文〔1〕对一道不等式再思考后提出了四个猜想,现对这四个猜想逐一加以探究.为了证明猜想,需用到下面的引理.引理设pq0,x1,x2,…,xn为正实数则x1p+x2p+…+xnpn1P≥x1q+x2q+…+xnqn1q,当且仅当x1=x2=…=xn时等号成立见文〔2〕.猜想1若an+bn=2,a,b∈R+,n≥2,n∈N,则a+b≤2,ab≤1 In [1], after rethinking an inequality, four conjectures are proposed. Now these four conjectures are explored one by one. In order to prove the conjecture, the following lemma is used. Lemma designation pq0,x1,x2,...,xn For positive real numbers x1p+x2p+...+xnpn1P≥x1q+x2q+...+xnqn1q, if and only if x1=x2=...=xn, the equal sign is established as in [2]. Conjecture 1 if an+bn=2,a, b∈R+,n≥2,n∈N, then a+b≤2, ab≤1