parmenides51 wrote: Let $a, b, c$ be real numbers such that $0 < a, b, c < 1/2$ and $a + b + c= 1$. Prove that for all real numbers $x,y,z$, $$abc(x + y + z)^2 > ayz( 1- 2a) + bxz( 1 - 2b) + cxy( 1 - 2c)$$.When does equality hold? Maybe Let $a, b, c$ be real numbers such that $0 < a, b, c < 1/2$ and $a + b + c= 1$. Prove that for all real numbers $x,y,z$, $$abc(x + y + z)^2 \geq ayz( 1- 2a) + bxz( 1 - 2b) + cxy( 1 - 2c)$$.When does equality hold?

By symmetry, we may assume that $\frac{x}{a}\geq \frac{y}{b}\geq \frac{z}{c}$. Let $\frac{y}{b}=q$, $\frac{x}{a}=q+\alpha$ and $\frac{z}{c}=q-\beta$, where $\alpha,\beta\geq 0$. Thus $x=a(q+\alpha)$, $y=bq$, $z=c(q-\beta)$. We have \begin{align*} &abc(x+y+z)^2 \geq ayz(1-2a)+bxz(1-2b)+cxy(1-2c) \\ \Leftrightarrow\quad & (aq+a\alpha+bq+cq-c\beta)^2 \geq q(q+\alpha)(1-2c)+q(q-\beta)(1-2a)+(q-\beta)(q+\alpha)(1-2b) \\ \Leftrightarrow\quad & [q+(a\alpha-c\beta)]^2 \geq q^2 + 2q(a\alpha-c\beta)-\alpha\beta (1-2b) \\ \Leftrightarrow \quad & (a\alpha-c\beta)^2 \geq -\alpha\beta(1-2b). \end{align*}Since $\alpha,\beta\geq 0$ and $b\leq 0.5$, the last inequality holds. Also, equality holds iff $x/a=y/b=z/c$.

I was talking to a friend about inequalities that use geometry and thought of this problem, and then I looked it up and the solution I came up with during the contest wasn't here, so I'm writing this out for the sake of posterity.

It maybe isn't shorter than the intended solution (see the post above) but I do think it's easier to motivate.