Tuesday, March 8, 2016

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].

My solution:

I know there are students, who upon seeing the equality wanted to expand the cube inside the parentheses and get started from there.

I won't say that is not a good start to prove the intended inequality, but, if we could, we should always look for ways to simplify things, rather than to complicate the problem or given data/information at hand.

Therefore, our next effort should be channeled into simplifying the information that we're given:


When the right situation arises, we could always use the help of a substitution.

Let $x=a-b,\,y=b-c$, then if we add up $x$ and $y$, we get another relation that says:


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





Also, the intended inequality becomes

[MATH]\frac{1}{x^2}+\frac{1}{y^2}+\frac{1}{(-(x+y))^2}\ge \sqrt[3]{3}[/MATH]

[MATH]\frac{1}{x^2}+\frac{1}{y^2}+\frac{1}{(x+y)^2}\ge \sqrt[3]{3}---(2)[/MATH]

If we replace [MATH]\frac{1}{x+y}=-\frac{xy}{3}---(1)[/MATH] into the LHS of the expression in the inequality (2), we see that we have:



[MATH]\ge 3\left(\frac{1}{x^2} \cdot \frac{1}{y^2} \cdot \frac{x^2y^2}{9}\right)^{\frac{1}{3}}[/MATH] (by the AM-GM inequality)

[MATH]\ge 3\left(\frac{1}{3^2}\right)^{\frac{1}{3}}[/MATH]

[MATH]\ge 3^{\frac{1}{3}}[/MATH]

and we're hence done.

I will do another post for expanding the given equality to see where that leads us, will it lead us to the simplicity or complexity? I bet you have already an answer for it...

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