How to prove the inequality $abc(a+b+c)^2≤(a^3+b^3+c^3)(ab+bc+ca)$? [closed]

I need to prove something like that:

For a,b,c>0 prove: $$abc(a+b+c)^2≤(a^3+b^3+c^3)(ab+bc+ca)$$.

I know that $$3abc≤(a^3+b^3+c^3)$$, but then I derived $$3(ab+bc+ca) ≤ (a+b+c)^2$$, I can't move on.

Can anyone help me?

• Please add the Mathematica code you've tried to your question. Commented Nov 28, 2020 at 16:13

For a proof, in the mathematical sense, ask on math.SE. For a Mathematica verification, here's a way:

(a > 0 && b > 0 && c > 0) \[Implies]
Reduce[a*b*c*(a + b + c)^2 <= (a^3 + b^3 + c^3)*(a*b + b*c + c*a),
{a, b, c}, PositiveReals] // Simplify

(*  True  *)


It's easy to disprove by finding a counter-example:

FindInstance[a*b*c*(a + b + c)^2 > (a^3 + b^3 + c^3)*(a*b + b*c + c*a), {a, b, c}]

(*    {{a -> -1, b -> -1, c -> 0}}    *)

• Sorry, I forgot to add conditions that a,b,c>0
– musk
Commented Nov 28, 2020 at 16:05

This is not true on the reals as

NMinimize[-a*b*c *(a + b + c)^2 + (a^3 + b^3 + c^3)* (a*b + b*c + c*a), {a, b, c}]


says. The inequality is valid on the positive reals:

Minimize[-a*b*c*(a + b + c)^2 + (a^3 + b^3 + c^3)*(a*b + b*c + c*a), {a, b, c}, PositiveReals]
(*{0, {a -> 1, b -> 1, c -> 1}}*)


Addition. Here is the proof by logic tools without bells and jingles

Resolve[ForAll[{a, b, c},a*b*c*(a + b + c)^2 <= (a^3 + b^3 + c^3)*(a*b + b*c +
c*a)], PositiveReals]
(*True*)

• Sorry, I forgot to add conditions that a,b,c>0. I want to prove it on the positive reals
– musk
Commented Nov 28, 2020 at 16:05

We can prove the result directly.

Expand[(a^3 + b^3 + c^3)*(a*b + b*c + c*a) -
a*b*c*(a + b + c)^2] // Factor


(a + b) (a + c) (b + c) (a^2 - a b + b^2 - a c - b c + c^2)

and $$a^2 - a b + b^2 - a c - b c + c^2=((a-b)^2+(b-c)^2+(c-a)^2)/2\geq 0$$

ForAll[{a, b, c}, a^2 - a b + b^2 - a c - b c + c^2 >= 0] // Resolve


True

So if $$a,b,c\geq 0$$,the inequality is true.

If we depend all of this by MMA, it also work.

ForAll[{a, b, c},
a >= 0 && b >= 0 &&
c >= 0 , (a^3 + b^3 + c^3)*(a*b + b*c + c*a) -
a*b*c*(a + b + c)^2 >= 0] // Resolve


True

• Maybe your last step can be made with Mathematica too, using for instance this approach mathematica.stackexchange.com/questions/26283/… Commented Nov 28, 2020 at 23:30
• @yarchik Thanks for your links. Commented Nov 28, 2020 at 23:34

Or similar to the version of @MichaelE2

Reduce[a*b*c*(a + b + c)^2 <= (a^3 + b^3 + c^3)*(a*b + b*c + c*a), {a,
b, c}, Reals] //
Simplify[#, a > 0 && b > 0 && c > 0] &

(*   True   *)