có $\frac{1}{a}=\frac{a+b+c}{a}=1+\frac{b}{a}+\frac{c}{a}$tương tự có $\frac{1}{b}=1+\frac{a}{b}+\frac{c}{b}$ $\frac{1}{c}=1+\frac{a}{c}+\frac{b}{c}$$\frac{ab}{a^{2}+b^{2}}+\frac{bc}{b^{2}+c^{2}}+\frac{ca}{c^{2}+a^{2}}=\frac{1}{\frac{a}{b}+\frac{b}{a}}+\frac{1}{\frac{b}{c}+\frac{c}{b}}+\frac{1}{\frac{c}{a}+\frac{a}{c}}\geq\frac{9}{\frac{1}{a}+\frac{1}{b}+\frac{1}{c}-3}$ta cần chứng minh $\frac{9}{t-3}+\frac{t}{4}\geq\frac{15}{4}$ (*) với $t=\frac{1}{a}+\frac{1}{b}+\frac{1}{c}$thật vậy (*) $\Leftrightarrow(t-9)^{2}\geq0$ "=" xayr ra khi $a=b=c=\frac{1}{3}$

có $\frac{1}{a}=\frac{a+b+c}{a}=1+\frac{b}{a}+\frac{c}{a}$tương tự có $\frac{1}{b}=1+\frac{a}{b}+\frac{c}{b}$ $\frac{1}{c}=1+\frac{a}{c}+\frac{b}{c}$$\frac{ab}{a^{2}+b^{2}}+\frac{bc}{b^{2}+c^{2}}+\frac{ca}{c^{2}+a^{2}}=\frac{1}{\frac{a}{b}+\frac{b}{a}}+\frac{1}{\frac{b}{c}+\frac{c}{b}}+\frac{1}{\frac{c}{a}+\frac{a}{c}}\geq\frac{9}{\frac{1}{a}+\frac{1}{b}+\frac{1}{c}-3}$ta cần chứng minh $\frac{9}{t-3}+\frac{t}{4}\geq\frac{15}{4}$ (*) với $t=\frac{1}{a}+\frac{1}{b}+\frac{1}{c}$thật vậy (*) $\Leftrightarrow(t-9)^{2}\geq0$ "=" xảy ra khi $a=b=c=\frac{1}{3}$