Mathematical Reflections 2014, Issue 1 - Problem J290

Problem:
Let $a, b, c$ be nonnegative real numbers such that $a + b + c = 1$. Prove that
$$\sqrt[3]{13a^3+14b^3}+\sqrt[3]{13b^3+14c^3}+\sqrt[3]{13c^3+14a^3} \geq 3.$$

Proposed by Titu Andreescu.

Solution:

The second Minkowski's Inequality (see \emph{Zdravko Cvetkovski - Inequalities. Theorems, Techniques and Selected Problems, page 99}) states that if $a_1,a_2,\ldots,a_n$ and $b_1,b_2,\ldots,b_n$ are positive real numbers and $p>1$, then
$$\left(\left(\sum_{i=1}^n a_i\right)^p+\left(\sum_{i=1}^n b_i\right)^p \right)^{1/p} \leq \sum_{i=1}^n (a^p_i+b^p_i)^{1/p}.$$
Equality occurs if and only if $\dfrac{a_1}{b_1}=\dfrac{a_2}{b_2}=\ldots=\dfrac{a_n}{b_n}$.\\
Now, take $$(a_1,a_2,a_3)=(\sqrt[3]{13}a,\sqrt[3]{13}b,\sqrt[3]{13}c),$$ $$(b_1,b_2,b_3)=(\sqrt[3]{14}b, \sqrt[3]{14}c,\sqrt[3]{14}a)$$ and $p=3$. Then,
$$\begin{array}{lll} \sqrt[3]{13a^3+14b^3}+\sqrt[3]{13b^3+14c^3}+\sqrt[3]{13c^3+14a^3} & \geq & \sqrt[3]{\left(\sqrt[3]{13(a+b+c)}\right)^3+\left(\sqrt[3]{14(a+b+c)}\right)^3} \\ & = & \sqrt[3]{(\sqrt[3]{13})^3+(\sqrt[3]{14})^3} \\ &=& 3. \end{array}$$
Equality occurs if and only if $\dfrac{a}{b}=\dfrac{b}{c}=\dfrac{c}{a}$.