Question

Cho các số thực dương a,b,c thoả mãn ab+bc+ca=3. Chứng minh rằng: \(4\left(a+b+c\right)+abc\ge13.\)

Answer 1

cảm ơn. Nghĩ hộ mình nhé!

Answer 2

Ta có 

\(\frac{\left(a+b+c\right)^2}{3}\)> ab + bc + ca =3 => a + b + => 3

ta có abc > ( a+b+c) ( b + c -a ) ( c + a -b)

=   ( a+b+c+ 2c) ( b + c -a +2a) ( c + a -b+2b)

> ( 3 -2c ) ( 3 - 2 a ) ( 3 - 2 b ) ( do a+b + c)> 3

= 12 ( xy + yz + zx ) -8 xyz - 18 ( x + y + z ) + 27

= 12 .3 - 8xyz - 18 .3 +27

9 - 8 xyz

ta có : xyz > 9 - 8 xyz + 8 xyz > 9 => xyz > 1

do đó : 4 ( a + b + c ) + abc > 4.3 + 1 = 13 (^dpcm)

hok tốt

Answer 3

Ta có 

\(\frac{\left(a+b+c\right)^2}{3}\)> ab + bc + ca =3 => a + b + => 3

ta có abc > ( a+b+c) ( b + c -a ) ( c + a -b)

=   ( a+b+c+ 2c) ( b + c -a +2a) ( c + a -b+2b)

> ( 3 -2c ) ( 3 - 2 a ) ( 3 - 2 b ) ( do a+b + c)> 3

= 12 ( xy + yz + zx ) -8 xyz - 18 ( x + y + z ) + 27

= 12 .3 - 8xyz - 18 .3 +27

9 - 8 xyz

ta có : xyz > 9 - 8 xyz + 8 xyz > 9 => xyz > 1

do đó : 4 ( a + b + c ) + abc > 4.3 + 1 = 13 (^dpcm)

hok tốt

Answer 4

Có: \(a+b+c\ge3\)

\(VT=4\left(a+b+c\right)-\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)+\left(abc+\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)\)

\(\ge4\left(a+b+c\right)-\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)+4=4\left(a+b+c\right)-\frac{3}{abc}+4\)

\(=\frac{a+b+c-3}{abc}+3\left(a+b+c\right)+4\ge\frac{3-3}{abc}+3.3+4=13\)

Dấu "=" xảy ra khi \(a=b=c=1\)