Show that if a+b...........?

Show that if a+b>=0a+b≥0 then a^3 + b^3 >= a^2b + ab^2a3+b3≥a2b+ab2.

2 Answers
Cesareo R.
May 8, 2017

See below.

Explanation:

If a+b ge 0a+b≥0 then a+b=delta^2 ge 0a+b=δ2≥0

Calling f(a,b)=a^3 + b^3 - a^2 b - a b^2f(a,b)=a3+b3−a2b−ab2 and substituting a = delta^2-ba=δ2−b we have after simplifications

(f@(a+b=delta^2)) = delta^2(4b^2-4b delta^2+delta^4)= 4delta^2(b-delta^2/2)^2 ge 0(f∘(a+b=δ2))=δ2(4b2−4bδ2+δ4)=4δ2(b−δ22)2≥0 so this proves that if

a+b ge 0a+b≥0 then f(a,b) ge 0f(a,b)≥0

Ratnaker Mehta
May 8, 2017

The Proof is given in the Explanation Section.

Explanation:

If a+b=0,a+b=0, then

a^3+b^3=(a+b)(a^2-ab+b^2)=(0)(a^2-ab+b^2)=0, a3+b3=(a+b)(a2−ab+b2)=(0)(a2−ab+b2)=0, and,

a^2b+ab^2=ab(a+b)=ab(0)=0.a2b+ab2=ab(a+b)=ab(0)=0.

This proves that, incase, a+b=0, then, a^3+b^3gea^2b+ab^2.a+b=0,then,a3+b3≥a2b+ab2.

Therefore, we need prove this Result for a+b>0.a+b>0.

Now, consider, (a^2-ab+b^2)-(ab)=a^2-2ab+b^2=(a-b)^2 ge 0.(a2−ab+b2)−(ab)=a2−2ab+b2=(a−b)2≥0.

:. a^2-ab+b^2 ge ab.

Multiplying by (a+b) > o, the inequality remains unaltered, and

becomes, (a+b)(a^2-ab+b^2) ge ab(a+b).

This is the same as, a^3+b^3 ge a^2b+ab^2.

Hence, the Proof.

Enjoy Maths.!

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