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)

Steven Zheng Steven Zheng posted 1 month ago


Answer

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

Steven Zheng Steven Zheng posted 1 month ago

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