# Show that $\frac{1}{a^n+b^n+c^n} = \frac{1}{a^n} + \frac{1}{b^n} + \frac{1}{c^n}$ [closed]

Let $n \in N, n=2k+1, and \text{ } \frac{1}{a+b+c} = \frac{1}{a} + \frac{1}{b} + \frac{1}{c}$.

Show that $$\frac{1}{a^n+b^n+c^n} = \frac{1}{a^n} + \frac{1}{b^n} + \frac{1}{c^n}$$

I have tried, but I don't get anything. Can you please give me a hint?

From the original equation, we get:

$$abc=(a+b+c)(ab+bc+ca)$$ which is equivalent to $$a^2(b+c)+bc(b+c)+ab(b+c)+ca(b+c)=0$$ $$\implies(b+c)(a+c)(a+b)=0$$ Then, obviously any one of the following must hold: $$a=-b\\b=-c\\c=-a$$

In any case we can prove the equation $$\frac{1}{a^n+b^n+c^n}=\frac{1}{a^n}+\frac{1}{b^n}+\frac{1}{c^n}$$ with odd $n$. Since if we take $a=-b$ we get $$\frac{1}{(-b)^n+b^n+c^n}=\frac{1}{(-b)^n}+\frac{1}{b^n}+\frac{1}{c^n}$$ which is equivalent to $$\frac{1}{c^n}=\frac{1}{c^n}$$ and this is true.....

Hint

• At the first step, show that $(a+b)(a+c)(b+c)=0$
• Next, show that two of the three numbers are opposite.

The condition gives $$(a+b+c)(ab+ac+bc)=abc$$ or $$\sum_{cyc}(a^2b+a^2c+abc)=abc$$ or $$\sum_{cyc}\left(a^2b+a^2c+\frac{2}{3}abc\right)=0$$ or $$\prod_{cyc}(a+b)=0.$$ We need to prove that $$\prod_{cyc}(a^n+b^n)=0$$ or $$\prod_{cyc}(a+b)\prod_{cyc}\left(a^{2k}-a^{2k-1}b+...+b^{2k}\right)=0,$$ which is obvious.

We can use also the following reasoning.

Since the condition is true for $a=-b$ and it's symmetric and degree $3$, we obtain that the condition is equal to $$(a+b)(a+c)(b+c)=0$$ and we need to prove that $$\left(a^{2k+1}+b^{2k+1}\right)\left(a^{2k+1}+c^{2k+1}\right)\left(b^{2k+1}+c^{2k+1}\right)=0.$$

This method of infinite ascent is probably wrong.. but...

Suppose none of $a = -b, b = -c, c = -a$ is true.

Then multiplying both sides by $abc$ and inverting we get:

$\frac{a+b+c}{abc} = \frac{1}{ab + bc + ca} \implies \frac{1}{ab + bc + ca}= \frac{1}{ab} + \frac{1}{bc} + \frac{1}{ca}$

We can now write $x = ab, y = bc, z = ca$ and do this all over again which is probably absurd (needs better logic here).