# How is 2(a^3+b^3+c^3)>=ab(a+b)+(b+c)bc+(a+c)ac?a>0 b>0 c>0

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You need to use the nextÂ inequality `(a-b)^2 gt= 0` .

Expanding the binomial yields: `a^2 - 2ab - b^2 gt= 0`

Hence, `a^2 - ab - ab- b^2 gt= 0 =gt a^2 - ab - b^2 gt= ab`

Multiplying both sides by (a+b) yields:

`(a+b)(a^2 - ab - b^2) gt= (a+b)ab`

The product from the left side yields the sum of cubes: `a^3 + b^3` .

`a^3 + b^3 gt= (a+b)ab`

You need to follow the same procedure for b^3 + c^3 and `a^3 + c^3 ` such that:

`b^3 + c^3 gt= (b+c)bc`

``Â `a^3 + c^3 gt= (a+c)ac`

Adding these three inequalities yields:

`a^3 + b^3 + b^3 + c^3 + a^3 + c^3 gt= (a+b)ab + (b+c)bc + (a+c)ac`

`2(a^3+b^3+c^3) gt= (a+b)ab + (b+c)bc + (a+c)ac`

**The last line proves the inequality **

`2(a^3+b^3+c^3) gt= (a+b)ab + (b+c)bc + (a+c)ac`