a) Xét
√a+√a+5<√a+2+√a+3,a>0
⇔a+a+5+2√a(a+5)<a+2+a+3+2√(a+2)(a+3)
⇔a(a+5)<(a+2)(a+3)⇔0<6, đúng
Thay
a=2000⇒√2000+√2005<√2002+√2003
b) Giả sử √a+2+√a+4≤√a+√a+6
⇔(√a+2+√a+4)2≤(√a+√a+6)2
⇔a+2+a+4+2√(a+2)(a+4)≤a+a+6+2√a(a+6)
⇔√(a+2)(a+4)≤√a(a+6)
⇔(a+2)(a+4)≤a(a+6)⇔a2+6a+8≤a2+6a⇔8≤0 Sai
Vậy √a+2+√a+4>√a+√a+6 với a>0