Cho a, b, c là các số thực dương thỏa mãn: ab + bc + ca = 3abc.

* Hướng dẫn giải

Ta có: $$\begin{gathered} K=\frac{a^2}{c\left(c^2+a^2\right)}+\frac{b^2}{a\left(a^2+b^2\right)}+\frac{c^2}{b\left(b^2+c^2\right)} \\ =\frac{a^2+c^2-c^2}{c(c^2+a^2)}+\frac{b^2+a^2-a^2}{a\left(a^2+b^2\right)}+\frac{c^2+b^2-b^2}{b\left(b^2+c^2\right)} \\ =\left(\frac{1}{c}-\frac{c}{c^2+a^2}\right)+\left(\frac{1}{a}-\frac{a}{a^2+b^2}\right)+\left(\frac{1}{b}-\frac{b}{b^2+c^2}\right) \\ {\ge }^{Cosi}\left(\frac{1}{c}-\frac{c}{2\sqrt{c^2{a}^2}}\right)+\left(\frac{1}{a}-\frac{a}{2\sqrt{a^2{b}^2}}\right)+\left(\frac{1}{b}-\frac{b}{2\sqrt{b^2{c}^2}}\right) \\ =\left(\frac{1}{c}-\frac{1}{2a}\right)+\left(\frac{1}{a}-\frac{1}{2b}\right)+\left(\frac{1}{b}-\frac{1}{2c}\right) \\ =\frac{1}{2}\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)=\frac{ab+bc+ac}{2abc}=\frac{3}{2} \\ \Rightarrow K\ge \frac{3}{2} \end{gathered}$$

Vậy Min K = $\frac{3}{2}$ khi a = b = c.

Mà ab + bc + ac = 3abc

$\Rightarrow a=b=c=1$.