Prove that for all $a,b,c \in \mathbb{R}^+$, $\sqrt[3]{\dfrac ab} + \sqrt[5]{\dfrac bc} + \sqrt[7]{\dfrac ca} > \dfrac 52$.
My thought process along with the proof: Since we're dealing with positive real numbers, I decided to use the AM-GM inequality in a way which lets us get rid of the exponents in the product. That is: $$\sqrt[3]{\dfrac ab} + \sqrt[5]{\dfrac bc} + \sqrt[7]{\dfrac ca} = \sum^3 \dfrac 13\sqrt[3]{\dfrac ab} + \sum^5 \dfrac 15\sqrt[5]{\dfrac bc} + \sum^7 \dfrac 17\sqrt[7]{\dfrac ca},$$ and we apply the AM-GM inequality to get that $$\sum^3 \dfrac 13\sqrt[3]{\dfrac ab} + \sum^5 \dfrac 15\sqrt[5]{\dfrac bc} + \sum^7 \dfrac 17\sqrt[7]{\dfrac ca} \ge 15\sqrt[15]{\frac{1}{3^3 5^5 7^7}}.$$ Then one can notice that it suffices to show that $15\sqrt[15]{\frac{1}{3^3 5^5 7^7}} > \frac 52$, and so I tried to manipulate the right-hand side.
Notice that \begin{align*} 15\sqrt[15]{\dfrac{1}{3^3 5^5 7^7}} &> \frac{15}{\sqrt[15]{3^1 5^7 7^7}} \\ &= \frac{15}{\sqrt[15]{3^1 35^7}} \\ &> \frac{15}{\sqrt[15]{3^1 36^7}} \\ &= \frac{15}{\sqrt[15]{3^1 6^{14}}} \\ &> \frac{15}{\sqrt[15]{6^{15}}} \\ &= \frac{15}{6} \\ &= \frac{5}{2}, \end{align*} and so we should be done.