MHB Prove Inequality for Positive Reals a, b, c

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The discussion focuses on proving the inequality involving positive reals a, b, and c, specifically that (a^8 + b^8 + c^8) / (a^3b^3c^3) > (1/a + 1/b + 1/c) when not all variables are equal. An initial incorrect formulation was corrected, demonstrating the importance of precise mathematical expressions. Participants noted that using the Arithmetic Mean-Geometric Mean (AM-GM) inequality simplifies the proof. The conversation highlights the significance of careful problem setup in mathematical discussions. The corrected inequality is now the focus for further exploration and proof.
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Given that a, b, c are positive reals and not all equal, show that

$\dfrac{a^3 + b^3 + c^3}{a^3b^3c^3}\ge\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}$
 
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anemone said:
Given that a, b, c are positive reals and not all equal, show that

$\dfrac{a^3 + b^3 + c^3}{a^3b^3c^3}\ge\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}$
[sp]
it is not true
exam:$a=1,b=2,c=3$
left side=$\dfrac{1+8+27}{1\times 8\times 27}<1$
right side=$1+\dfrac {1}{2}+\dfrac{1}{3}>1$
[/sp]
 
Ops...I am so sorry:o...the problem should read:

anemone said:
Given that a, b, c are positive reals and not all equal, show that

$\dfrac{a^8 + b^8 + c^8}{a^3b^3c^3}\gt\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}$
 
anemone said:
Ops...I am so sorry:o...the problem should read:
$\dfrac {a^8+b^8+c^8}{a^3b^3c^3}>\dfrac{1}{a}+\dfrac{1}{b}+\dfrac {1}{c}$
using AP>GP is easier
for :$a^2+b^2+c^2>ab+bc+ca$
$\dfrac{a^8+b^8+c^8}{a^3b^3c^3}>\dfrac{a^4b^4+b^4c^4+c^4a^4}{a^3b^3c^3}>\dfrac{a^4b^2c^2+b^4a^2c^2+c^4a^2b^2}{a^3b^3c^3}\\
=\dfrac{a^2+b^2+c^2}{abc}>\dfrac{ab+bc+ca}{abc}=\dfrac {1}{a}+\dfrac {1}{b}+\dfrac {1}{c}$
 
Last edited:
Albert said:
using AP>GP is easier
for :$a^2+b^2+c^2>ab+bc+ca$
$\dfrac{a^8+b^8+c^8}{a^3b^3c^3}>\dfrac{a^4b^4+b^4c^4+c^4a^4}{a^3b^3c^3}>\dfrac{a^4b^2c^2+b^4a^2c^2+c^4a^2b^2}{a^3b^3c^3}\\
=\dfrac{a^2+b^2+c^2}{abc}>\dfrac{ab+bc+ca}{abc}=\dfrac {1}{a}+\dfrac {1}{b}+\dfrac {1}{c}$

Well done Albert!(Cool) Thanks for participating!
 
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