Alternative approaches to prove the following inequality 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.

We can assume that $a+b+c = 1$ and so with $$ f(x) = \left(\frac{2x}{1-x}\right)^{2/3}$$ the inequality becomes $$ f(a) + f(b) + f(c) \geq 3.$$ Now observe that $f(x) \geq 3x$ for $x \in [0,1]$.
