ge 4B^3,$ which is equivalent to
$(27r^3+99r^2+r+1)^2\ge 32r(9r-1)(3r+1)^3,$
$729r^6-2430r^5+2943r^4-1476r^3+199r^2+34r+1\ge 0,$
$(r-1)^2(27r^2-18r-1)^2\ge 0.$

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