Ta có: $\dfrac{a^3}{(2a^2+b^2)(2a^2+c^2)}=\dfrac{1}{a}\dfrac{1}{(2+\dfrac{b^2}{a^2})(2+\dfrac{c^2}{a^2})}$
Áp dụng BĐT Bunhia ta có:
$(2+\dfrac{b^2}{a^2})(2+\dfrac{c^2}{a^2})=(1+1+\dfrac{b^2}{a^2})(1+\dfrac{c^2}{a^2}+1)\ge(1+\dfrac{c}{a}+\dfrac{b}{a})^2=\dfrac{(a+b+c)^2}{a^2}$
Suy ra: $\dfrac{a^3}{(2a^2+b^2)(2a^2+c^2)}\le\dfrac{a}{(a+b+c)^2}$
$\Rightarrow \dfrac{a^3}{(2a^2+b^2)(2a^2+c^2)}+\dfrac{b^3}{(2b^2+a^2)(2b^2+c^2)}+\dfrac{c^3}{(2c^2+a^2)(2c^2+a^2)}\le\dfrac{1}{a+b+c}=\dfrac{1}{3}$
Dấu bằng xảy ra khi: $a=b=c=1$