To prove $(a^2+b^2+c^2)(ab+bc+ca)^2 \ge (a^2+2bc)(b^2+2ca)(c^2+2ab)$.

Question

I think that there is not a simple proof of this inequality. (for all real numbers)

My Attempt $1$:
Equality occurs for $a=b=c$.
It is quit sure that the inequality is homogeneous.
So normalize with $ab+bc+ca = 1$. $$ (a^2+b^2+c^2)(ab+bc+ca)^2 \ge (a^2+2bc)(b^2+2ca)(c^2+2ab) $$ $$ a^2+b^2+c^2 \ge (a^2+2bc)(b^2+2ca)(c^2+2ab) $$

We're done if $$ 1 \ge 2\sum_{cyc}{a^3b^3}+4\sum_{cyc}{a^4bc}+9a^2b^2c^2 $$ By the hypothesis, $$ (ab+bc+ca)^3 $$$$= a^3 b^3 + 3 a^3 b^2 c + 3 a^3 b c^2 + a^3 c^3 + 3 a^2 b^3 c$$$$ + 6 a^2 b^2 c^2 + 3 a^2 b c^3 + 3 a b^3 c^2 + 3 a b^2 c^3 + b^3 c^3 $$$$= 1= \sum_{cyc}{a^3b^3} + 3\sum_{cyc}{a^3b^2c} + 3\sum_{cyc}{a^3bc^2}+6a^2b^2c^2 $$

$$ 3\sum_{cyc}{a^3b^2c} + 3\sum_{cyc}{a^3bc^2}\ge \sum_{cyc}{a^3b^3}+4\sum_{cyc}{a^4bc}+3a^2b^2c^2 $$ As @Michael Rozenberg's feedback in one of my recent previous posts, I have carefully preserved the Equality case for 5 steps. Hope this helps!

I think that this inequality is sharp, as I have already used a blunt inequality. So I think SOS could work.

But this attempt is surely incorrect. Try $a=b=1$ and $c=0$.

My Attempt $2$:
Expand everything. $$ (a^2+b^2+c^2)(ab+bc+ca)^2 \ge (a^2+2bc)(b^2+2ca)(c^2+2ab) $$ Here, $$ (ab+bc+ca)^2 = a^2b^2+b^2c^2+c^2a^2 + 2abc(a+b+c) $$ $$ = \sum_{cyc}{a^2b^2}+2abc\sum_{cyc}{a} $$ Multiplying this thing to $\sum_{cyc}{a^2}$ to get:

$$ = \sum_{cyc}{a^4b^2}+\sum_{cyc}{a^2b^4}+3a^2b^2c^2 $$ $$ + 2abc (\sum_{cyc}{a^3}+\sum_{cyc}{a^2b}+\sum_{cyc}{ab^2}) $$ As before, $$ (a^2+2bc)(b^2+2ca)(c^2+2ab) $$ $$ = 2\sum_{cyc}{a^3b^3}+4abc\sum_{cyc}{a^3}+9a^2b^2c^2 $$ We just need to prove that: $$ \sum_{cyc}{a^4b^2}+\sum_{cyc}{a^2b^4}+ 2abc (\sum_{cyc}{a^2b}+\sum_{cyc}{ab^2}) $$ $$ \ge 2\sum_{cyc}{a^3b^3}+2abc\sum_{cyc}{a^3}+6a^2b^2c^2 $$

From here, it's very easy to get a wrong inequality again: As $(4,2)\succ(3,3)$. $$ \sum_{cyc}{a^4b^2}+\sum_{cyc}{a^2b^4} \ge 2\sum_{cyc}{a^3b^3} $$ It rests to prove that: $$ \sum_{cyc}{a^2b}+\sum_{cyc}{ab^2} \ge \sum_{cyc}{a^3} + 3abc $$ Supplied $$ \sum_{cyc}{ab^2} \ge 3abc $$ The inequality is obviously wrong: $$ \sum_{cyc}{a^2b} \ge \sum_{cyc}{a^3} $$

I already have Vask's solution, so I don't need that proof.
I just ask if someone has another good solution, Thanks!

Answer 1

Note that \begin{eqnarray*} (a^2b+b^2c+c^2a-a^2c-b^2a-c^2b)^2 \geq 0 \end{eqnarray*} and this can be rearranged to give the required inequality.

Edit:

I first calculated that the inequality is equivalent to \begin{eqnarray*} & & (a^2+b^2+c^2)(ab+bc+ca)^2 -(a^2+2bc)(b^2+2ca)(c^2+2ab) \\ &=& \sum_{cyc} a^4b^2 -2 \sum_{cyc} a^3 b^3 +2 \sum _{sym}a^3b^2c -6 a^2 b^2 c^2 \geq 0. \end{eqnarray*} The first two terms suggest the form above ... and luckily they also give the next two terms as well! ... $ \ddot \smile $

Answer 2

Hint: Both sides are symmetric polynomials. Express them in terms of elementary symmetric functions.

