Đặt: $P=\dfrac{a^4}{(a+b)(a^2+b^2)}+\dfrac{b^4}{(b+c)(b^2+c^2)}+\dfrac{c^4}{(c+d)(c^2+d^2)}+\dfrac{d^4}{(d+a)(d^2+a^2)}$
$Q=\dfrac{b^4}{(a+b)(a^2+b^2)}+\dfrac{c^4}{(b+c)(b^2+c^2)}+\dfrac{d^4}{(c+d)(c^2+d^2)}+\dfrac{a^4}{(d+a)(d^2+a^2)}$
Ta có:
$P-Q=\dfrac{a^4-b^4}{(a+b)(a^2+b^2)}+\dfrac{b^4-c^4}{(b+c)(b^2+c^2)}+\dfrac{c^4-d^4}{(c+d)(c^2+d^2)}+\dfrac{d^4-a^4}{(d+a)(d^2+a^2)}$
$=(a-b)+(b-c)+(c-d)+(d-a)=0$
Suy ra:
$2P=\dfrac{a^4+b^4}{(a+b)(a^2+b^2)}+\dfrac{b^4+c^4}{(b+c)(b^2+c^2)}+\dfrac{c^4+d^4}{(c+d)(c^2+d^2)}+\dfrac{d^4+a^4}{(d+a)(d^2+a^2)}$
$\ge\dfrac{(a^2+b^2)^2}{2(a+b)(a^2+b^2)}+\dfrac{(b^2+c^2)^2}{2(b+c)(b^2+c^2)}+\dfrac{(c^2+d^2)^2}{2(c+d)(c^2+d^2)}+\dfrac{(d^2+a^2)^2}{2(d+a)(d^2+a^2)}$
$\ge\dfrac{a^2+b^2}{2(a+b)}+\dfrac{b^2+c^2}{2(b+c)}+\dfrac{c^2+d^2}{2(c+d)}+\dfrac{d^2+a^2}{2(d+a)}$
$\ge\dfrac{(a+b)^2}{4(a+b)}+\dfrac{(b+c)^2}{4(b+c)}+\dfrac{(c+d)^2}{4(c+d)}+\dfrac{(d+a)^2}{4(d+a)}$
$\ge\dfrac{a+b}{4}+\dfrac{b+c}{4}+\dfrac{c+d}{4}+\dfrac{d+a}{4}=\dfrac{a+b+c+d}{2}$
Suy ra: $P\ge\dfrac{a+b+c+d}{4}$
Dấu bằng xảy ra khi: $a=b=c=d$