For $a,b,c \in \mathbb{R^+},$ prove that $$\left(\dfrac{2a}{b+c} \right)^{\frac{2}{3}} + \left(\dfrac{2b}{c+a}\right)^{\frac{2}{3}} + \left(\dfrac{2c}{a+b}\right)^{\frac{2}{3}} \geq 3.$$ I managed to prove this problem using the technique of isolated fudging. In particular, one can prove that $\dfrac{1}{3}\left(\frac{2a}{b+c} \right)^{\frac{2}{3}} \geq \dfrac{a}{a+b+c} $ (motivation is explained below) using AM-GM, as follows: \begin{align*} \dfrac{1}{3}\left(\dfrac{2a}{b+c} \right)^{\frac{2}{3}} \geq \dfrac{a}{a+b+c} & \iff \left(\dfrac{2a}{b+c} \right)^{\frac{2}{3}} \geq \dfrac{3a}{a+b+c} \\ & \iff \left(\dfrac{2a}{b+c} \right)^{2} \geq \left(\dfrac{3a}{a+b+c}\right)^3 \\ & \iff 4a^2(a+b+c)^3 \geq 27a^3(b+c)^2. \end{align*} The preceding inequality is homogeneous, hence W.L.O.G. set $a+b+c=3$. It thus suffices for us to prove that $4 \geq a(3-a)^2 \iff 8 \geq 2a(3-a)^2$. But this is obvious from AM-GM: $2a(3-a)^2 \leq \left(\dfrac{2a+(3-a)+(3-a)}{3}\right)^3=8.$ Similarly, we have $\dfrac{1}{3}\left(\dfrac{2b}{a+c} \right)^{\frac{2}{3}} \geq \dfrac{b}{a+b+c} $ and $\dfrac{1}{3}\left(\dfrac{2c}{a+b} \right)^{\frac{2}{3}} \geq \dfrac{c}{a+b+c} $. Thus, summing cyclically, we obtain: $$\dfrac{1}{3}\left(\dfrac{2a}{b+c} \right)^{\frac{2}{3}} + \dfrac{1}{3}\left(\dfrac{2b}{a+c} \right)^{\frac{2}{3}} + \dfrac{1}{3}\left(\dfrac{2c}{a+b} \right)^{\frac{2}{3}} \geq 1 \Rightarrow \left(\dfrac{2a}{b+c} \right)^{\frac{2}{3}} + \left(\dfrac{2b}{c+a}\right)^{\frac{2}{3}} + \left(\dfrac{2c}{a+b}\right)^{\frac{2}{3}} \geq 3 .$$ And we are done.
Some motivation:
Let $f(a,b,c)=\dfrac{1}{3}\left(\dfrac{2a}{b+c} \right)^{\frac{2}{3}} - \dfrac{a^r}{a^r+b^r+c^r}$, for some $r \in \mathbb{R}$. Ideally, we want $f(a,b,c) \geq 0$ for all values of $a,b,c$. Note that $f(1,1,1)=0$, which suggests that we should set $(1,1,1)$ as a local minimum of $f$. Hence, take the partial derivative of $f$ with respect to $a$, and set it to zero at $(1,1,1)$. By solving the resulting equation, we find a corresponding value of $r$:
$$\dfrac{\partial f}{\partial a}= \dfrac{1}{3} \cdot \sqrt[^3]{4} \cdot \left(\dfrac{1}{b+c} \right)^{\frac{2}{3}} \cdot \dfrac{2}{3} \cdot a^{\frac{-1}{3}} - \dfrac{ra^{r-1}(a^r+b^r+c^r)-a^r(ra^{r-1})}{(a^r+b^r+c^r)^2}$$ Hence, $$\frac{\partial f}{\partial a }\Bigr|_{(1,1,1)} =0 \Rightarrow \dfrac{2}{9}- \dfrac{2r}{9} =0 \Rightarrow r=1.$$
My question is, is there any other way to solve this inequality besides isolated fudging? I.e. would methods such as Cauchy / Holder / Jensen also work here? I would love to see any other alternative approach.
Let $p=a+b+c,\,q=ab+bc+ca,\,r=abc.$ Using the Holder inequality, we have $$\left[\sum \left(\dfrac{2a}{b+c} \right)^{\frac{2}{3}}\right]^3 \sum a^2(b+c)^2 \geqslant 4(a+b+c)^4.$$ It remain to show that $$4(a+b+c)^4 \geqslant 27 \sum a^2(b+c)^2,$$ or $$2p^4+27pr \geqslant 27q^2.$$ If $p^2 > 9q,$ then $$2p^4 > 2(9q)^2 > 27q^2.$$ If $p^2 \leqslant 9q,$ by the Schur inequality, we have $$r \ge \frac{4pq-p^3}{9}.$$ We will show that $$2p^4+3p^2(4q-p^2) \geqslant 27q^2,$$ equivalent to $$(9q - p^2)(p^2-3q) \geqslant 0.$$ This is true. The proof is completed.