Answer 3

It's better to learn the following identities: $$(a+b)^2=a^2+2ab+b^2,$$ $$(a-b)^2=a^2-2ab+b^2,$$ $$.$$ $$.$$ $$.$$ $$\left(\sum_{sym}a^2b\right)^2=\sum_{sym}(a^4b^2+a^3b^3+a^4bc+2a^3b^2c+a^2b^2c^2)$$ and $$\prod_{cyc}(a-b)^2=\sum_{sym}(a^4b^2-a^3b^3-a^4bc+2a^3b^2c-a^2b^2c^2).$$ Now, we can solve your problem during one minute in one line by hand: $$\sum_{cyc}a^2\left(\sum_{cyc}ab\right)^2-\prod_{cyc}(a^2+2bc)=$$ $$\sum_{cyc}a^2\sum_{cyc}(a^2b^2+2a^2bc)-\sum_{cyc}(2a^3b^3+4a^4bc+3a^2b^2c^2)=$$ $$\sum_{cyc}(a^4b^2+a^4c^2+a^2b^2c^2+2a^4bc+2a^3b^2c+2a^3c^2b)-\sum_{cyc}(2a^3b^3+4a^4bc+3a^2b^2c^2)=$$ $$=\sum_{cyc}(a^4b^2+a^4c^2-2a^3b^3-2a^4bc+2a^3b^2c+2a^3c^2b-2a^2b^2c^2)=\prod_{cyc}(a-b)^2\geq0.$$ The identity $$\sum_{cyc}(a^4b^2+a^4c^2-2a^3b^3-2a^4bc+2a^3b^2c+2a^3c^2b-2a^2b^2c^2)=\prod_{cyc}(a-b)^2\geq0$$ we can get by the following way.

Let $a=b$.

Thus, we obtain: $$2a^6+2a^4c^2+2c^4a^2-2a^6-4a^3c^3-4a^5c-2c^4a^2+4a^5c+4a^3c^3+4a^4c^2-6a^4c^2=0\cdot(a-c)^4$$ or $$0=0,$$ which since our statement is symmetric and $a-b$ is not symmetric, we got a factor $(a-b)^2.$

Now, since our statement is symmetric, we have also factors $(a-c)^2$ and $(b-c)^2$ and since the left expression is sixth degree, we obtain: $$\sum_{cyc}(a^4b^2+a^4c^2-2a^3b^3-2a^4bc+2a^3b^2c+2a^3c^2b-2a^2b^2c^2)=C\prod_{cyc}(a-b)^2,$$ for some real number $C$.

Now, let $c=0$ and $b=1$.

We obtain for any real $a$: $$a^4+a^2-2a^3=C(a-1)^2a^2,$$ which gives $C=1$.

Answer 4

The first, it's easy to check $$(x+a)(x+b)(x+c) = x^3+(a+b+c)x^2+(ab+bc+ca)x+abc.$$ So, setting $X=ab+bc+ca,$ we have $$\prod (a^2+2bc) = \prod \left[X +(a-b)(a-c)\right]$$ $$=X^3+(a^2+b^2+c^2-ab-bc-ca)X^2-(a-b)^2(b-c)^2(c-a)^2$$ $$=X^2\left[(a^2+b^2+c^2-ab-bc-ca)+X\right]-(a-b)^2(b-c)^2(c-a)^2$$ $$=(ab+bc+ca)^2(a^2+b^2+c^2)-(a-b)^2(b-c)^2(c-a)^2$$ $$ \leqslant (ab+bc+ca)^2(a^2+b^2+c^2).$$

Answer 5

Let $A = a^2 + b^2 + c^2$, $x = a- b$, $y = b - c$ and $z = c-a$. We have \begin{align} \mathrm{RHS} &= (A - y^2)(A - z^2)(A - x^2)\\ &= A^3 - A^2(x^2 + y^2 + z^2) + A(x^2y^2 + y^2z^2 + z^2x^2) - x^2y^2z^2\\ &= A^3 + 2A^2(xy+yz+zx) + A(xy+yz+zx)^2 - x^2y^2z^2\\ &= A(A + xy + yz + zx)^2 - x^2y^2z^2\\ &= (a^2 + b^2 + c^2)(ab+bc+ca)^2 - (a-b)^2(b-c)^2(c-a)^2 \end{align} where we have used $$x+y+z = 0,$$ $$x^2+y^2+z^2 = (x+y+z)^2 - 2(xy+yz+zx) = -2(xy+yz+zx),$$ $$x^2y^2 + y^2z^2 + z^2 x^2 = (xy + yz + zx)^2 - 2xyz(x+y+z) = (xy + yz + zx)^2.$$

We are done.