# The solution for $a+b+c+d = 4$ and $\left( \frac{1}{a^{12}} + ... + \frac{1}{d^{12}} \right) \left( 1 + 3abcd \right) = 16$

$$a,b,c,d > 0$$, then find the solution for $$a+b+c+d = 4$$ and $$\left( \frac{1}{a^{12}} + \frac{1}{b^{12}} + \frac{1}{c^{12}} + \frac{1}{d^{12}} \right) \left( 1 + 3abcd \right) = 16$$

Attempt:

We know that $$a=b=c=d=1$$ is a solution. Are there any other solutions? By AM-GM, $$\frac{4}{4} \ge (abcd)^{1/4} \implies abcd \le 1$$

Also by HM-GM we have

$$\frac{4}{ \frac{1}{a^{12}} +\frac{1}{b^{12}} + \frac{1}{c^{12}} + \frac{1}{d^{12}} } \le \left( a^{12} b^{12} c^{12} d^{12} \right)^{1/4}$$

$$\frac{4}{(abcd)^{3}} \le \frac{1}{a^{12}} +\frac{1}{b^{12}} + \frac{1}{c^{12}} + \frac{1}{d^{12}}$$

then

$$4 \le \frac{4}{(abcd)^{3}} \le \frac{1}{a^{12}} +\frac{1}{b^{12}} + \frac{1}{c^{12}} + \frac{1}{d^{12}}$$

and of course $$1 + 3abcd \le 4$$.

Next,

$$\frac{1}{a^{12}} +\frac{1}{b^{12}} + \frac{1}{c^{12}} + \frac{1}{d^{12}} = \frac{ (bcd)^{12} + (acd)^{12} + (abd)^{12} + (abc)^{12} }{(abcd)^{12}}$$

so we have $$\left( \frac{1}{a^{12}} + \frac{1}{b^{12}} + \frac{1}{c^{12}} + \frac{1}{d^{12}} \right) \left( 1 + 3abcd \right) = \frac{ (bcd)^{12} + (acd)^{12} + (abd)^{12} + (abc)^{12} }{(abcd)^{12}} + \frac{ 3(bcd)^{12} + 3(acd)^{12} + 3(abd)^{12} + 3(abc)^{12} }{(abcd)^{11}} = 16$$

## 1 Answer

Note that \begin{align*} \left(\frac{1}{a^{12}} + \frac{1}{b^{12}} + \frac{1}{c^{12}} + \frac{1}{d^{12}}\right)(1 + 3abcd) &\geq \frac{4}{(abcd)^3}(1 + 3abcd) \\ &= \frac{4}{(abcd)^3} + \frac{12}{(abcd)^2} \\ &\geq 4 + 12 \\ &= 16 \end{align*} where the first inequality holds by AM-GM, and the last inequality holds since $$abcd \leq 1$$ (again established by AM-GM). But by the equality conditions of AM-GM, equality is attained in each of these inequalities only when $$a, b, c, d$$ are equal.