+ + +  ( )( ) ( ) ( ) 1 1 1A B C B C A C A B+ + ≤ + +  + + A B C...

Bài47:Lời Giải: + + +  

( )

( ) ( ) ( ) 1 1 1a b c b c a c a b+ + ≤ + +  + + a b c+ + +  

2

2

2

a bc b ca c ab a b c

2

 + + +   a b c b c a c a b⇔ + + + + +  ≤ + +  + + + + + + ≤ + +a b c ab a c b c b c( ) ( )( )

∑ ∑ ∑

2 3 .(*)

( )( )

+ + +a bc a bc b ca bc+ + + + − − + ≤2 3 0.+ ≤ + = +a b c a b c b cSử dụng bsst AM-GM ta có: (

₂

) ( ) ( )

∑

+

∑ ∑

a bc a bc bc2 2Mặt khác

(

^a

²

^{+bc b}

)(

²

^+ca

)

^{−ab a( +c b c)( + )=c a b( − ) (}

²

^{a b+ )≥0.}⇒ ≤ ⇒ ≤

(

^{aab a}

²

^{(bc b+}

)(

^{c b c)(}

²

^+ca)

)

^1.

(

^{aab a}

²

^{( bc b+}

)(

^{c b c)(}

²

^+ca)

)

^3.+ +

∑

+ +

( )

²

+ +  +  −

b c

b c b c b c

∑ ∑

⇒ ≤ + − − =  −  = − ≤

LHS bc bc bc bc

(*) 6 3 1 0.

 

2 2 2

Suy ra (*)ñúng. Đẳng thức xảy ra tại a= =b c.