Let $a, b, c, m, n, p, k$ be positive real numbers that satisfy $a+m = b+n = c+p=k$. Show that $an+bp+cm < k^2$.
The given condition is equivalent to $$\frac{a}{k}+\frac{m}{k}=\frac{b}{k}+\frac{n}{k}=\frac{c}{k}+\frac{p}{k}=1$$
Then all of these fraction numbers between $0$ and $1$. Let $P(A) = \frac{a}{k}$, $P(B) = \frac{b}{k}$, and $P(C) = \frac{c}{k}$ where $A$, $B$, and $C$ are three indepedent events. Then we have $$\begin{align}&P(A+B+C)\\=&P(A)+P(B)+P(C)-P(AB)-P(BC)-P(CA)+P(ABC)\\=&\frac{a}{k}+\frac{b}{k}+\frac{c}{k}-\frac{ab}{k^2}-\frac{bc}{k^2}-\frac{ca}{k^2}+\frac{abc}{k^3}\\ > & \left(\frac{a}{k}-\frac{ab}{k^2}\right)+\left(\frac{b}{k}-\frac{bc}{k^2}\right)+\left(\frac{c}{k}-\frac{ca}{k^2}\right)\\=& \frac{a}{k}\left(1-\frac{b}{k}\right)+\frac{b}{k}\left(1-\frac{c}{k}\right)+\frac{c}{k}\left(1-\frac{a}{k}\right)\\=&\frac{a}{k}\cdot\frac{n}{k}+\frac{b}{k}\cdot\frac{p}{k}+\frac{c}{k}\cdot\frac{m}{k}\end{align}$$
However, we know $P(A+B+C)\le 1$ which means $$\frac{a}{k}\cdot\frac{n}{k}+\frac{b}{k}\cdot\frac{p}{k}+\frac{c}{k}\cdot\frac{m}{k}\le 1\implies an + b + cm < k^2$$