## Thursday, March 10, 2016

### Second Attempt: Let $a,\,b$ and $c$ be real numbers such that $(a-b)^3+(b-c)^3+(c-a)^3=9$. Prove that [MATH]\frac{1}{(a-b)^2}+\frac{1}{(b-c)^2}+\frac{1}{(c-a)^2}\ge \sqrt[3]{3}[/MATH].

Let $a,\,b$ and $c$ be real numbers such that $(a-b)^3+(b-c)^3+(c-a)^3=9$.

Prove that [MATH]\frac{1}{(a-b)^2}+\frac{1}{(b-c)^2}+\frac{1}{(c-a)^2}\ge \sqrt[3]{3}[/MATH].

Second attempt:

Recall from the previous blog post that I have mentioned trying to prove the intended inequality by expanding the given equality and see what that leads us?

Now, we will do that and you willsee in a minute why that is not a great start to work with this problem:

$(a-b)^3+(b-c)^3+(c-a)^3=9$

$a^3-3a^2b+3ab^2-b^3+b^3-3b^2c+3bc^2-c^3+c^3-3c^2a+3ca^2-a^3=9$

$(a^3-a^3)-3a^2b+3ab^2+(-b^3+b^3)-3b^2c+3bc^2+(-c^3+c^3)-3c^2a+3ca^2=9$

$-3a^2b+3ab^2-3b^2c+3bc^2-3c^2a+3ca^2=9$

Here comes the first hurdle, as we don't know how to group and factor out the common factor wisely, do we need to factor out $ab,\,bc,\,ca$ or $a^2+b^2+c^2$ respectively?

$3ab(b-a)+3bc(c-b)+3ca(a-c)=9$ [MATH]\color{yellow}\bbox[5px,purple]{\implies\,\,\,ab(b-a)+bc(c-b)+ca(a-c)=3}[/MATH]

or

$3a^2(c-b)+3b^2(a-c)+3c^2(b-a)=9$ [MATH]\color{yellow}\bbox[5px,green]{\implies\,\,\,a^2(c-b)+b^2(a-c)+c^2(b-a)=3}[/MATH]

The following hurdle that comes along the way is we don't know for sure if

$b-a,\,c-b,\,a-c$ are positive or negative...

And we couldn't use the AM-GM inequality to squeeze for more useful summary from those equality...

Yes, I hear you, we could assume, WLOG, that $a\ge b \ge c$ so that we can tell

$b-a\ge 0,\,c-b\le 0,\,a-c\ge 0$

But that doesn't worth the effort as what we're intended to prove is:

[MATH]\frac{1}{(a-b)^2}+\frac{1}{(b-c)^2}+\frac{1}{(c-a)^2}\ge \sqrt[3]{3}[/MATH]

Which, upon expanding on the LHS we see the expression:

[MATH]\frac{1}{a^2-2ab+b^2}+\frac{1}{b^2-2bc+c^2}+\frac{1}{c^2-2ca+a^2}\ge \sqrt[3]{3}[/MATH]

Up to this junction, we just don't see how

[MATH]\color{yellow}\bbox[5px,purple]{ab(b-a)+bc(c-b)+ca(a-c)=3}[/MATH]

or

[MATH]\color{yellow}\bbox[5px,green]{a^2(c-b)+b^2(a-c)+c^2(b-a)=3}[/MATH]

are going to help...

When we've encountered an impasse or we've found that nothing that we've worked so far would help us out in carrying out our next step, it's always best to leave the approach and look for a new one, yes, there always exist multitude of ways to solve a good math competition problem, but not every approach that we adopted will work like charm, we've to improvise ourselves and quit it or continue it as we see fit!