Problem: Let $a$, $b$, $c$ be positive real numbers.
Prove that
\[ \frac{(2a+b+c)^2}{2a^2+(b+c)^2}
+ \frac{(2b+c+a)^2}{2b^2+(c+a)^2}
+ \frac{(2c+a+b)^2}{2c^2+(a+b)^2} \le 8.  \]

Solution: This is a canonical example of tangent line trick.
Homogenize so that $a + b + c = 3$.
The desired inequality reads
\[ \sum_{\text{cyc}} \frac{(a+3)^2}{2a^2+(3-a)^2} \le 8. \]
This follows from
\[ f(x) = \frac{(x+3)^2}{2x^2+(3-x)^2}
  \le \frac{1}{3} (4x + 4) \]
which can be checked as
$\frac 13 (4x+4)(2x^2+(3-x)^2) - (x+3)^2
= (x-1)^2 (4x+3) \ge 0$.