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

Question

Show that if #a+b>=0# then #a^3 + b^3 >= a^2b + ab^2#.

1045 views around the world

You can reuse this answer
Creative Commons License

Answer 1 · 2017-05-08T16:04:57

See below.

If #a+b ge 0# then #a+b=delta^2 ge 0#

Calling #f(a,b)=a^3 + b^3 - a^2 b - a b^2# and substituting #a = delta^2-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# so this proves that if

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

Answer 2 · 2017-05-08T16:55:20

The Proof is given in the Explanation Section.

If #a+b=0,# then

#a^3+b^3=(a+b)(a^2-ab+b^2)=(0)(a^2-ab+b^2)=0, # and,

#a^2b+ab^2=ab(a+b)=ab(0)=0.#

This proves that, incase, #a+b=0, then, a^3+b^3gea^2b+ab^2.#

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

Now, consider, #(a^2-ab+b^2)-(ab)=a^2-2ab+b^2=(a-b)^2 ge 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.!