Prove $2(a^2+b^2)(b^2+c^2)(c^2+a^2)\geq (a-b)^2(b-c)^2(c-a)^2$

Question

I found this problem in an olympic book and it doens't have the solution, here is the problem:

Let: $a,b,c$ are real numbers. Prove that: $2(a^2+b^2)(b^2+c^2)(c^2+a^2)\geq (a-b)^2(b-c)^2(c-a)^2$ with all $a,b,c$

I try to find the short and natural solution but it seem impossible, I found the solution on AoPS and it said the inequality is rewritten as: $(\sum_{sym}^{}a^2b-2abc)^2\geq 0$ And it is found and solved by computer, even if do it by hand I will not have enough time in test I tried this:

Suppose: $ab \geq 0$

$ \rightarrow2(a^2+b^2)\geq (a-b)^2$

So we need to prove: $(b^2+c^2)(c^2+a^2)\geq (b-c)^2(c-a)^2$ which is hard and complicated!

Can anyone help me with this? Thank you a lots!

Answer 1

Let $a+b+c=3u$, $ab+ac+bc=3v^2,$ where $v^2$ may be negative and $abc=w^3$.

Thus, we need to prove that: $$2\prod_{cyc}(a^2+b^2)\geq\prod_{cyc}(a^2+b^2-2ab)$$ or $$\prod_{cyc}(a^2+b^2)\geq\sum_{cyc}(-2ab(a^2+c^2)(b^2+c^2)+4a^2bc(b^2+c^2))-8a^2b^2c^2$$ $$\prod_{cyc}(a^2+b^2+c^2-c^2)\geq$$ $$\geq\sum_{cyc}(-2a^3b^3-2a^3b^2c-2a^3c^2b-2a^4bc+4a^3b^2c+4a^3c^2b)-8a^2b^2c^2$$ $$\prod_{cyc}(9u^2-6v^2-a^2)\geq$$ $$\geq-2(27v^6-27uv^2w^3+3w^6)-2w^3(27u^3-27uv^2+3w^3)+2w^3(9uv^2-3w^3)-8w^6$$ or $$(9u^2-6v^2)^3-(9u^2-6v^2)^3+(9u^2-6v^2)(9v^4-6uw^3)-w^6\geq$$ $$\geq-26w^6-54u^3w^3+126uv^2w^3-54v^6$$ or $$25w^6-90uv^2w^3+81u^2v^4\geq0$$ or $$(5w^3-9uv^2)^2\geq0$$ and we are done!

Answer 2

Let $$\sum_{cyc}a^2b = X \text{ and } \sum_{cyc}ab^2 = Y \text{ and } abc=Z.$$ Then we have: $$\prod_{cyc}(a-b) = ab^2+bc^2+ca^2-a^2b-b^2c-c^2a = Y-X$$ and using $(x^2+y^2)(z^2+t^2) = (xz-yt)^2 + (xt+yz)^2$ $$\prod_{cyc}(a^2+b^2) = (a^2+b^2)(b^2+c^2)(c^2+a^2) = ((ab-bc)^2 + (ac+b^2)^2)(c^2+a^2) = $$ $$ = (abc-bc^2-a^2c-ab^2)^2 + (a^2b-abc+ac^2+b^2c)^2 = (Z-Y)^2 + (Z-X)^2.$$ So we want to prove: $$2(Z-X)^2+2(Z-Y)^2\geq (X-Y)^2\iff X^2+Y^2+4Z^2-4ZX-4ZY+2XY\geq 0$$ $$\iff (X+Y-2Z)^2\geq 0.$$

I am only writing this answer to show that once you are experienced enough, you can get away with things like this without brute-force expanding.

Answer 3

A solution using Cauchy-Schwarz inequality.

Because: \begin{align} \left(a^2+b^2 \right) \left(a^2+c^2 \right) = \left(a^2+bc \right)^2 + \left(ac-ab \right)^2 \end{align} \begin{align} 2\left(b^2+c^2\right) = \left(b-c\right)^2 + \left(b+c\right)^2 \end{align} Applying Cauchy-Schwarz inequality, we have: \begin{align} \left[ \left(a^2+bc \right)^2 + \left(ac-ab\right)^2 \right] \left[ \left(b-c\right)^2 + \left(b+c\right)^2\right] & \ge \left[\left( a^2+bc \right) \left(b-c \right) + \left(ac-ab\right) \left(b+c \right) \right]^2 \\ & = \left(a-b \right)^2 \left(b-c \right)^2 \left(c-a \right)^2\end{align} Thus, we have Q.E.D.

