Prove Inequality for Positive Reals a, b, c

[anemone]

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

$\dfrac{a^3 + b^3 + c^3}{a^3b^3c^3}\ge\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}$

 

[Albert1]

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]

 

[anemone]

Ops...I am so sorry...the problem should read:

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

$\dfrac{a^8 + b^8 + c^8}{a^3b^3c^3}\gt\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}$

 

[Albert1]

Ops...I am so sorry...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

Spoiler

for :$a^2+b^2+c^2>ab+bc+ca$
$\dfrac{a^8+b^8+c^8}{a^3b^3c^3}>\dfrac{a^4b^4+b^4c^4+c^4a^4}{a^3b^3c^3}>\dfrac{a^4b^2c^2+b^4a^2c^2+c^4a^2b^2}{a^3b^3c^3}\\
=\dfrac{a^2+b^2+c^2}{abc}>\dfrac{ab+bc+ca}{abc}=\dfrac {1}{a}+\dfrac {1}{b}+\dfrac {1}{c}$

 

[anemone]

using AP>GP is easier

Spoiler

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!