Let x1,x2,…,x5 be nonnegative real numbers with ∑i=151+xi1=1. Prove that ∑i=154+xi2xi≤1. (posed by Li Shenghong)
Solution — click to reveal
Let yi=1+xi1, i=1,2,…,5, then xi=yi1−yi, i=1,2,…,5 and ∑i=15yi=1. We have i=1∑54+xi2xi≤1⇔i=1∑55yi2−2yi+1−yi2+yi≤1⇔i=1∑55yi2−2yi+1−5yi2+5yi≤5⇔i=1∑5(−1+5yi2−2yi+13yi+1)≤5⇔i=1∑55(yi−51)2+543yi+1≤10. Furthermore, i=1∑55(yi−51)2+543yi+1≤i=1∑5543yi+1=45i=1∑5(3yi+1)=45×(3+5)=10. This completes the proof.