Question

设a>0,b>0,且a+b=1求证(a+1/a)^2+(b+1/b)^2>=25/2

Answer 1

(a+ 1/a)^2+(b+ 1/b)^2
=4+a^2+b^2+1/(a^2)+1/(b^2)
=4+(a^2+b^2)[1+1/(a^2*b^2)]
=4+(1-2ab)[1+(1/ab)^2]

显然，随着ab值的增大，值会减小；
即ab取最大值时，(a+ 1/a)^2+(b+ 1/b)^2有最小值；

2ab<=a^2+b^2=1-2ab,所以，ab<=1/4,此时a=b=1/2;

将a,b带入原式，所以
(a+ 1/a)^2+(b+ 1/b)^2
≥(2+1/2)^2+(2+1/2)^2=25/2

Answer 2

先求(a+1/a)(b+1/b)
左式=ab+a/b+1/ab+b/a
=(a2b2+a2+1+b2)/ab
=[a2b2+(1-2ab)+1]/ab
=[(ab-1)2+1]/ab

a+b=1
ab<=[(a+b)/2]²=1/4

所以(ab-1)^2+1≥25/16,0<ab≤1/4,1/ab≥4
相乘得到，左式≥25/4
因为原式=(a+1/a)^2+(b+1/b)^2≥2(a+1/a)(b+1/b)≥25/2