Inequality

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#1 h Apr 28, 2020, 4:26 PM

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If $abc=1$ prove that $8(a^3+b^3+c^3) \geq 3(a^2+bc)(b^2+ac)(c^2+ab)$

This post has been edited 1 time. Last edited by Marinchoo, Nov 17, 2020, 2:54 PM

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TuZo

#2 h Apr 28, 2020, 4:44 PM • 2 Y

Y by Zorger74, Kgxtixigct

Marinchoo wrote:

If $abc=1$ prove that $8(a^3+b^3+c^3)=>3(a^2+bc)(b^2+ac)(c^2+ab)$

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arqady

#3 h Apr 28, 2020, 8:30 PM

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Marinchoo wrote:

If abc=1 prove that 8(a^3+b^3+c^3)=>3(a^2+bc)(b^2+ac)(c^2+ab)

It's wrong. Try $c\rightarrow0^+$ .

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#4 h Apr 29, 2020, 12:12 AM

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If a, b and c can be negative, the inequality is wrong ( a=1 b = -1/2. c = -2)

If a, b, c are positive reals, we have by hypothesis:
cc = 1/a. ac = 1/b ab = 1/c. So
(a^2 + bc)(b^2 + ac)(c^2 + ab) =
(a^2 + 1/a)(b^2 + 1/b)(c^2 + 1/b) =
(a^3 + 1)(b^3 + 1)(c^3 + 1) because abc=1
And
(a^3 + 1)/2. (b^3 + 1)/2. (c^3 + 1) /2
=< (a^3 . b^3. c^3 + 1)/2 =1
by using Tchebychev

Also, (a^3 + b^3 + c^3)/3 >= abc=1>=
(a^3 + 1)/2. (b^3 + 1)/2. (c^3 + 1) /2
That finishes the proof. Egality hold if a=b=c= 1.

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Victoria_Discalceata1

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Victoria_Discalceata1

#5 h Apr 29, 2020, 12:24 AM

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Taking $a=b=2,\ c=\frac{1}{4}$ shows that it cannot hold.

In general, taking $a=b=x,\ c=\frac{1}{x^2}$ shows that it is false for sufficiently large $x$ while the reverse is false for sufficiently small $x$ .

This post has been edited 1 time. Last edited by Victoria_Discalceata1, Apr 29, 2020, 12:35 AM

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#6 h May 28, 2020, 6:49 AM

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Is $8(a^3+b^3+c^3)<=9(a^2+bc)(b^2+ac)(c^2+ab)$ true for all positive reals $a, b, c$ , $abc=1$ ?

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sqing

#7 h Mar 19, 2025, 2:46 PM

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Marinchoo wrote:

If $abc=1$ prove that $8(a^3+b^3+c^3) \geq 3(a^2+bc)(b^2+ac)(c^2+ab)$

Let $a,b, c>0 .$ Prove that $8(a^3+b^3+c^3) \geq 9(a^2+bc)(b^2+ca)(c^2+ab)$

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