$a$, $b$, $c$ are three distinct real numbers, given $\lambda >0$. Proof that
\[\frac{1+ \lambda ^2a^2b^2}{(a-b)^2}+\frac{1+ \lambda ^2b^2c^2}{(b-c)^2}+\frac{1+ \lambda ^2c^2a^2}{(c-a)^2} \geq \frac 32 \lambda.\]
HIDE: Remark Old problem, can be found here. Double post to have a cleaner thread for collection (as the original one contains a messy quote)
Define $x, y, z$ as $1/a, 1/b,$ and $1/c$ respectively. Denote $f_x(t) = \frac{t-yz}{y-z}$. Define $f_y, f_z$ similarly. Consider the polynomial $P(t) = t$. From Lagrange Interpolation on $P$ with points $yz, zx, $ and $xy$ we obtain the identity: \[f_x(t)f_y(t)+f_y(t)f_z(t)+f_z(t)f_x(t) = t\]We now note that \[\frac{1+\lambda^2a^2b^2}{(a-b)^2} = \frac{1}{2}[f_z(\lambda)^2+f_z(-\lambda)^2]\]We now use the key estimate $u^2+v^2+w^2 \geq \text{max}(uv+vw+wu, -2(uv+vw+wu))$ on the (cyclic) expressions $f_z(\lambda)$ and $f_z(-\lambda)$ giving \[\frac{1+ \lambda ^2a^2b^2}{(a-b)^2}+\frac{1+ \lambda ^2b^2c^2}{(b-c)^2}+\frac{1+ \lambda ^2c^2a^2}{(c-a)^2} \geq \frac{1}{2}[\lambda+2\lambda] = \frac{3}{2}\lambda\]Done.