Prove Triangle Inequality: $\frac{a}{\sqrt[3]{4b^3+4c^3}}+...<2$

[anemone]

Let $a,\,b$ and $c$ be the side lengths of a triangle. Prove that $\dfrac{a}{\sqrt[3]{4b^3+4c^3}}+\dfrac{c}{\sqrt[3]{4a^3+4b^3}}+\dfrac{a}{\sqrt[3]{4b^3+4c^3}}<2$.

 

[anemone]

Spoiler: Solution of other

Since $\dfrac{b^3+c^3}{2}\ge \left(\dfrac{b+c}{2}\right)^2$, we have

$\sqrt[3]{4(b^3+c^3)}\ge b+c$.

From $b+c>a$, it follows that $2(b+c)>a+b+c$. Thus

$\dfrac{a}{\sqrt[3]{4(b^3+c^3)}}<\dfrac{a}{b+c}<\dfrac{2a}{a+b+c}$

Therefore

$\displaystyle \sum_{\text{cyclic}}\dfrac{a}{\sqrt[3]{4(b^3+c^3)}}<\sum_{\text{cyclic}}\dfrac{2a}{a+b+c}=2$