Question
Given positive integers a,b,c, prove
a^3+b^3+c^3≥\dfrac{1}{3} (a^2+b^2+c^2)(a+b+c)
Given positive integers a,b,c, prove
a^3+b^3+c^3≥\dfrac{1}{3} (a^2+b^2+c^2)(a+b+c)
Times 3 at two sides of the inequality
3(a^3+b^3+c^3)≥a^3+b^3+c^3+a^2(b+c)+b^2(a+c)+c^2(b+c)
2(a^3+b^3+c^3)≥ a^2(b+c)+b^2(a+c)+c^2(b+c)
=ab(a+b)+bc(b+c)+ac(a+c)
On the other hand,
a^3+b^3=(a+b)(a^2-ab+b^2)
≥(a+b)ab
Similarly,
b^3+c^3≥bc(b+c)
a^3+c^3≥ac(a+c)
Therefore
2(a^3+b^3+c^3)≥ ab(a+b)+bc(b+c)+ac(a+c)
Now we’ve completed the proof