Problem: For $a,b,c > 0$ prove that
\[ \frac{a^3+3b^3}{5a+b}
  + \frac{b^3+3c^3}{5b+c}
  + \frac{c^3+3a^3}{5c+a}
  \ge \frac23(a^2+b^2+c^2).  \]

Solution: Here are two possible approaches.

\paragraph{Cauchy-Schwarz approach.}
Apply Titu lemma to get
\[ \sum_{\text{cyc}} \frac{a^3}{5a+b}
  = \sum_{\text{cyc}} \frac{a^4}{5a^2+ab}
  \ge \frac{(a^2+b^2+c^2)^2}{\sum_{\text{cyc}} (5a^2+ab)}
  \ge \frac{a^2+b^2+c^2}{6} \]
where the last step follows
from the identity $\sum_{\text{cyc}} (5a^2+ab) \le 6(a^2+b^2+c^2)$.

Similarly,
\[ \sum_{\text{cyc}} \frac{b^3}{5a+b}
  = \sum_{\text{cyc}} \frac{b^4}{5ab+b^2}
  \ge \frac{(a^2+b^2+c^2)^2}{\sum_{\text{cyc}} (5ab+b^2)}
  \ge \frac{a^2+b^2+c^2}{6} \]
using the fact that $\sum_{\text{cyc}} 5ab+b^2 \le 6(a^2+b^2+c^2)$.

Therefore, adding the first display to
three times the second display implies the result.

\paragraph{Second linearization approach.}
The main magical claim is:
\begin{claim*}
  We have
  \[ \frac{a^3+3b^3}{5a+b} \geq \frac{25}{36}b^2 - \frac{1}{36}a^2. \]
\end{claim*}
\begin{proof}
  Let $x=a/b > 0$. The desired inequality is equivalent to
  \[ \frac{x^3+3}{5x+1} \ge \frac{25-x^2}{36}. \]
  However,
  \begin{align*}
    36(x^3+3) - (5x+1)(25-x^2)
    &= 41x^3 + x^2 - 125x + 83 \\
    &= (x-1)^2(41x+83) \ge 0. \qedhere
  \end{align*}
\end{proof}
Sum the claim cyclically to finish.
\begin{remark*}
  [Derivation of the main claim]
  The overall strategy is to hope for a constant $k$ such that
  \[ \frac{a^3+3b^3}{5a+b} \geq ka^2 + \left(\frac{2}{3}-k\right)b^2. \]
  is true.
  Letting $x=a/b$ as above and expanding,
  we need a value $k$ such that the cubic polynomial
  \[ P(x) \coloneq (x^3+3)-(5x+1)\left( kx^2 + \left( \frac 23 - k \right) \right)
    = (1-5k)x^3 - kx^2 + \left( 5k-\frac{10}{3} \right)x + \left( k + \frac{7}{3} \right) \]
  is nonnegative everywhere.
  Since $P(1) = 0$ necessarily, in order for $P(1-\eps)$ and $P(1+\eps)$
  to both be nonnegative (for small $\eps$),
  the polynomial $P$ must have a double root at $1$, meaning
  the first derivative $P'(1) = 0$ needs to vanish.
  In other words, we need
  \[ 3(1-5k) - 2k + \left( 5k - \frac{10}{3} \right) = 0. \]
  Solving gives $k = -1/36$.
  One then factors out the repeated root $(x-1)^2$ from the resulting $P$.
\end{remark*}