Đáp án:

$M_{min}=504$ khi $a=b=c=d=504$

Giải thích các bước giải:

 Đặt $N=\dfrac{b^4}{(a^2+b^2)(a+b)}+\dfrac{c^4}{(b^2+c^2)(b+c)}+\dfrac{d^4}{(c^2+d^2)(c+d)}+\dfrac{a^4}{(d^2+a^2)(d+a)}$

Ta có:

$x^4-y^4=(x^2-y^2)(x^2+y^2)=(x-y)(x+y)(x^2+y^2)$

Từ đó:

$M-N=\dfrac{a^4-b^4}{(a^2+b^2)(a+b)}+\dfrac{b^4-c^4}{(b^2+c^2)(b+c)}+\dfrac{c^4-d^4}{(c^2+d^2)(c+d)}+\dfrac{d^4-a^4}{(d^2+a^2)(d+a)}$

$=\dfrac{(a-b)(a+b)(a^2+b^2)}{(a^2+b^2)(a+b)}+\dfrac{(b-c)(b+c)(b^2+c^2)}{(b^2+c^2)(b+c)}+\dfrac{(c-d)(c+d)(c^2+d^2)}{(c^2+d^2)(c+d)}+\dfrac{(d-a)(d+a)(d^2+a^2)}{(d^2+a^2)(d+a)}$

$=a-b+b-c+c-d+d-a=0$

$⇒M=N$

$⇒2M=M+N=\dfrac{a^4+b^4}{(a^2+b^2)(a+b)}+\dfrac{b^4+c^4}{(b^2+c^2)(b+c)}+\dfrac{c^4+d^4}{(c^2+d^2)(c+d)}+\dfrac{d^4+a^4}{(d^2+a^2)(d+a)}$

Ta có:

$a^4+b^4 \geq \dfrac{1}{2}(a^2+b^2)^2=\dfrac{1}{2}(a^2+b^2)(a^2+b^2) \geq \dfrac{1}{4}(a+b)^2(a^2+b^2)$

$⇒\dfrac{a^4+b^4}{(a+b)(a^2+b^2)} \geq \dfrac{(a+b)^2(a^2+b^2)}{4(a+b)(a^2+b^2)}=\dfrac{a+b}{4}$

Tương tự:

$\dfrac{b^4+c^4}{(b+c)(b^2+c^2)} \geq \dfrac{b+c}{4}$

$\dfrac{c^4+d^4}{(c+d)(c^2+d^2)} \geq \dfrac{c+d}{4}$

$\dfrac{d^4+a^4}{(d+a)(d^2+a^2)} \geq \dfrac{d+a}{4}$

Cộng vế:

$2M \geq \dfrac{a+b+b+c+c+d+d+a}{4}=\dfrac{a+b+c+d}{2} \geq \dfrac{2016}{2}=1008$

$⇒M \geq \dfrac{1008}{2}=504$

Vậy $M_{min}=504$ khi $a=b=c=d=504$