Let $ a,b,c >0 $ and $\sqrt[3]{a}+\sqrt[3]{b}+\sqrt[3]{c}=3. $ Find the maximum value of $abc(a+b)(b+c)(c+a) .$

sqing wrote: Let $ a,b,c d>0 $ and $\sqrt[3]{a}+\sqrt[3]{b}+\sqrt[3]{c}+\sqrt[3]{d}=4. $ Prove that$$abcd(a+b)(b+c)(c+d)(d+a)\leq 16$$

apparently equality case does not give maximum as $(0.729,1.331,0.729,1.331)$ gives a larger solution for $abcd(a+b)(b+c)(c+d)(d+a)$

The answer is $\boxed{\tfrac{2^{28}}{5^{10}}}$, and the equality case is $$(a,b,c,d) = \left(\left(1-\tfrac 1{\sqrt 5}\right)^3, \left(1+\tfrac 1{\sqrt 5}\right)^3, \left(1-\tfrac 1{\sqrt 5}\right)^3, \left(1+\tfrac 1{\sqrt 5}\right)^3\right).$$To prove the bound, we note the following lemma, which is a special case of the problem. Lemma. For any positive real numbers $a,b$, we have $$ab(a+b)^2 \leq \frac{2^2}{5^5}(\sqrt[3]a + \sqrt[3]b)^{12}.$$Proof. Set $a=x^3, b=y^3$. Also, since the inequality is homogeneous, WLOG, assume that $x+y=2$. Then, we have \begin{align*} ab(a+b)^2 &= x^3y^3(x^3+y^3)^2 \\ &= x^3y^3\left((x+y)^3 - 3xy(x+y)\right) \\ &= (xy)^3 (8-6xy)^2. \end{align*}To maximize this, use AM-GM, $$(4xy)^3(8-6xy)^2 \leq \left(\frac{(4xy) + (4xy) + (4xy) + (8-6xy) + (8-6xy)}{5}\right)^5 = \frac{2^{20}}{5^5},$$implying that $ab(a+b)^2 \leq \tfrac{2^{14}}{5^5}$, so we are done. $\blacksquare$ To finish the problem, we use the lemma four times: \begin{align*} ab(a+b)^2 &\leq \frac{2^2}{5^5}(\sqrt[3]a + \sqrt[3]b)^{12} \\ bc(b+c)^2 &\leq \frac{2^2}{5^5}(\sqrt[3]b + \sqrt[3]c)^{12} \\ cd(c+d)^2 &\leq \frac{2^2}{5^5}(\sqrt[3]c + \sqrt[3]d)^{12} \\ da(d+a)^2 &\leq \frac{2^2}{5^5}(\sqrt[3]d + \sqrt[3]a)^{12}, \end{align*}Moreover, we note that $(\sqrt[3]a + \sqrt[3]b)(\sqrt[3]c + \sqrt[3]d) \leq 4$, and similarly, $(\sqrt[3]b + \sqrt[3]c)(\sqrt[3]d + \sqrt[3]a)\leq 4$. Thus, multiplying all four above together gives $$\big(abcd(a+b)(b+c)(c+d)(d+a)\big)^2 \leq \frac{2^{56}}{5^{20}},$$giving the desired bound.

