$ a^3/(a^2+b^2) + b^3/(b^2+c^2) + c^3/(c^2+a^2) \ge \frac{1}{2}$ is equivalent to $\sum \frac{ab^2}{a^2+b^2} \le \sum \frac{ab^2}{2ab} \le \sum \frac{b}{2}=\frac{1}{2}$

vralex wrote: How do I prove that, for every $a, b, c$ positive real numbers such that $a+b+c = 1$ the following inequality holds: $\frac{a^3}{a^2+b^2} +\frac{b^3}{b^2+c^2} +\frac {c^3}{c^2+a^2} \geq \frac{1}{2}$? What year is it？ Thanks.

sqing wrote: vralex wrote: How do I prove that, for every $a, b, c$ positive real numbers such that $a+b+c = 1$ the following inequality holds: $\frac{a^3}{a^2+b^2} +\frac{b^3}{b^2+c^2} +\frac {c^3}{c^2+a^2} \geq \frac{1}{2}$? What year is it？ Thanks. 2019

Let $n \in \mathbb{N}, n \ge 2$ and the positive real numbers $a_1,a_2,…,a_n.$ Prove that $$\sum\limits_{k=1}^n \frac{a_k^2}{a_k+b_k} \ge \frac{a_1+a_2+…+a_n}{2}.$$

Similar: Let $a, b, c >0$ be reals such that $ab + bc + ca = 1$. Prove that $ \sum_{cyc} \frac{b^3c}{a^2+b^2} \geq \frac{1}{2}$.

vralex wrote: Similar: Let $a, b, c >0$ be reals such that $ab + bc + ca = 1$. Prove that $ \sum_{cyc} \frac{b^3c}{a^2+b^2} \geq \frac{1}{2}$. See also here https://artofproblemsolving.com/community/c6h463728p2599550

vralex wrote: How do I prove that, for every $a, b, c$ positive real numbers such that $a+b+c = 1$ the following inequality holds: $$\frac{a^3}{a^2+b^2} +\frac{b^3}{b^2+c^2} +\frac {c^3}{c^2+a^2} \geq \frac{1}{2}$$? we have: $$ \frac{a^3}{a^2+b^2} =a-\frac{ab^2}{a^2+b^2} \ge a-\frac{b}{2}, ...$$

Credits for the proof: Kwok Choy Yue via Quora.

Let $ a,b,c$ be positive real numbers . Prove that $$ \frac {a^3}{ka^2 + b^2} + \frac {b^3}{kb^2 + c^2} + \frac {c^3}{kc^2 + a^2}\geqslant \frac {a + b + c}{k+1}$$Where $k>0.$

sqing wrote: Let $ a,b,c$ be positive real numbers . Prove that $$ \frac {a^3}{ka^2 + b^2} + \frac {b^3}{kb^2 + c^2} + \frac {c^3}{kc^2 + a^2}\geqslant \frac {a + b + c}{k+1}$$Where $k>1.$ Using weighted A.M.-G.M. inequality twice, \begin{align*} & \sum_{cyc} \frac {a^3}{ka^2 + b^2}= \sum_{cyc} \frac{1}{k}\left(a-\frac{ab^2}{ka^2+b^2} \right) \\& \\& \geq \sum_{cyc} \enspace \frac{1}{k} \left(a-\frac{ab^2}{(k+1) a^{\frac{2k}{k+1}}b^{\frac{2}{k+1}}} \right) \\& \\& = \sum_{cyc} \enspace \frac{1}{k}\left(a-\frac{a^{\frac{1-k}{1+k}} b^{\frac{2k}{k+1}}}{k+1}\right) \\& \\& \geq \frac{1}{k} \enspace \sum_{cyc} \left(a - \frac{ (1-k)a+2kb }{(k+1)^2}\right) \\& \\& = \frac{1}{k} \enspace \sum_{cyc} \left( \frac{ (k^2+3k)a-2kb}{(1+k)^2} \right) \\& \\& = \frac{1}{k} \enspace \sum_{cyc} \left( \frac{ (k^2+k)a}{(1+k)^2} \right) \\& \\& = \sum_{cyc} \enspace \frac{a}{k+1} \\& \\& =\frac{a+b+c}{k+1} \end{align*}

An easier solution is to say WLOG, $a^3+b^3+c^3=1$ and then use Nesbitt's inequality.