Solve for $a,b,c,d$ over $a^4+b^4+c^4+d^4=48, abcd=12$

Question

Find the number of ordered quadruples $(a,b,c,d)$ of real numbers such that \begin{align*} a^4 + b^4 + c^4 + d^4 &= 48, \\ abcd &= 12. \end{align*}

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I think I should apply some inequalities, so I tried using Cauchy-Schwarz on the first equation. However, this didn't really get me anywhere. I also tried applying AM-GM on the second one, but that also didn't get me anywhere. I would appreciate it if anyone could be be some guidance!

Also, I know that some of these problems can be solved by calc, but I don't know any calc yet, so hopefully, someone can if some non-calc hints!

Thanks in advance!!!!!!!!!!

Answer 1

Alright, I'm going to use an inequality-based approach to solve this problem.

First of all, we know that $(a,b,c,d)$ is a quadruple such that; \begin{align*} a^4+b^4+c^4+d^4 &= 48\\ abcd &= 12. \end{align*}

Now, we know that; \begin{align*} a^4+b^4+c^4+d^4 &= 48 \\ \implies \frac{a^4+b^4+c^4+d^4}{4} &= \frac{48}{4}\\ & = 12. \end{align*} Now, we know that the Arithmetic mean of the numbers $a^4,$ $b^4,$ $c^4,$ and $d^4$ is $12.$ However, the Geometric Mean of $a^4,$ $b^4,$ $c^4,$ and $d^4$ is also $12$; \begin{align*} \sqrt[4]{a^4b^4c^4d^4} &= abcd\\ &= 12. \end{align*}

But since the Arithmetic and the Geometric Mean are both equal, we know that the four numbers being compared must also be equal. Recall that the AM-GM inequality states that the Arithmetic Mean is always greater than equal to the Geometric mean, and the two means are only equal whenever the $4$ numbers being compared are all equal.

So, obviously $a^4=b^4=c^4=d^4,$ and since these numbers add up to $48,$ they are all equal to $12.$ So we know that; \begin{align*} a^4&=12\\ b^4&=12\\ c^4&=12\\ d^4&=12. \end{align*}

Since $a, b, c, d$ are all real numbers with a $4^{\text{th}}$ power equal to $12,$ we know that they can have only two possible values; \begin{align*} &\sqrt[4]{12}=1.8612\cdots \hspace{2mm}\text{ or }\hspace{2mm} -\sqrt[4]{12}=-1.8612\cdots \end{align*}

So, we know that each one of $a,b,c,$ and $d$ are either $\sqrt[4]{12}$ or $-\sqrt[4]{12}.$ So we need to choose for each of the variables whether it is $\sqrt[4]{12}$ or $-\sqrt[4]{12},$ but we also need to make sure that their product is positive (equal to $12$).

Since we need to make sure That the product of the $4$ numbers is positive (and equal to $12$), as soon as we choose any three of the numbers,the fourth one would get automatically chosen. So the total number of ordered quadruples of solutions $(a,b,c,d)$ depends on making $3$ decisions, in $2\times 2\times 2=8$ total ways.

There are total $8$ such quadruples. See how we used a little bit of combo in the end. You never know what to expect in a solution lol.

Answer 2

Hints: Apply Tchebychev inequality:

Let the sets $(a_1, a_2, a_3, ..)$ , $(b_1, b_2, b_3, . . .)$ and $(c_1, c_2, c_3, ...)$ be psitive real numbers and the members of each set are in ascending order of magnitude, then:

$$\frac{\Sigma a_r}n\cdot\frac{\Sigma b_r}n\cdot\frac{\Sigma c_r}n\cdot\cdot\cdot\leq\frac{\Sigma a_r\cdot b_r\cdot c_r\cdot\cdot\cdot}n$$

Let $a<b<c<d$ then we have:

$$\frac 14(a+b+c+d)\cdot \frac 14(a+b+c+d)\cdot \frac 14(a+b+c+d)\cdot\frac 14(a+b+c+d)\leq \frac 14 (a^4+b^4+c^4+d^4)$$

Or:

$$\frac 1{256} (a+b+c+d)^4\leq \frac 14 (a^4+b^4+c^4+d^4)$$

$$(a+b+c+d)^4\leq 64(a^4+b^4+c^4+d^4)$$

Putting given values we get:

$$(a+b+c+d)^4\leq 64\times 48$$

Taking 4th root of both sides we get:

$$a+b+c+d\leq 4\times 12^{\frac 14}\approx 7$$

For rough estimation let (a, b, c, d)=(1, 2, 2, 2) which gives:

$a^4+b^4+c^4+d^4=49$; $abcd=8$

By try and error I found:

$(a, b, c, d)=(1.35=\frac{135}{100}, 1.9=\frac{19}{10}, 2, 2)$

such that:

$a^4+b^4+c^4+d^4= 48.35..$

$abcd=10.26$

You can try other numbers for more accuracy.