Tuymaada 2022 Senior/4 wrote: For every positive $a_1, a_2, \dots, a_6$, prove the inequality \[ \sqrt[4]{\frac{a_1}{a_2 + a_3 + a_4}} + \sqrt[4]{\frac{a_2}{a_3 + a_4 + a_5}} + \dots + \sqrt[4]{\frac{a_6}{a_1 + a_2 + a_3}} \ge 2 \] This is a surprisingly weak inequality. Wishful thinking gives us the following claim. Claim 01. Let us consider the indices modulo $6$. For any $a_1, \dots, a_6$, we have \[ \sqrt[4]{\frac{a_i}{a_{i + 1} + a_{i + 2} + a_{i + 3}}} > \frac{2 \sqrt{a_i}}{\sqrt{a_i} + \sqrt{a_{i + 1}} + \sqrt{a_{i + 2}} + \sqrt{a_{i + 3}}} > \frac{2\sqrt{a_i}}{\sqrt{a_1} + \dots + \sqrt{a_6}} \]Proof. The last inequality is clearly obvious. It suffices to prove that for any $a,b,c,d \in \mathbb{R}^+$, we have \[ \sqrt[4]{\frac{a^2}{b^2 + c^2 + d^2}} > \frac{2a}{a + b + c + d} \]However, \begin{align*} \sqrt[4]{\frac{a^2}{b^2 + c^2 + d^2}} &> \sqrt[4]{\frac{a^2}{(b + c + d)^2}} = \sqrt{\frac{a}{b + c + d}} > \frac{a}{\sqrt{a(b + c + d)}} \stackrel{\text{AM-GM}}{\ge} \frac{2a}{a + b + c + d} \end{align*}By using the above claim, we thus conclude that \[ \sum_{cyc} \sqrt[4]{\frac{a_i}{a_{i + 1} + a_{i + 2} + a_{i + 3}}} > 2 \]as required.

IndoMathXdZ wrote: For every positive $a_1, a_2, \dots, a_6$, prove the inequality \[ \sqrt[4]{\frac{a_1}{a_2 + a_3 + a_4}} + \sqrt[4]{\frac{a_2}{a_3 + a_4 + a_5}} + \dots + \sqrt[4]{\frac{a_6}{a_1 + a_2 + a_3}} \ge 2 \] It seems that $$\sum_{cyc}\sqrt[4]{\frac{a_1}{a_2 + a_3 + a_4}}\geq3$$is also true. No! $a_2=a_5=a_6\rightarrow0^+$ and $a_1=a_3=a_4=1$.