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.

• Nicely done though instead of doing all that i woud do $$3{(\frac{b+c}{2a})}^{2/3}\le 2\cdot \frac{b+c}{2a}+1=\frac{a+b+c}{a}$$ by AM-GM Jan 23, 2021 at 16:00
• Oooh, nice observation of looking at the reciprocal! Jan 23, 2021 at 16:03

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.
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 $$x \in [0,1]$$.
• This method is more or less the same approach taken by OP except that you added $a+b+c=1$ Jan 23, 2021 at 15:48
• How is it obvious that $f(x) \geq 3x$ for $x \in [0,1]$? $\left(\frac{2x}{1-x}\right)^{\frac{2}{3}} \geq 3x \iff 4 \geq 27x(1-x)^2$, but this isn't obvious at all. Jan 23, 2021 at 15:56
• I didn't say obvious. :) If found that by looking for a coefficient $c$ such that the line $y = c x$ touches the graph of $f$.
• @SeeHai $$4\ge 27x({(1-x)}^2$$ $$\iff -(3x-4){(3x-1)}^2 \ge 0$$ whcih is obvious Jan 23, 2021 at 16:15