Do [tex]a^2 ; b^2 \geq 0[/tex]
Áp dụng BĐT Cô- si , ta có :
[tex]a^2 + b^2 \geq 2ab ; b^2 + 1 \geq 2\sqrt{b^2} = 2b[/tex]
[tex]\Rightarrow a^2 + b^2 + b^2 + 1 + 2 \geq 2ab + 2b + 2[/tex]
[tex]\Rightarrow 1/a^2 + 2b^2 + 3 \leq 1/2ab + 2b + 2[/tex] ( 1 )
CMTT , ta có :
1/b^2 + 2c^2 + 3 [tex]\leq[/tex] 1/2bc + 2c + 2 ( 2 )
1/c^2 + 2a^2 + 3 [tex]\leq[/tex] 1/2ac + 2a + 2 ( 3 )
Từ ( 1 ) ; ( 2 ) ; ( 3 )
=> 1/a^2 + 2b^2 + 3 + 1/b^2 + 2c^2 + 3 1/c^2 + 2a^2 + 3 [tex]\leq[/tex] 1/2ab + 2b + 2 + 1/2bc + 2c + 2 + 1/2ac + 2a + 2
=> 1/a^2 + 2b^2 + 3 + 1/b^2 + 2c^2 + 3 1/c^2 + 2a^2 + 3[tex]\leq[/tex] 1/2(1/ab + b + 1 + 1/bc + c + 1 + 1/ac + a + 1) (*)
Lại có: 1/ab + b + 1 + 1/bc + c + 1 + 1/ac + a + 1
= 1/ab + b + 1 + ab/ab.bc + abc + ab + b/abc + ab + b
= 1/ab + b + 1 + ab/b + 1 + ab + b/1 + ab + b
= 1+ab+b/ab + b + 1
= 1
( *')
Từ (*); (*') , có :
1/a^2 + 2b^2 + 3 + 1/b^2 + 2c^2 + 3 + 1/c^2 + 2a^2 + 3 [tex]\leq[/tex] 1/2
Dấu " = " xảy ra <=> a = b = c = 1