Consider the following expression:
$$S = \sqrt[3]{\frac{a}{b+7}}+\sqrt[3]{\frac{b}{c+7}}+\sqrt[3]{\frac{c}{d+7}} +\sqrt[3]{\frac{d}{a+7}}$$
If $a+b+c+d=100$, what is the maximum value of $S$?
Note: $a,b,c,d$ are non-negative reals
I have a strong suspicion that the maximum amount is $4$, but I don't know how to prove it. It is easy to find values of $a,b,c,d$ such that $S=4$, but it is apparently very difficult to prove that $S\leq 4$
I tried using inequality of means, but I got nothing.For the cube roots I used the geometric mean, I got some inequalities, but they don't help much, for example:
$$\sqrt[3]{\frac{a}{b+7}} \leq \frac{1+a+\frac{1}{b+7}}{3}$$
I also tried to make some variable changes, like for example:
$$x =b+7, \quad y = c+7, \quad z=d+7, \quad k=a+7$$
Does anyone have a solution for this problem?
Solution from my students Lior Hadassi and Shvo Regavim:
By using Holder and AM-GM we obtain: $$\sum_{cyc}\sqrt[3]{\tfrac{a}{b+7}}=\sum_{cyc}\sqrt[3]{\tfrac{a}{a+7}\cdot\tfrac{1}{b+7}\cdot(a+7)}\leq\sqrt[3]{\sum_{cyc}\tfrac{a}{a+7}\sum_{cyc}\tfrac{1}{b+7}\sum_{cyc}(a+7)}=$$ $$=\sqrt[3]{128\left(4-7\sum_{cyc}\tfrac{1}{a+7}\right)\sum_{cyc}\tfrac{1}{a+7}}=\sqrt[3]{128\cdot7\left(\tfrac{4}{7}-\sum_{cyc}\tfrac{1}{a+7}\right)\sum_{cyc}\tfrac{1}{a+7}}\leq$$ $$\leq\sqrt[3]{128\cdot7\cdot\left(\tfrac{2}{7}\right)^2}=\tfrac{8}{\sqrt[3]7}.$$ The equality occurs for $(a,b,c,d)=(49,1,49,1),$ which says that we got a maximal value.