More powerful then AM-GM for 3 variables.

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More powerful then AM-GM for 3 variables.

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Ovchinnikov Denis

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Ovchinnikov Denis

#1 h Oct 15, 2010, 9:18 AM • 2 Y

Y by Adventure10, Mango247

Prove, that for all nonnegative reals $a,b,c$ holds next inequality: $a^3+b^3+c^3 \geq 4|a-b| \cdot |b-c| \cdot |c-a|+3abc$

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#2 h Oct 15, 2010, 12:01 PM • 1 Y

Y by Adventure10

We may assume $a \ge b \ge c$ ,Let
$f(a,b,c)=a^3+b^3+c^3-4(a-b)(b-c)(a-c)-3abc.$

We can easily show $f(a,b,c) \ge f(a-c,b-c,0)$ .

Then we only need to prove the inequality when $c=0$ ,this become
$a^3+b^3 \ge 4(a-b)ab \iff a(a-2b)^2+b^3 \ge 0.$

This post has been edited 1 time. Last edited by zhaobin, Oct 15, 2010, 3:49 PM

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kuing

#3 h Oct 15, 2010, 12:28 PM • 1 Y

Y by Adventure10

Ovchinnikov Denis wrote:

Prove, that for all nonnegative reals $a,b,c$ holds next inequality: $a^3+b^3+c^3 \geq 4|a-b||b-c||c-a|+3abc$

$a^3+b^3+c^3 \geq k\left|a-b\right|\left|b-c\right|\left|c-a\right|+3abc$

$k_{\max}=\sqrt{9+6\sqrt{3}}$

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Ovchinnikov Denis

470 posts

Ovchinnikov Denis

#4 h Oct 15, 2010, 3:04 PM • 1 Y

Y by Adventure10

kuing wrote:

Ovchinnikov Denis wrote:

Prove, that for all nonnegative reals $a,b,c$ holds next inequality: $a^3+b^3+c^3 \geq 4|a-b||b-c||c-a|+3abc$

$a^3+b^3+c^3 \geq k\left|a-b\right|\left|b-c\right|\left|c-a\right|+3abc$

$k_{\max}=\sqrt{9+6\sqrt{3}}$

Click to reveal hidden text

Yes, but for $k=4$ solution is nice

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Abdek

#5 h Oct 15, 2010, 4:27 PM • 2 Y

Y by Adventure10, Mango247

Ovchinnikov Denis wrote:

Prove, that for all nonnegative reals $a,b,c$ holds next inequality: $a^3+b^3+c^3 \geq 4|a-b| \cdot |b-c| \cdot |c-a|+3abc$

let $c=\min\{a,b,c\}$
According to Am-Gm inequality we have
$\left( (a-b)^2+(b-c)(a-c) \right)^2 \ge 4(a-b)^2(b-c)(a-c)$
and
$(a+b+c)^2 \ge (a+b-2c)^2 \ge 4(a-c)(b-c)$
Multiplying the above inequalities we get
$(a+b+c)^2\left( (a-b)^2+(b-c)(a-c) \right)^2 \ge 16(a-b)^2(b-c)^2(c-a)^2$
which yields to the desired result because
$(a+b+c)\left( (a-b)^2+(b-c)(a-c) \right)=a^3+b^3+c^3-3abc$
The proof is completed equality occurs when $a=b=c$

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BaronShadeNight

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BaronShadeNight

#6 h Oct 16, 2010, 5:46 AM • 2 Y

Y by Adventure10, Mango247

kuing wrote:

Ovchinnikov Denis wrote:

Prove, that for all nonnegative reals $a,b,c$ holds next inequality: $a^3+b^3+c^3 \geq 4|a-b||b-c||c-a|+3abc$

$a^3+b^3+c^3 \geq k\left|a-b\right|\left|b-c\right|\left|c-a\right|+3abc$

$k_{\max}=\sqrt{9+6\sqrt{3}}$

Click to reveal hidden text

http://www.artofproblemsolving.com/Forum/viewtopic.php?p=858831&sid=f460d7e5a2786a34fed4aaa8db23582b#p858831

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kuing

#7 h Oct 16, 2010, 7:04 AM • 1 Y

Y by Adventure10

Abdek wrote:

Ovchinnikov Denis wrote:

Prove, that for all nonnegative reals $a,b,c$ holds next inequality: $a^3+b^3+c^3 \geq 4|a-b| \cdot |b-c| \cdot |c-a|+3abc$

let $c=\min\{a,b,c\}$
According to Am-Gm inequality we have
$\left( (a-b)^2+(b-c)(a-c) \right)^2 \ge 4(a-b)^2(b-c)(a-c)$
and
$(a+b+c)^2 \ge (a+b-2c)^2 \ge 4(a-c)(b-c)$
Multiplying the above inequalities we get
$(a+b+c)^2\left( (a-b)^2+(b-c)(a-c) \right)^2 \ge 16(a-b)^2(b-c)^2(c-a)^2$
which yields to the desired result because
$(a+b+c)\left( (a-b)^2+(b-c)(a-c) \right)=a^3+b^3+c^3-3abc$
The proof is completed equality occurs when $a=b=c$

so nice prove

BaronShadeNight wrote:

kuing wrote:

Ovchinnikov Denis wrote:

Prove, that for all nonnegative reals $a,b,c$ holds next inequality: $a^3+b^3+c^3 \geq 4|a-b||b-c||c-a|+3abc$

$a^3+b^3+c^3 \geq k\left|a-b\right|\left|b-c\right|\left|c-a\right|+3abc$

$k_{\max}=\sqrt{9+6\sqrt{3}}$

Click to reveal hidden text

http://www.artofproblemsolving.com/Forum/viewtopic.php?p=858831&sid=f460d7e5a2786a34fed4aaa8db23582b#p858831

thank you.

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