Question

已知a>0,b>0,c>0,abc=1,试证明:1/a2(b+c)+1/b2(a+c)+1/c2(a+b)≥3/2

Answer 1

令x=ab,y=ac,z=bc,则xyz=1

不妨设x≥y≥z,则x+y≥x+z≥y+z

∴1/(y+z)≥1/(x+z)≥1/(x+y)

由顺序和≥乱序和,得

x/(y+z)+y/(x+z)+z/(x+y)≥y/(y+z)+z/(x+z)+x/(x+y)

x/(y+z)+y/(x+z)+z/(x+y)≥z/(y+z)+x/(x+z)+y/(x+y)

上面两式相加得x/(y+z)+y/(x+z)+z/(x+y)≥3/2

而x/(y+z)=ab/(ac+bc)=1/c²(a+b)

y/(x+z)=ac/(ab+bc)=1/b²(a+c)

z/(y+x)=bc/(ac+ab)=1/a²(c+b)

即1/a²(c+b)+1/b²(a+c)+1/c²(a+b)≥3/2