inequality: If a,b,c>=0 are real, prove a^3+b^3+c^3-3abc>=[(b+c)/2 - a]^3

[akramzine]

\(\displaystyle \mbox{Let }\, a,\, b\, c\, \geq\, 0\, \mbox{ be real numbers.}\)

\(\displaystyle \mbox{Prove that }\, a^3\, +\, b^3\, +\, c^3\, -\, 3abc\, \geq\, \left(\dfrac{b\, +\, c}{2}\, -\, a\right)^3\)

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i proved that a^3+b^3+c^3-3abc was positive but couldn't advance any further

 

[akramzine]

after proving that a^3+b^3+c^3-3abc=0.5(a+b+c)((a-b)²+(b-c)²+(c-a)²)
i neglected the first two cases wherea the right side of the inequality would be either negatif or equaling 0
now there is one case left and that's when the right side is positif
implies b+c>2a
can anyone help me progress please

 

[akramzine]

is there something i can use in this problem i can't progress that much

 

[akramzine]

and sorry in the question redaction it's ; being real numbers rather than :reel

 

[stapel]

\(\displaystyle \mbox{Let }\, a,\, b\, c\, \geq\, 0\, \mbox{ be real numbers.}\)

\(\displaystyle \mbox{Prove that }\, a^3\, +\, b^3\, +\, c^3\, -\, 3abc\, \geq\, \left(\dfrac{b\, +\, c}{2}\, -\, a\right)^3\)

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i proved that a^3+b^3+c^3-3abc was positive but couldn't advance any further

Does the discussion here help at all?

 

[akramzine]

thank you mr stapel i appreciate your help