Answer 4

We have that

$$2(a^2+b^2)(b^2+c^2)(c^2+a^2)\geq (a-b)^2(b-c)^2(c-a)^2$$

which is trivially true for the cases with $abc=0$, then assume $abc \neq 0$ and dividing by $(abc)^2$ we obtain

$$2\left(1+\frac{b^2}{a^2}\right)\left(1+\frac{c^2}{b^2}\right)\left(1+\frac{a^2}{c^2}\right)\geq \left(1-\frac ba\right)^2\left(1-\frac cb\right)^2\left(1-\frac ac\right)^2$$

and by $x=\frac b a\neq 0$, $y=\frac c b\neq 0$

$$2(1+x^2)(1+y^2)\left(1+\frac1{x^2y^2}\right)\ge (1-x)^2(1-y)^2\left(1-\frac1{xy}\right)^2$$

which simplifies to

$$1 + 2x + x^2 + 2y - 2xy - 2x^2y + 2x^3y + y^2 - 2xy^2 + 10x^2y^2 - 2x^3y^2 + x^4y^2 + 2xy^3 - 2x^2y^3 - 2x^3y^3 + 2x^4y^3 + x^2y^4 + 2x^3y^4 + x^4y^4 \ge 0$$

which can be reordered as

$$\overbrace{1 + x^2 + y^2 + 4x^2y^2 + x^4y^2 + x^2 y^4 + x^4 y^4}^{\text{squared terms}} \\ \overbrace{+2x + 2y - 4xy + 2x^2y + 2xy^2 + 2x^2y^2}^{\text{cross terms from} \;1^{st} \to 7^{th}}\\ \overbrace{+ 2xy - 4 x^2 y + 2 x^3 y + 2x^2 y^2 + 2 x^3 y^2}^{\text{cross terms from} \;2^{nd} \to 7^{th}}\\ \overbrace{- 4 x y^2 + 2x^2 y^2 + 2 x y^3 + 2 x^2 y^3}^{\text{cross terms from} \;3^{rd} \to 7^{th}}\\ \overbrace{- 4 x^3y^2- 4 x^2y^3 - 4 x^3 y^3}^{\text{cross terms from} \;4^{th} \to 7^{th}}\\ \overbrace{+ 2 x^3 y^3 + 2 x^4 y^3}^{\text{cross terms from} \;5^{th} \to 7^{th}}\\ \overbrace{+2x^3y^4}^{\text{cross terms from} \;6^{th} \to 7^{th}} \ge0$$

$$(1 + x + y - 2 x y + x^2 y + x y^2 + x^2 y^2)^2\ge 0$$

Answer 5

A solution using only complex numbers and their properties.

$2\left(a^2+b^2\right)\left(b^2+c^2\right)\left(c^2+a^2\right)=$

$=\bigg|\big(1+i\big)\big(a+ib\big)\big(b+ic\big)\big(c+ia\big)\bigg|^{\,2}\!\!=$

$=\bigg|\big[(a-b)+i(a+b)\big]\big[\!-\!c(a-b)+i\left(ab+c^2\right)\!\big]\bigg|^{\,2}\!\!=$

$=\bigg|\!-\!c\big(a\!-\!b\big)^{\!2}\!\!-\!\big(a\!+\!b\big)\!\left(ab\!+\!c^2\right)\!+\!i\big(a\!-\!b\big)\!\left(ab\!+\!c^2\!-\!ac\!-\!bc\right)\!\bigg|^{\,2}\!\!\!\geqslant$

$\geqslant\big(a-b\big)^{\!2}\left(ab+c^2-ac-bc\right)^2=$

$=\big(a-b\big)^{\!2}\big[\big(b-c\big)\big(a-c\big)\big]^2=$

$=\big(a-b\big)^{\!2}\big(b-c\big)^{\!2}\big(c-a\big)^{\!2}.$