Finding the fraction $\frac{a^5+b^5+c^5+d^5}{a^6+b^6+c^6+d^6}$ when knowing the sums $a+b+c+d$ to $a^4+b^4+c^4+d^4$

One way to approach such questions is to view $a, b, c, d$ as roots of a single quartic equation $x^4-s_1x^3+s_2x^2-s_3x+s_4=0$, when we have Vieta's relations $$s_1=a+b+c+d$$$$s_2=ab+ac+ad+bc+bd+cd=(a+b)(c+d)+ab+cd$$$$s_3=abc+abd+acd+bcd$$$$s_4=abcd$$

We then let $p_k=a^k+b^k+c^k+d^k$ with: $$p_0=4$$$$p_1-s_1=0$$$$p_2-p_1s_1+2s_2=0$$$$p_3-p_2s_1+p_1s_2-3s_3=0$$

With the given values of $p_1, p_2, p_3$ these determine $s_1, s_2, s_3$.

For higher powers $(k\ge4)$ we have $$a^k-s_1a^{k-1}+s_2a^{k-2}-s_3a^{k-3}+s_4a^{k-4}=0$$ using the quartic equation. Adding together the corresponding equations for $a , b, c, d$ we obtain the recurrence for $p_k$$$p_k-s_1p_{k-1}+s_2p_{k-2}-s_3p_{k-3}+s_4p_{k-4}=0$$

Since we know $p_4$ we can use the recurrence to find $s_4$, and this is then sufficient to compute $p_5, p_6$ from the recurrence.

If this looks complicated, this is misleading. The relations between the elementary symmetric functions $s_r$ and the sums of powers $p_r$ are useful to know, and render this kind of problem essentially mechanical. You can see that you can solve the problem without solving explicitly for $a, b, c, d$.

Following this scheme we find successively $s_1=2, s_2=-13, s_3=-14, s_4=24$. Then the recurrence gives:$$p_5-2p_4-13p_3+14p_2+24p_1=0$$ and$$p_6-2p_5-13p_4+14p_3+24p_2=0$$

And this gives $p_5=812, p_6=4890$

Here's an approach which involves AM-GM inequality(To bound the numbers).

I have considered the positive values of $(a,b,c,d)$. I can consider $a^2,b^2,c^2$ and $d^2$, since they are all strictly positive.

$\dfrac{a^2+b^2+c^2+d^2}{4} \ge \sqrt{abcd}$

$\dfrac{30}{4} \ge \sqrt{abcd} \implies 56.25 \ge abcd$

Since $a^2+b^2+c^2+d^2=30$ and $56 >|abcd|$, One of the $|a|,|b|,|c|,|d|$ is at most $5$.

Considering $a^4+b^4+c^4+d^4$, one of the $|a|,|b|,|c|,|d|$ is atmost $4$, since $5^4=625$ .

Now we have $a^2+b^2+c^2+d^2=30$, the value of $(|a|,|b|,|c|,|d|)$ is from set $(1,2,3,4)$.

And also:

Considering $4th$ degree polynomial such that $a,b,c,d$ are the roots of the equation.

$x^4+px^3+qx^2+rx+s=0$

$a+b+c+d=2=-p$

$(a+b+c+d)^2-a^2+b^2+c^2+d^2=2(ab+bc+cd+ad+ac+bd)=q=-13$

Finding $\sum abc$ and $abcd$ with the help of other inequalities and finding the roots of equation is another way to go. (Not quite sure whether constructing polynomial is a great way to go. )

Edit: I have considered the absolute values of $(a,b,c,d)$ . I didn't assume them to be $+ve$.

I do not know how to solve it, but $a=-3,b=4,c=2,d=-1$