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

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*}

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!

• while you can use AM-GM, a fun way of solving is noting that $0=a^4 + b^4 + c^4 + d^4 -4abcd=(a^2-b^2)^2+(c^2-d^2)^2+2(ab-cd)^2$ (and of course any permutations thereof) Mar 16, 2022 at 2:38
• When we apply AM-GM(we can, since forth powers are positive) we get $a^4+b^4+c^4+d^4\geq 4abcd$, where $4abcd$ is the smallest possible value of $a^4+b^4+c^4+d^4$. Equality happens since $48= 4\cdot 12$. Equality can happen only if all terms are equal hence $a^4=b^4=c^4=d^4$ . Mar 16, 2022 at 20:38

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.

• You can choose the sign for only $3$ out of the $4$ variables, since the last one is constrained by $abcd=12$, so in the end there are only $2^3$ solutions. Also, $\sqrt[4]{a^4b^4c^4d^4} = abcd$ should rather be $\sqrt[4]{a^4b^4c^4d^4} = |abcd|$.
– dxiv
May 11, 2022 at 17:23
• @dxiv Thanks so much for pointing out. I just made an edit to fix the answer according to your suggestion. May 11, 2022 at 17:46

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 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.

• If the question asks for equality, what is the reason to use inequality? Why is 8 a rough estimate for 12? Mar 21, 2022 at 14:35