# Prove that: $\sum_{i=1}^{5}x_i\sum_{i=1}^{5}\frac {1}{x_i}\leq 25$

Let $$a,c\neq 0$$ and $$P(x)=ax^5+bx^2+c$$ be a polynomial with the real coefficients, where $$x_1,x_2,...,x_5$$ are the complex roots of $$P(x)$$. Prove that: $$\sum_{i=1}^{5}x_i^5\sum_{i=1}^{5}\frac {1}{x_i^5}\leq 25$$

Some trivial thoughts.

This is my textbook contest problem . I've been trying to solve this question for a while. I asked a partial question earlier.

I tried to use the Cauchy–Schwarz inequality

$$(a^5+b^5+c^5+d^5+e^5)\left(\frac 1{a^5}+\frac 1{b^5}+\frac 1{c^5}+\frac {1}{d^5}+\frac {1}{e^5}\right)\geq 5^2=25$$

However, not all roots of the equation $$ax^5+bx^2+c=0$$ can even be real numbers!

I've never heard of an inequality that works with complex numbers.

• The title doesn't quite match the question. I presume that fifth powers are summed, for the sum of the roots is trivially zero. Dec 18, 2022 at 13:48
• @JyrkiLahtonen yes the OP corrected his typo in the text and forgot to correct it in the title. Dec 18, 2022 at 13:53

Start by finding $$\sum_{i=1}^5 x_i^5$$ . Notice that $$P(x_1)=P(x_2)=...=P(x_5)=0$$

Therefore $$P(x_1)+P(x_2)+...+P(x_5)=0$$ which implies $$a\sum_{i=1}^5 x_i^5+b\sum_{i=1}^5 x_i^2+5c=0$$

Notice that $$\sum_{i=1}^5 x_i^2=\left(\sum_{i=1}^5 x_i\right)^2-2\left(\sum_{i\not{=}j}x_ix_j\right)$$. Using Vieta's formulas we know that $$\left(\sum_{i=1}^5 x_i\right)^2=0$$ and $$\sum_{i\not{=}j}x_ix_j=0$$ and therefore $$\sum_{i=1}^5 x_i^2=0$$.

This means that $$\sum_{i=1}^5 x_i^5=-\frac{5c}{a}$$

To find $$\sum_{i=1}^5 \frac{1}{x_i^5}$$ requires a lot more effort. Notice once again that $$\frac{P(x_1)}{x_1^5}=\frac{P(x_2)}{x_2^5}=...=\frac{P(x_5)}{x_5^5}=0$$ since all our roots are clearly non-zero.

Doing the same summation process as above gives us this time $$5a+b\sum_{i=1}^5 \frac{1}{x_i^3}+c\sum_{i=1}^5 \frac{1}{x_i^5}=0$$ We will come back to this equation so it will be equation $$(1)$$.

Now we want to notice that $$\frac{P(x_1)}{x_1^2}=\frac{P(x_2)}{x_2^2}=...=\frac{P(x_5)}{x_5^2}=0$$. We will now use $$x_1$$ as an example to derive equations for all the roots.

Using $$\frac{P(x_1)}{x_1^2}=0$$ we get $$ax^3+b+\frac{c}{x^2}=0$$. By squaring both sides, we arrive at $$a^2x_1^6+2abx_1^3+2acx_1+b^2+\frac{2bc}{x_1^2}+\frac{c^2}{x_1^4}=0$$

Multiplying both sides by $$\frac{1}{x_1}$$ we get $$a^2x_1^5+2abx_1^2+2ac+\frac{b^2}{x_1}+\frac{2bc}{x_1^3}+\frac{c^2}{x_1^5}=0$$

Doing the exact same process for the other roots and summing the equations up gives $$a^2\sum_{i=1}^5 x_i^5+2ab\sum_{i=1}^5 x_i^2+10ac+b^2\sum_{i=1}^5 \frac{1}{x_i}+2bc\sum_{i=1}^5 \frac{1}{x_i^3}+c^2\sum_{i=1}^5 \frac{1}{x_i^5}=0$$

Note that we have already established $$\sum_{i=1}^5 x_i^2=0, \sum_{i=1}^5 x_i^5=-\frac{5c}{a}$$ and we know that $$\sum_{i=1}^5 \frac{1}{x_i}=\frac{\sum_{i\not{=}j}x_ix_j}{x_1x_2x_3x_4x_5}=0$$.

Therefore the chunky equation simplifies to $$5ac+2bc\sum_{i=1}^5 \frac{1}{x_i^3}+c^2\sum_{i=1}^5 \frac{1}{x_i^5}=0$$ Dividing both sides by $$c$$ gives $$5a+2b\sum_{i=1}^5 \frac{1}{x_i^3}+c\sum_{i=1}^5 \frac{1}{x_i^5}=0$$

Now if we compare $$5a+2b\sum_{i=1}^5 \frac{1}{x_i^3}+c\sum_{i=1}^5 \frac{1}{x_i^5}=0$$ and equation $$(1)$$ $$5a+b\sum_{i=1}^5 \frac{1}{x_i^3}+c\sum_{i=1}^5 \frac{1}{x_i^5}=0$$, it is clear that these can only simultaneously hold true if $$\sum_{i=1}^5 \frac{1}{x_i^3}=0$$.

We therefore get $$5a+c\sum_{i=1}^5 \frac{1}{x_i^5}=0\Longrightarrow\sum_{i=1}^5 \frac{1}{x_i^5}=-\frac{5a}{c}$$

$$\therefore \sum_{i=1}^5 x_i^5\sum_{i=1}^5 \frac{1}{x_i^5}=\frac{-5c}{a}\frac{-5a}{c}=25≤25$$

To be honest I cannot see why there is an inequality in the question but this result does trivially fulfill the inequality as well.

For the $$x_i$$'s, by Viete's formulas, the values of the elementary symmetric polynomials are $$e_1=0,\quad e_2=0,\quad e_3=-\frac ba,\quad e_4=0,\quad e_5=-\frac ca$$ so that the Newton identity $$p_5 =e_1^5 - 5 e_2 e_1^3 + 5 e_3 e_1^2 + 5 e_2^2 e_1 - 5 e_4 e_1 - 5 e_3e_2 + 5 e_5$$ boils down to $$p_5=5e_5$$ i.e. $$\sum_{i=1}^{5}x_i^5=-5\frac ca.$$

For the $$y_i:=\frac1{x_i}$$'s, which are roots of $$cy^5+by^3+a=0,$$ we get similarly: $$e_1=0,\quad e_2=\frac bc,\quad e_3=0,\quad e_4=0,\quad e_5=-\frac ac$$ hence $$\sum_{i=1}^5\frac1{x_i^5}=-5\frac ac.$$ I guess the $$\le$$ in the problem was intentionally misleading. In fact we have an equality.