Rewrite $\sqrt[3]{a}=w$,$\sqrt[3]{b}=x$,$\sqrt[3]{c}=y$,$\sqrt[3]{d}=z$. We want to minimise \[w^3x^3y^3z^3(w^3+x^3)(x^3+y^3)(y^3+z^3)(z^3+w^3)\]given $w+x+y+z=4$. Main claim: For any positive reals $p,q$, we have \[p^3+q^3\leq\sqrt{\frac{2^2(p+q)^{12}}{5^5p^3q^3}}.\]Proof: \begin{align*} (p^2-pq+q^2)^2(2pq)^3&\leq\left(\frac{2(p^2-pq+q^2)+3(2pq)}{5}\right)^5\\ &=\left(\frac{2(p+q)^2}{5}\right)^5\\ &=\frac{2^5(p+q)^{10}}{5^5}\\ \implies p^2-pq+q^2&\leq\sqrt{\frac{2^5(p+q)^{10}}{2^35^5p^3q^3}}\\ \implies p^3+q^3&\leq\sqrt{\frac{2^2(p+q)^{12}}{5^5p^3q^3}} \end{align*}Now we plug this back into the original equation, to get \begin{align*} &\quad w^3x^3y^3z^3(w^3+x^3)(x^3+y^3)(y^3+z^3)(z^3+w^3)\\ &\leq (wxyz)^3\sqrt{\frac{2^8((w+x)(x+y)(y+z)(z+x))^{12}}{5^{20}(wxyz)^6}}\\ &=\frac{2^4((w+x)(x+y)(y+z)(z+x))^6}{5^{10}}\\ &\leq \frac{2^4\left(\frac{2w+2x+2y+2z}{4}\right)^{24}}{5^{10}}\\ &=\frac{2^{28}}{5^{10}}. \end{align*}Equality is achieved when $w=y=\frac{5-\sqrt{5}}{5}$ and $x=z=\frac{5+\sqrt{5}}{5}$ (or vice versa).

Here is a solution which is obviously less elegant than the direct approaches given above, but which I found pretty straightforward (whereas the solutions in #6 and #7 appear as a miracle to me), using Lagrange multipliers: As above, we substitute third powers, so that we have $a+b+c+d=4$ and want to maximize $a^3b^3c^3d^3(a^3+b^3)(b^3+c^3)(c^3+d^3)(d^3+a^3)$. Clearly, the maximum is not achieved at the boundary, so it suffices to consider the derivatives and find the solutions to \begin{align*} 1+\frac{a^3}{d^3+a^3}+\frac{a^3}{a^3+b^3}&=\lambda a\\ 1+\frac{b^3}{a^3+b^3}+\frac{b^3}{b^3+c^3}&=\lambda b\\ 1+\frac{c^3}{b^3+c^3}+\frac{c^3}{c^3+d^3}&=\lambda c\\ 1+\frac{d^3}{c^3+d^3}+\frac{d^3}{d^3+a^3}&=\lambda d \end{align*}Summing up the equations we get that $\lambda=2$. We then also easily get that $\frac{1}{2} \le a,b,c,d \le \frac{3}{2}$. Now consider the function \[f(x)=\frac{x^3}{d^3+x^3}+\frac{x^3}{b^3+x^3}-2x.\]If $f$ is monotonic on an interval which includes $a$ and $c$, it follows that $a=c$, similarly for $b$ and $d$. Computing, we see that this amounts to \[\frac{x^2b^3}{(x^3+b^3)^2}+\frac{x^2d^3}{(x^3+d^3)^2} \le \frac{2}{3}\]in this range. This will certainly be true if $3x^2b^3 \le 2(x^3+b^3)^2$ and the same with $d$, which is certainly true if $x\ge \frac{3}{4}$ by AM-GM. To summarize the discussion so far, if $a,c \ge \frac{3}{4}$, then $a=c$ and similarly for $b,d$. But adding the first two equations, we get that $a+b \ge \frac{3}{2}$ so that either $a,c \ge \frac{3}{4}$ or $b,d \ge \frac{3}{4}$ certainly holds. Let's suppose that $b,d \ge \frac{3}{4}$ so that $b=d$. Moreover, suppose that $a<\frac{3}{4}$. Then the first equation says $\frac{a^3}{a^3+b^3}<\frac{1}{4}$ and hence $b^3>3a^3$. But then the AM-GM argument from above can be improved to hold already for $x>\frac{9}{16}$. So we are done unless also $a<\frac{9}{16}$. But then the first equation becomes $\frac{a^3}{a^3+b^3}<\frac{1}{16}$ and hence $b^3>15a^3$. Repeating this, the AM-GM argument can be improved to already hold for $x>\frac{45}{256}$, which is well beyond the range $x \ge \frac{1}{2}$, that we always have. In any case, we also get $a=c$, and remember that we already have $b=d$. The rest is easy: We have $a+b=2$ and $1+\frac{2a^3}{a^3+b^3}=2a$ from where we get the solutions $a=b=1$ and $a,b=1 \pm \frac{1}{\sqrt{5}}$ leading to the optimal value.

did anyone else find this problem to be overly bashy and not require much insight?