Giải giúp em bài bất đẳng thức này vs

[ledinhtu]

cho 3 số thực.a,b,c.và abc=1cmr:
1/(a+b+1)+1/(b+c+1)+1/(a+c+1)<=1

 

[chibao]

Đặt a=x[SUP]3[/SUP],b=y[SUP]3[/SUP],c=z[SUP]3[/SUP]
Xyz=1
=> x[SUP]3[/SUP]+y[SUP]3[/SUP]+1=> x[SUP]2[/SUP]y+xy[SUP]2[/SUP]+1
=>1/ (x[SUP]3[/SUP]+y[SUP]3[/SUP]+1)=<1/ (x[SUP]2[/SUP]y+xy[SUP]2[/SUP]+1)=1/ ( x[SUP]2[/SUP] y+xy[SUP]2[/SUP] +xyz)=1/ ( xy(X+Y+Z)
Tương tự ta có
1/ (x[SUP]3[/SUP]+y[SUP]3[/SUP]+1) =< 1/ ( xy(X+Y+Z)
1/ (z[SUP]3[/SUP]+y[SUP]3[/SUP]+1)=< 1/ ( zy(X+Y+Z)
1/ (x[SUP]3[/SUP]+z[SUP]3[/SUP]+1)=< 1/ ( xz(X+Y+Z)

1/ (x[SUP]3[/SUP]+y[SUP]3[/SUP]+1)+ 1/ (z[SUP]3[/SUP]+y[SUP]3[/SUP]+1)+ 1/ (x[SUP]3[/SUP]+z[SUP]3[/SUP]+1)=<[1/(x+Y+z)].[1/xy+1/zx+1/yz]
=>1/ (x[SUP]3[/SUP]+y[SUP]3[/SUP]+1)+ 1/ (z[SUP]3[/SUP]+y[SUP]3[/SUP]+1)+ 1/ (x[SUP]3[/SUP]+z[SUP]3[/SUP]+1=<[1/(x+Y+z)][(x+y+z)/xyz]=1
óđpcm
Chúc bạn làm bài tốt nha!