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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.

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

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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.

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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.

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