3 var inequality

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3 var inequality

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#1 h May 17, 2025, 1:05 PM • 1 Y

Y by PikaPika999

Let $a,b,c \in R$ ,such that $a^2+b^2+c^2=4(ab+bc+ca)$ Prove that : $\frac{7-2\sqrt{14}}{48} \leq \frac{a^3b+b^3c+c^3a}{(a^2+b^2+c^2)^2} \leq \frac{7+2\sqrt{14}}{48}$

This post has been edited 3 times. Last edited by SunnyEvan, May 18, 2025, 6:04 AM

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sqing

#2 h May 17, 2025, 2:55 PM • 1 Y

Y by PikaPika999

Let $a,b,c \in R$ ,such that $a^2+b^2+c^2=4(ab+bc+ca).$ Prove that $\frac{7-2\sqrt{14}}{48} \leq \frac{a^3b+b^3c+c^3a}{(a^2+b^2+c^2)^2} \leq \frac{7+2\sqrt{14}}{48}$
Maybe...

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sqing

#3 h May 17, 2025, 3:23 PM

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Let $a,b,c$ be reals such that $a^2+b^2+c^2=4(ab+bc+ca).$ Prove that $\frac{2-\sqrt 2}{18} \leq \frac{a^2b+b^2c+c^2a}{(a+b+c)^3} \leq \frac{2+\sqrt 2}{18}$

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#4 h May 17, 2025, 5:51 PM

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SunnyEvan wrote:

Let $a,b,c \in R$ ,such that $a^2+b^2+c^2=4(ab+bc+ca)$ Prove that : $\frac{53}{2}-9\sqrt{14} \leq \frac{8(a^3b+b^3c+c^3a)}{27(a^2+b^2+c^2)^2} \leq \frac{53}{2}+9\sqrt{14}$

We have the identity
$a^3b + b^3c + c^3a = (a+b+c)(a^2b + b^2c + c^2a) - abc(a+b+c) - (a^2b^2 + b^2c^2 + c^2a^2)$ $\Rightarrow a^3b + b^3c + c^3a = e_1(a^2b + b^2c + c^2a) + e_1e_3 - e_2^2$
WLOG $a^2 + b^2 + c^2 = 1$

Enough to show that
$\frac{1432 - 486\sqrt{14}}{16} \geq \sqrt{\frac{3}{2}} (a^2b + b^2c + c^2a + abc)$
We have,
$abc = (a+b+c)(ab+bc+ca) - (a+b)(b+c)(c+a) = \sqrt{\frac{3}{2}}\left(\frac{1}{4}\right) - (a+b)(b+c)(c+a)$
Also,
$a^2b + b^2c + c^2a + ab^2 + bc^2 + ca^2 = (a+b)(b+c)(c+a) - 2abc = -(a+b)(b+c)(c+a) - 2\sqrt{\frac{3}{2}}\left(\frac{1}{4}\right)$
Let $(a+b)(b+c)(c+a) = x$

Hence the inequality becomes $x \geq -10$ (approx). But we have
$x \geq 8(abc)^{2/3}$
Hence it is enough to show that
$(abc)^{2/3} \geq -\frac{5}{4} \Rightarrow \text{True since } (abc)^{2/3} \geq 0 \text{ for real } a,b,c$

This post has been edited 3 times. Last edited by JARP091, May 18, 2025, 2:38 PM
Reason: Typo

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#6 h May 18, 2025, 6:05 AM

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Thank you very much.@Sqing

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#9 h May 18, 2025, 6:04 PM

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Ok... I give up tell me the solutions

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#10 h May 21, 2025, 6:22 PM

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sqing wrote:

Let $a,b,c \in R$ ,such that $a^2+b^2+c^2=4(ab+bc+ca).$ Prove that $\frac{7-2\sqrt{14}}{48} \leq \frac{a^3b+b^3c+c^3a}{(a^2+b^2+c^2)^2} \leq \frac{7+2\sqrt{14}}{48}$
Maybe...

@sqing please tell your solutions

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#11 h May 28, 2025, 12:55 PM

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Let $a,b,c \in R$ ,such that : $(a+b+c)^2=-2(ab+bc+ca) .$ Prove that :
$-\frac{41+10\sqrt{70}}{432} \leq \frac{a^3b+b^3c+c^3a}{(a^2+b^2+c^2)^2} \leq \frac{-41+10\sqrt{70}}{432}$

This post has been edited 2 times. Last edited by SunnyEvan, May 28, 2025, 1:27 PM

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sqing

#12 h May 28, 2025, 1:24 PM

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SunnyEvan wrote:

Let $a,b,c \in R$ ,such that : $(a+b+c)^2=2(ab+bc+ca) .$ Prove that :
$-\frac{41+10\sqrt{70}}{432} \leq \frac{a^3b+b^3c+c^3a}{(a^2+b^2+c^2)^2} \leq \frac{-41+10\sqrt{70}}{432}$

$(a+b+c)^2=2(ab+bc+ca) \iff a^2+b^2+c^2=0$

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#13 h May 28, 2025, 1:27 PM

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My problem $-2(ab+bc+ca)$

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sqing

#14 h May 28, 2025, 2:56 PM

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Let $a,b,c$ be reals such that $2(a+b+c)^2\geq -3(ab+bc+ca).$ Prove that
$\frac{1-6\sqrt 3}{9} \leq \frac{a^2b+b^2c+c^2a}{(a+b+c)^3} \leq \frac{1+6\sqrt 3}{9}$ Let $a,b,c$ be reals such that $4(a+b+c)^2\geq -3(ab+bc+ca).$ Prove that
$\frac{1-10\sqrt 5}{9} \leq \frac{a^2b+b^2c+c^2a}{(a+b+c)^3} \leq \frac{1+10\sqrt 5}{9}$

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mihaig

#15 h May 28, 2025, 3:32 PM

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SunnyEvan wrote:

Let $a,b,c \in R$ ,such that : $(a+b+c)^2=-2(ab+bc+ca) .$ Prove that :
$-\frac{41+10\sqrt{70}}{432} \leq \frac{a^3b+b^3c+c^3a}{(a^2+b^2+c^2)^2} \leq \frac{-41+10\sqrt{70}}{432}$

Who is the author?

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sqing

#16 h May 29, 2025, 12:44 AM

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Let $a,b,c$ be reals such that $5(a+b+c)^2\geq -(ab+bc+ca).$ Prove that
$-\frac{127}{9} \leq \frac{a^2b+b^2c+c^2a}{(a+b+c)^3} \leq \frac{43}{3}$

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#19 h May 29, 2025, 1:00 AM

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mihaig wrote:

SunnyEvan wrote:

Let $a,b,c \in R$ ,such that : $(a+b+c)^2=-2(ab+bc+ca) .$ Prove that :
$-\frac{41+10\sqrt{70}}{432} \leq \frac{a^3b+b^3c+c^3a}{(a^2+b^2+c^2)^2} \leq \frac{-41+10\sqrt{70}}{432}$

Who is the author?

I remember Professor Vasile proved the general inequality in his series of papers on three-variable degree four inequalities.
https://artofproblemsolving.com/community/c6h457867p2582175

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