$a+b+c+d \leq 8$ $abcd \geq 2a+2b+2c+2d \geq 8\sqrt[4]{abcd}$ $(abcd)^{\frac{3}{4}} \geq 8$ $4=ab+cd \geq \sqrt{abcd}$ $16 \geq abcd \geq 16$ thus $abcd=16$ $a+b+c+d=8$ $8 \leq \frac{16}{ab}+ab = 8$ by am gm $ab=4$ ,$cd=4$ $8=a+b+c+d $ $8 =\frac{4}{a}+a+\frac{4}{c}+c \geq 8$ which give $(a,b,c,d) = (2,2,2,2)$