一道不等式竞赛题设a,b,c∈R+且a+b+c=3,求证(a²+3b²)/ab²(4-ab

首先有(a²+3b²)/(ab²(4-ab)) ≥ (2ab+2b²)/(ab²(4-ab)) = 2(a+b)/(ab(4-ab)).
同理(b²+3c²)/(bc²(4-bc)) ≥ 2(b+c)/(bc(4-bc)),(c²+3a²)/(ca²(4-ca)) ≥ 2(c+a)/(ca(4-ca)).
只需证明:36(a+b)/(ab(4-ab))+36(b+c)/(bc(4-bc))+36(c+a)/(ca(4-ca)) ≥ 72.
而36(a+b)/(ab(4-ab)) = 9(a+b)(1/(ab)+1/(4-ab))
= 9/b+9/a+9a/(4-ab)+9b/(4-ab)
= 9a/(4-ab)+(4-ab)/a+9b/(4-ab)+(4-ab)/b+5/a+5/b+a+b
≥ 6+6+5/a+5/b+a+b
= 12+5(1/a+1/b)+(a+b).
同理36(b+c)/(bc(4-bc)) ≥ 12+5(1/b+1/c)+(b+c),36(c+a)/(ca(4-ca)) ≥ 12+5(1/c+1/a)+(c+a).
于是36(a+b)/(ab(4-ab))+36(b+c)/(bc(4-bc))+36(c+a)/(ca(4-ca))
≥ 36+10(1/a+1/b+1/c)+2(a+b+c)
= 42+10(1/a+1/b+1/c)
而由Cauchy不等式得(a+b+c)(1/a+1/b+1/c) ≥ 9,即1/a+1/b+1/c ≥ 3.
代入即得:36(a+b)/(ab(4-ab))+36(b+c)/(bc(4-bc))+36(c+a)/(ca(4-ca)) ≥ 72.