Vì \(ab+bc+ac=3\)  =>   \(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}=\frac{3}{abc}\)

Đặt \(\frac{1}{a}=x\):  \(\frac{1}{b}=y\):  \(\frac{1}{c}=z\)=> x+y+z=3xyz

Ta có   \(4\left(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}\right)+\frac{1}{xyz}\ge13\)

AD BĐT  \(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\ge\frac{9}{a+b+c}\) dấu = khi a=b=c ta có 

  \(4\left(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}\right)\ge\frac{36}{x+y+z}\)=\(\frac{36}{3xyz}=\frac{12}{xyz}\)

=> \(\frac{12}{xyz}+\frac{1}{xyz}\ge13\)

=>  \(\frac{13}{xyz}\ge13\)

mà \(3xyz=x+y+z\ge3\sqrt[3]{xyz}\)dấu = khi x=y=z 

=> xyz\(\le1\)

=> đpcm 

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

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

Ta có:\(\sqrt{abc}=a+b+c\ge3\sqrt[3]{abc}\)\(\Rightarrow\left(\sqrt{abc}\right)^6\ge\left(3\sqrt[3]{abc}\right)^6\Leftrightarrow\left(abc\right)^3\ge3^6\left(abc\right)^2\)

\(\Leftrightarrow abc\ge3^6\)(1).Lại có:\(ab+bc+ca\ge3\sqrt[3]{\left(abc\right)^2}\)

BĐT cần chứng minh tương đương với:\(3\sqrt[3]{\left(abc\right)^2}\ge9\sqrt{abc}\Leftrightarrow\sqrt[3]{\left(abc\right)^2}\ge3\sqrt{abc}\)

\(\Leftrightarrow\left(\sqrt[3]{\left(abc\right)^2}\right)^6\ge\left(3\sqrt{abc}\right)^6\)\(\Leftrightarrow\left(abc\right)^4\ge3^6\left(abc\right)^3\Leftrightarrow abc\ge3^6\).Điều này luôn đúng theo (1)
Suy ra:\(ab+bc+ca\ge9\sqrt{abc}=9\left(a+b+c\right)\).Hoàn tất chứng minh
Dấu "=" xảy ra khi \(a=b=c=9\)
 

Thanks bạn nhiều nhé!

Bạn có cách nào mà ko dùng BĐT Cauchy 3 số ko?

\(ab+1\le b\Rightarrow a+\dfrac{1}{b}\le1\)

Đặt \(\left(a;\dfrac{1}{b}\right)=\left(x;y\right)\Rightarrow x+y\le1\)

Gọi vế trái của BĐT cần chứng minh là P:

\(P=x+\dfrac{1}{x^2}+y+\dfrac{1}{y^2}=\left(\dfrac{1}{x^2}+8x+8x\right)+\left(\dfrac{1}{y^2}+8y+8y\right)-15\left(x+y\right)\)

\(P\ge3\sqrt[3]{\dfrac{64x^2}{x^2}}+3\sqrt[3]{\dfrac{64y^2}{y^2}}-15.1=9\) (đpcm)

Dấu "=" xảy ra khi \(\left(x;y\right)=\left(\dfrac{1}{2};\dfrac{1}{2}\right)\) hay \(\left(a;b\right)=\left(\dfrac{1}{2};2\right)\)