Any answer?

CSMO 2021 Grade 11/3 wrote: Let $a,b,c\geq 0$ and $a^2+b^2+c^2\leq 1.$ Prove that$$\frac{a}{a^2+bc+1}+\frac{b}{b^2+ca+1}+\frac{c}{c^2+ab+1}+3abc<\sqrt 3$$ A really weak inequality. Claim 01. For any nonnegative $a,b,c$, we have \[ \frac{a}{a^2 + bc + 1} + \frac{b}{b^2 + ca + 1} + \frac{c}{c^2 + ab + 1} + 3abc \le a + b + c \]Proof. Note that \begin{align*} (a + b + c) - \left( \frac{a}{a^2 + bc + 1} + \frac{b}{b^2 + ca + 1} + \frac{c}{c^2 + ab + 1} \right) &= \sum \frac{a(a^2 + bc)}{a^2 + bc + 1} \\ &\ge \sum \frac{a(a^2 + bc)}{a^2 + b^2 + c^2 + 1} \\ &\ge \frac{a^3 + b^3 + c^3 + 3abc}{2} \\ &\ge 3abc \end{align*}We are done, because, $a + b + c \le \sqrt{3(a^2 + b^2 + c^2)} \le \sqrt{3}$. Equality cannot hold because the last line forces equality if and only if $a = b = c = 1$, which is not the case.

sqing wrote: Let $a,b,c\geq 0$ and $a^2+b^2+c^2\leq 1.$ Prove that$$\frac{a}{a^2+bc+1}+\frac{b}{b^2+ca+1}+\frac{c}{c^2+ab+1}+3abc<\sqrt 3$$ Let $a,b,c\geq 0.$ Prove that$$\frac{a}{a^2+bc+1}+\frac{b}{b^2+ca+1}+\frac{c}{c^2+ab+1}+\frac{18}{5}abc\leq a+b+c.$$

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Let $a,b,c\geq 0$ and $a^2+b^2+c^2\leq 1.$ Prove that$$\frac{a}{a^2+bc+1}+\frac{b}{b^2+ca+1}+\frac{c}{c^2+ab+1}+3abc\le\frac{14\sqrt3}{15} $$

sqing wrote: Let $a,b,c\geq 0$ and $a^2+b^2+c^2\leq 1.$ Prove that$$\frac{a}{a^2+bc+1}+\frac{b}{b^2+ca+1}+\frac{c}{c^2+ab+1}+3abc\le\frac{14\sqrt3}{15} $$ WOW Sorry for not seeing this post. I searched for this year's CSMO but found nothing so I posted the problem.>_<

A stupid problem, I think.

minecraftfaq wrote: A stupid problem, I think. The competition itself is not a hard one so you cannot expect CSMO to give you extremely difficult problems. But the problem this year for Grade11 is indeed way too easy.

Hoto_Mukai wrote: sqing wrote: Let $a,b,c\geq 0$ and $a^2+b^2+c^2\leq 1.$ Prove that$$\frac{a}{a^2+bc+1}+\frac{b}{b^2+ca+1}+\frac{c}{c^2+ab+1}+3abc\le\frac{14\sqrt3}{15} $$ WOW Sorry for not seeing this post. I searched for this year's CSMO but found nothing so I posted the problem.>_< 2021 South East Mathematical Olympiad

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IndoMathXdZ wrote: CSMO 2021 Grade 11/3 wrote: Let $a,b,c\geq 0$ and $a^2+b^2+c^2\leq 1.$ Prove that$$\frac{a}{a^2+bc+1}+\frac{b}{b^2+ca+1}+\frac{c}{c^2+ab+1}+3abc<\sqrt 3$$ A really weak inequality. Claim 01. For any nonnegative $a,b,c$, we have \[ \frac{a}{a^2 + bc + 1} + \frac{b}{b^2 + ca + 1} + \frac{c}{c^2 + ab + 1} + 3abc \le a + b + c \]Proof. Note that \begin{align*} (a + b + c) - \left( \frac{a}{a^2 + bc + 1} + \frac{b}{b^2 + ca + 1} + \frac{c}{c^2 + ab + 1} \right) &= \sum \frac{a(a^2 + bc)}{a^2 + bc + 1} \\ &\ge \sum \frac{a(a^2 + bc)}{a^2 + b^2 + c^2 + 1} \\ &\ge \frac{a^3 + b^3 + c^3 + 3abc}{2} \\ &\ge 3abc \end{align*}We are done, because, $a + b + c \le \sqrt{3(a^2 + b^2 + c^2)} \le \sqrt{3}$. Equality cannot hold because the last line forces equality if and only if $a = b = c = 1$, which is not the case. I think if you use the Cauchy，the result will be better.You can get 3.6 instead.