İn BMO 2014 First problem was easy. But was not $A1$. $A1$ is this and very easy. I am waiting other solutions.

If you know BMO Shortlist can you please post the full shortlist? It would be a delight to have it in contest collection.

Seriously? Is it the Balkan MO I know? By QM-AM-GM, $\sqrt{\frac{a^2+b^2+c^2}{3}}\geq\frac13\geq\sqrt[3]{abc}$. Hence, $a^2+b^2+c^2\geq\frac13$ and $abc\leq\frac{1}{27}$. That's it, the inequality follows

Wow this is very easy: By cauchy we have $(a^2+b^2+c^2)(3)\ge 1$. Then we can manipulate this into $2(a^2+b^2+c^2)\ge 2/3$. Also by AM-GM we have $5/9\ge 15abc$. We add this inequality with the one preceding it to yield $2(a^2+b^2+c^2)\ge 15abc+1/9$ which is our desired result.

By Cauchy, we have $3(a^2+b^2+c^2) \geq 1$, so $a^2+b^2+c^2 \geq \frac{1}{3}$. By AM-GM, we have $abc \leq \frac{1}{27}$, so the minimum value of the LHS is $\frac{2}{3}$ and the maximum value of the RHS is $\frac{1}{9}+\frac{15}{27} = \frac{2}{3}$, implying the desired inequality. Equality is obtained at $a=b=c=\frac{1}{3}$.

$2(a^2+b^2+c^2)\ge \frac{1}{9}+15abc$ By well-known lemma $2(a^2+b^2+c^2) \ge \frac{6}{9}$ so it is equal to prove $1 \ge 27abc$ which is $(a+b+c)^3 \ge 27 abc$

The a b + c and a², b², c² make it look like something was squared but the abc kinda points you more in the way of the HM-GM-AM-QM inequality. Someone above me has already shown that so this'll just be the kinda intuition behind that.

Bruh QM-AM-GM kills this problem… This seems like a level 13 intermediate algebra problem

tenplusten wrote: $\boxed{\text{A1}}$Let $a,b,c$ be positive reals numbers such that $a+b+c=1$.Prove that $2(a^2+b^2+c^2)\ge \frac{1}{9}+15abc$ Why was this in the SL? $$2(a^2+b^2+c^2)\geq\frac{2}{3}(a+b+c)^2= \frac{2}{3}= \frac{1}{9}+\frac{5}{9}(a+b+c)^3\geq \frac{1}{9}+\frac{5}{9}\cdot27abc=\frac{1}{9}+15abc$$Equality for $a=b=c=\cfrac{1}{3}$.

the problem \(\iff2\sum a^2 \sum a \geq \frac{1}{9} \left(\sum a\right)^3 + 15abc\) \(\iff17\sum a^3 + 15\sum a^2b \geq 141abc\), which is true by AM-GM