急解高二不等式难题 谢谢

1.
有点委琐哈~
若a>b,ab>ad+bc ,<= a(b-d)>bc
若a=b,ab>ad+bc ,<= a*a>ad+ac, <= a(a-c)>ad
若a<b,ab>ad+bc ,<= b(a-c)>ad

2.
还是有点觉得不对劲阿~~~
2b+ab+a=30(a>0,b>0) => 2/b+1/a+1=30/(ab)
2/b+1/a >= 2(2/(ab))^(1/2)=(8/(ab))^(1/2)
=> (30/(ab)-1)^2 >= 8/(ab)
解得ab>=50或ab<=18
又y=1/ab的最小值，···还是不对哥~~~

(1)a>c+d>0===>(a-c)>d
b>c+d>0===>(b-d)>c
(a-c)(b-d)>cd>0
且ab-(ad+bc)
=ab-ad-bc+cd-cd
=a(b-d)-c(b-d)-cd
=(a-c)(b-d)-cd>0
所以ab>ad+bc
(2)a,b>0
2b+a>=2√2ab
ab+2√ab<=30,(2√ab)^2<=(30-ab)^2
a^2+b^2-68ab+900>=0
ab>=50(舍去)或ab<=18(当2b=a时取等号)
y=1/ab=1/18