Vietnamese Mathematical Olympiad

Let $a\to 0^{+}$ and $b=c=1$, one can get $$2+\frac{1}{k}\ge \frac{k+3}{2}$$ so $k\le 2$. For $k=2$, we need to prove that $$\frac{1}{2ab+c^2}+\frac{1}{2bc+a^2}+\frac{1}{2ca+b^2}\ge \frac{5}{a^2+b^2+c^2}$$ is true for all positive triples $(a,b,c)$ satisfying $a^2+b^2+c^2=2(ab+bc+ca)$. First, we will prove that $$\frac{1}{a^2+2bc}+\frac{1}{b^2+2ac}+\frac{1}{c^2+2ab}\ge \frac{2}{ab+bc+ac}+\frac{1}{a^2+b^2+c^2}$$ true for all positive real numbers $a,b,c$. Indeed, the above inequality can be rewritten as $$\frac{a^2+b^2+c^2}{a^2+2bc}+\frac{a^2+b^2+c^2}{b^2+2ac}+\frac{a^2+b^2+c^2}{c^2+2ab}\ge \frac{2(a^2+b^2+c^2)}{ab+bc+ca}+1,$$ thus $$\frac{(b-c{)}^2}{a^2+2bc}+\frac{(c-a{)}^2}{b^2+2ca}+\frac{(a-b{)}^2}{c^2+2ab}\ge \frac{(a-b{)}^2+(b-c{)}^2+(c-a{)}^2}{ab+bc+ac}.$$ If $(a-b)(b-c)(c-a)=0$ then the above inequality is true. Now consider the case $(a-b)(b-c)(c-a)\ne 0$, then $$\begin{array}{rl}LHS& \ge \frac{{\left[\sum (b-c{)}^2\right]}^2}{\sum (a^2+2bc)(b-c{)}^2}=\frac{{\left[\sum (b-c{)}^2\right]}^2}{(ab+bc+ca)\left[\sum (b-c{)}^2\right]}\\ & =\frac{(a-b{)}^2+(b-c{)}^2+(c-a{)}^2}{ab+bc+ac}=RHS.\end{array}$$ Back to the original problem, applying the above inequality, combined with $a^2+b^2+c^2=2(ab+bc+ca)$, we get $$\begin{array}{rl}\frac{1}{a^2+2bc}+\frac{1}{b^2+2ac}+\frac{1}{c^2+2ab}& \ge \frac{2}{ab+bc+ac}+\frac{1}{a^2+b^2+c^2}\\ & =\frac{5}{a^2+b^2+c^2}.\end{array}$$ So the maximum positive real number $k$ is 2. $\square$