Cho $x,y,z>0$  và $xy+yz+xz=3xyz.$  Tính giá trị nhỏ nhất của: $A=\frac{x^2}{z\left(z^2+x^2\right)}+\frac{y^2}{x\left(x^2+y^2\right)}+\frac{z^2}{y\left(y^2+z^2\right)}$

Phương pháp:

- Chia cả hai vế của đẳng thức đã cho cho $xyz.$

- Đặt $a=\frac{1}{x},b=\frac{1}{y},c=\frac{1}{z}$  đưa về tìm GTNN theo $a,b,c.$

- Sử dụng bất đẳng thức $a^2+b^2\ge 2ab.$

Cách giải:

Ta có: $xy+yz+zx=3xyz$

Chia cả hai vế cho $xyz\ne 0$  ta được:$\frac{1}{x}+\frac{1}{y}+\frac{1}{z}=3.$

Đặt $a=\frac{1}{x},b=\frac{1}{y},c=\frac{1}{z}\left(a,b,c>0\right)$  thì $a+b+c=3.$

Khi đó $\frac{x^2}{z\left(z^2+x^2\right)}=\frac{{\left(\frac{1}{a}\right)}^2}{\frac{1}{c}.\left(\frac{1}{c^2}+\frac{1}{a^2}\right)}=\frac{c^3}{a^2+c^2}=\frac{c^3+c{a}^2-c{a}^2}{c^2+a^2}=c-\frac{c{a}^2}{c^2+a^2}$

$$\begin{gathered} \frac{y^2}{x\left(x^2+y^2\right)}=\frac{{\left(\frac{1}{b}\right)}^2}{\frac{1}{a}\left(\frac{1}{a^2}+\frac{1}{b^2}\right)}=\frac{a^3}{a^2+b^2}=\frac{a^3+a{b}^2-a{b}^2}{a^2+b^2}=a-\frac{a{b}^2}{a^2+b^2} \\ \frac{z^2}{y\left(y^2+z^2\right)}=\frac{{\left(\frac{1}{c}\right)}^2}{\frac{1}{b}\left(\frac{1}{b^2}+\frac{1}{c^2}\right)}=\frac{b^3}{b^2+c^2}=\frac{b^3+b{c}^2-b{c}^2}{b^2+c^2}=b-\frac{b{c}^2}{b^2+c^2} \\ \Rightarrow A=c-\frac{{\displaystyle c{a}^2}}{{\displaystyle c^2+a^2}}+a-\frac{{\displaystyle a{b}^2}}{{\displaystyle a^2+b^2}}+b-\frac{{\displaystyle b{c}^2}}{{\displaystyle b^2+c^2}} \\ =\left(a+b+c\right)-\left(\frac{{\displaystyle c{a}^2}}{{\displaystyle c^2+a^2}}+\frac{{\displaystyle a{b}^2}}{{\displaystyle a^2+b^2}}+\frac{{\displaystyle b{c}^2}}{{\displaystyle b^2+c^2}}\right) \\ =3-\left(\frac{{\displaystyle c{a}^2}}{{\displaystyle c^2+a^2}}+\frac{{\displaystyle a{b}^2}}{{\displaystyle a^2+b^2}}+\frac{{\displaystyle b{c}^2}}{{\displaystyle b^2+c^2}}\right) \end{gathered}$$

Mà $c^2+a^2\ge 2ca\Rightarrow \frac{c{a}^2}{c^2+a^2}\le \frac{c{a}^2}{2ca}=\frac{a}{2}$

Tương tự $\frac{a{b}^2}{a^2+b^2}\le \frac{b}{2}$  và $\frac{b{c}^2}{b^2+c^2}\le \frac{c}{2}$

$$\begin{gathered} \Rightarrow \frac{c{a}^2}{c^2+a^2}+\frac{a{b}^2}{a^2+b^2}+\frac{b{c}^2}{b^2+c^2}\le \frac{a}{2}+\frac{b}{2}+\frac{c}{2}=\frac{3}{2} \\ \Rightarrow 3-\left(\frac{c{a}^2}{c^2+a^2}+\frac{a{b}^2}{a^2+b^2}+\frac{b{c}^2}{b^2+c^2}\right)\ge 3-\frac{3}{2}=\frac{3}{2}. \end{gathered}$$

Vậy $A\ge \frac{3}{2}$  nên $\min A=\frac{3}{2}.$

Dấu “=” xảy ra khi $a=b=c=1.$