Question

设a，b，c，d为正数，求证(a+c/a+b)+（b+d/b+c）+（c+a/c+d)+(d+b/d+a)≥4

Answer 1

(a+c)(1/(a+b)+1/(c+d))+(b+d)(1/(b+c)+1/(a+d))
=(a+c)(a+b+c+d)/(a+b)(c+d)+(b+d)(a+b+c+d)/(b+c)(a+d) (1)
又因为 (a+b)(c+d)<=(a+b+c+d)^2/4,(b+c)(a+d)<=(a+b+c+d)^2/4
所以 (1)>=4(a+c)/(a=b+c+d)+4(b+d)/(a+b+c+d)=4

Answer 2

(a+c/a+b)+(c+a/c+d)≥（a+c)/根号下（a+b)（c+d)
（b+d/b+c）+)+(d+b/d+a)≥（b+d）/根号下（a+d)(c+b)
(a+c)/根号下（a+b)(c+d)+(b+d）/根号下（a+d)(c+b)≥(a+c)(b+d)/根号下（a+b)(c+d)（a+d)(c+b)
展开后≥4就可以了