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

Theo bđt Cauchy - Schwart ta có:

\(\text{Σ}cyc\frac{c}{a^2\left(bc+1\right)}=\text{Σ}cyc\frac{\frac{1}{a^2}}{b+\frac{1}{c}}\ge\frac{\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)^2}{\frac{1}{a}+\frac{1}{b}+\frac{1}{c}+a+b+c}\)\(=\frac{\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)^2}{\frac{1}{a}+\frac{1}{b}+\frac{1}{c}+3}\)

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

Đặt \(ab+bc+ca=x;abc=y\).

Ta có: \(\frac{x^2}{xy+3y^2}\ge\frac{9}{x\left(1+y\right)}\Leftrightarrow x^3+x^3y\ge9xy+27y^2\)

\(\Leftrightarrow x\left(x^2-9y\right)+y\left(x^3-27y\right)\ge0\) ( luôn đúng )

Vậy BĐT đc CM. Dấu '=' xảy ra <=> a=b=c=1

sai rồi nhé bạn 

làm sao mà \(x\left(x^2-9y\right)+y\left(x^3-27y\right)\ge0\)lại luôn đúng

\(3=ab+bc+ca\ge3\sqrt[3]{\left(abc\right)^2}\Rightarrow abc\le1\)

\(\dfrac{1}{1+a^2\left(b+c\right)}=\dfrac{1}{1+a\left(ab+ac\right)}=\dfrac{1}{1+a\left(3-bc\right)}=\dfrac{1}{1+3a-abc}=\dfrac{1}{3a+\left(1-abc\right)}\le\dfrac{1}{3a}\)

Tương tự và cộng lại:

\(VT\le\dfrac{1}{3a}+\dfrac{1}{3b}+\dfrac{1}{3c}=\dfrac{ab+bc+ca}{3abc}=\dfrac{3}{3abc}=\dfrac{1}{abc}\)

Ta chứng minh BĐT sau cho các số dương:

\(x^5+y^5\ge xy\left(x^3+y^3\right)\)

\(\Leftrightarrow x^5-x^4y+y^5-xy^4\ge0\)

\(\Leftrightarrow\left(x^4-y^4\right)\left(x-y\right)\ge0\)

\(\Leftrightarrow\left(x-y\right)^2\left(x+y\right)\left(x^2+y^2\right)\ge0\) (đúng)

Áp dụng:

\(\dfrac{a^5+b^5}{ab\left(a+b\right)}\ge\dfrac{ab\left(a^3+b^3\right)}{ab\left(a+b\right)}=\dfrac{a^3+b^3}{a+b}=a^2-ab+b^2\)

Tương tự và cộng lại:

\(VT\ge2\left(a^2+b^2+c^2\right)-\left(ab+bc+ca\right)=2-\left(ab+ca+ca\right)\)

\(VT\ge4-\left(ab+bc+ca\right)-2=4\left(a^2+b^2+c^2\right)-\left(ab+bc+ca\right)-2\)

\(VT\ge4\left(ab+bc+ca\right)-\left(ab+bc+ca\right)-2=3\left(ab+bc+ca\right)-2\) (đpcm)

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 

Bất đẳng thức sai, chẳng hạn với \(a=b=10^{-4};c=0,5-a-b\).

Bất đẳng thức sai với [a = 35/256, b = 5/16, c = 3921/1840 ]

1)

Ta có: \(M=\Sigma_{cyc}\frac{\sqrt{3}\left(a+b+4c\right)}{\sqrt{3\left(a+b\right)\left(a+b+4c\right)}}\ge\Sigma_{cyc}\frac{\sqrt{3}\left(a+b+4c\right)}{\frac{3\left(a+b\right)+\left(a+b+4c\right)}{2}}=\Sigma_{cyc}\frac{\sqrt{3}\left(a+b+4c\right)}{2\left(a+b+c\right)}=3\sqrt{3}\)

Dấu "=" xảy ra khi a=b=c

2)

\(\Sigma_{cyc}\sqrt[3]{\left(\frac{2a}{ab+1}\right)^2}=\Sigma_{cyc}\frac{2a}{\sqrt[3]{2a\left(ab+1\right)^2}}\ge\Sigma_{cyc}\frac{2a}{\frac{2a+\left(ab+1\right)+\left(ab+1\right)}{3}}=3\Sigma_{cyc}\frac{a}{ab+a+1}\)

Ta có bổ đề: \(\frac{a}{ab+a+1}+\frac{b}{bc+b+1}+\frac{c}{ca+c+1}=1\left(abc=1\right)\)

\(\Rightarrow\Sigma_{cyc}\sqrt[3]{\left(\frac{2a}{ab+1}\right)^2}\ge3\)