SinaQane wrote: For positive real numbers $a$, $b$, and $c$ with $a+b+c=1$, prove that: $$ (a-b)^2 + (b-c)^2 + (c-a)^2 \geq \frac{1-27abc}{2}. $$ By well-know identity: $$a^3+b^3+c^3-3abc=(a+b+c)(a^2+b^2+c^2-ab-bc-ca)$$We have: $$a^3+b^3+c^3-3abc=(a^2+b^2+c^2-ab-bc-ca)$$Then: $$(a-b)^2 + (b-c)^2 + (c-a)^2=2(a^2+b^2+c^2-ab-bc-ca)=2(a^3+b^3+c^3-3abc)\ge \frac{1-27abc}{2}$$ P.S. watch @below to see explanation of last step

CROWmatician wrote: SinaQane wrote: For positive real numbers $a$, $b$, and $c$ with $a+b+c=1$, prove that: $$ (a-b)^2 + (b-c)^2 + (c-a)^2 \geq \frac{1-27abc}{2}. $$ By well-know identity: $$a^3+b^3+c^3-3abc=(a+b+c)(a^2+b^2+c^2-ab-bc-ca)$$We have: $$a^3+b^3+c^3-3abc=(a^2+b^2+c^2-ab-bc-ca)$$Then: $$(a-b)^2 + (b-c)^2 + (c-a)^2=2(a^2+b^2+c^2-ab-bc-ca)=2(a^3+b^3+c^3-3abc)\ge 6-6abc\ge \frac{1-27abc}{2}$$ P.S. You can prove that $6-6abc\ge \frac{1-27abc}{2}$ , by moving terms and noting that $15abc>-11$. $2(a^3+b^3+c^3-3abc)\ge 6-6abc$ isn't necessarily true. For example, consider $(\frac{1}{3}, \frac{1}{3}, \frac{1}{3})$. In general, for Olympiads if you get a very large bound (in your solution, $15abc>-11$ is a seriously generous bound), there's a good chance your solution is wrong somewhere. Edit: Anyways, here's my finish from @above's solution. We have that \begin{align*}& 2(a^3+b^3+c^3-3abc)\geq \frac{1-27abc}{2} \\ \iff & 4a^3+4b^3+4c^3\geq 1-15abc \\ \iff & 4a^3+4b^3+4c^3+15abc\geq (a+b+c)^3 \\ \iff & a^3+b^3+c^3+3abc\geq \sum_{cyc}a^2b \end{align*} This is just 1st degree Schur's, so we have $2(a^3+b^3+c^3-3abc)\geq \frac{1-27abc}{2}$ which finishes the result from above.