Ta có: $\frac{1}{4}(\frac{1}{a}+\frac{1}{b}+\frac{1}{c})=\frac{1}{4}(a+b+c)(\frac{1}{a}+\frac{1}{b}+\frac{1}{c})$
$=\frac{1}{4}[3+\sum(\frac{a}{b}+\frac{b}{a})]=\frac{1}{4}[3+\sum(\frac{a^2+b^2}{ab})]$
Theo bất đẳng thức AM - GM, ta có:
$\sum (\frac{ab}{a^2+b^2})+\frac{1}{4}(\frac{1}{a}+\frac{1}{b}+\frac{1}{c})=\sum (\frac{ab}{a^2+b^2})+\frac{1}{4}[3+\sum(\frac{a^2+b^2}{ab})]$
$= \sum(\frac{ab}{a^2+b^2}+\frac{1}{4}.\frac{a^2+b^2}{ab})+\frac{3}{4}$
$\geq 1+1+1+\frac{3}{4}=\frac{15}{4}$
Đẳng thức xảy ra tại $a=b=c=\frac{1}{3}.$