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I was trying to solve this inequality, but I wasn't able to do so:

Find the maximum number $k\in\mathbb R$ such that:

$$ \frac{a}{1+9bc+k(b-c)^2}+\frac{b}{1+9ca+k(c-a)^2}+\frac{c}{1+9ab+k(a-b)^2}\geq\frac{1}{2} $$

Holds for all $a,b,c\ge0$ with $a+b+c=1$

Any help is highly appreciated.

Redundant Aunt
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  • Note that at $a=b=c=\frac 13$ the inequality holds for every value of $k\in\Bbb R$. Another point of interest is $a=0,b=c=\frac12$ which means $k\le 4$. – abiessu Oct 07 '14 at 12:55

1 Answers1

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Plugging, $a=1/2,b=1/2,c=0 \implies k \le 4$. I will show that $k=4$. $$ \frac{a}{1+9bc+4(b-c)^2}+\frac{b}{1+9ca+4(c-a)^2}+\frac{c}{1+9ab+4(a-b)^2}\ge \frac{1}{2} \\ \Longleftrightarrow \frac{a^2}{a+9abc+4a(b-c)^2}+\frac{b^2}{b+9abc+4b(c-a)^2}+\frac{c^2}{c+9abc+4c(a-b)^2} \ge \frac{1}{2}$$ By cauchy it suffice to show that, $$ \frac{(a+b+c)^2}{a+b+c+3abc+4(a^2b+b^2c+c^2a+ab^2+bc^2+ca^2)} \ge \frac{1}{2} \\ \Longleftrightarrow \frac{1}{1+3abc+4(a^2b+b^2c+c^2a+ab^2+bc^2+ca^2)} \ge \frac{1}{2} \\ \Longleftrightarrow 1 \ge 3abc+4a^2b+b^2c+c^2a+ab^2+bc^2+ca^2 \\ \Longleftrightarrow (a+b+c)^3 \ge 3abc+a^2b+b^2c+c^2a+ab^2+bc^2+ca^2 \\ \Longleftrightarrow a(a-b)(a-c)+b(b-c)(b-a)+c(c-a)(c-b) \ge 0 $$

Last inequality is true by Schur. $\Box$

Shivang jindal
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