\(a^3+b^3\ge ab\left(a+b\right)\)

\(\Leftrightarrow\left(a+b\right)\left(a^2-ab+b^2\right)-ab\left(a+b\right)\ge0\)

\(\Leftrightarrow\left(a+b\right)\left(a-b\right)^2\ge0\) ( đúng )

Dấu "=" \(\Leftrightarrow a=b\)

a) Áp dụng BĐT trên ta có:

\(\Sigma\left(\frac{1}{a^3+b^3+abc}\right)\le\Sigma\left(\frac{1}{ab\left(a+b\right)+abc}\right)=\Sigma\left[\frac{1}{ab}\cdot\left(\frac{1}{a+b+c}\right)\right]=\frac{1}{a+b+c}\cdot\left(\frac{1}{ab}+\frac{1}{bc}+\frac{1}{ca}\right)=\frac{a+b+c}{\left(a+b+c\right)\cdot abc}=\frac{1}{abc}\)

Dấu "=" khi \(a=b=c\)

b) \(\Sigma\left(\frac{1}{a^3+b^3+1}\right)\le\Sigma\left(\frac{1}{ab\left(a+b\right)+abc}\right)=\Sigma\left[\frac{1}{ab}\cdot\left(\frac{1}{a+b+c}\right)\right]=\frac{1}{abc}=1\)

Dấu "=" khi \(a=b=c=1\)

c) \(\Sigma\left(\frac{1}{a+b+1}\right)\le\Sigma\left(\frac{1}{\sqrt[3]{ab}\left(\sqrt[3]{a}+\sqrt[3]{b}\right)+\sqrt[3]{abc}}\right)=\Sigma\left[\frac{1}{\sqrt[3]{ab}\left(\sqrt[3]{a}+\sqrt[3]{b}+\sqrt[3]{c}\right)}\right]\)

\(=\frac{1}{\sqrt[3]{a}+\sqrt[3]{b}+\sqrt[3]{c}}\cdot\left(\frac{1}{\sqrt[3]{ab}}+\frac{1}{\sqrt[3]{bc}}+\frac{1}{\sqrt[3]{ca}}\right)=\frac{\sqrt[3]{a}+\sqrt[3]{b}+\sqrt[3]{c}}{\left(\sqrt[3]{a}+\sqrt[3]{b}+\sqrt[3]{c}\right)\cdot\sqrt[3]{abc}}=\frac{1}{\sqrt[3]{abc}}=1\)

Dấu "=" khi \(a=b=c=1\)

BĐT\(\Leftrightarrow\frac{abc}{a^3\left(b+c\right)}+\frac{abc}{b^3\left(a+c\right)}+\frac{abc}{c^3\left(a+b\right)}\ge\frac{1}{2}\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)\)

\(\Leftrightarrow\frac{\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2}}{\frac{1}{b}+\frac{1}{c}.\frac{1}{a}+\frac{1}{c}.\frac{1}{a}+\frac{1}{b}}\ge\frac{1}{2}\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)\)

Đặt \(x=\frac{1}{a};y=\frac{1}{b};z=\frac{1}{c}\). Áp dụng BĐT: AM-GM ta có:

\(\frac{a^2}{b+c}+\frac{b+c}{4}\ge2\sqrt{\frac{a^2}{b+c}.\frac{b+c}{4}}=a\)

\(\frac{b^2}{a+b}+\frac{a+c}{4}\ge2\sqrt{\frac{b^2}{a+b}.\frac{a+b}{4}}=b\)

\(\frac{c^2}{a+b}+\frac{a+b}{4}\ge2\sqrt{\frac{c^2}{a+b}+\frac{a+b}{4}}=c\)

Cộng theo vế 3 BĐT trên ta có:

\(\frac{a^2}{b+c}+\frac{b^2}{a+c}+\frac{c^2}{a+b}\ge\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\)

hay \(\frac{a^2}{b+c}+\frac{b^2}{a+c}+\frac{c^2}{a+b}\ge\frac{3}{2}\)

Dấu bằng = xảy ra khi a = b = c = 1

Đặt  \(x=\frac{1}{a};y=\frac{1}{b};z=\frac{1}{c}\Rightarrow xyz=1;x>0;y>0;z>0\)

Ta cần chứng minh bất đẳng thức sau : \(A=\frac{x^2}{y+z}+\frac{y^2}{x+z}+\frac{z^2}{x+y}\ge\frac{3}{2}\)

Sử dụng bất đẳng thức Bunhiacopxki cho 2 bộ số :

\(\left(\sqrt{y+z};\sqrt{z+x};\sqrt{x+y}\right);\left(\frac{x}{\sqrt{y+z}};\frac{y}{\sqrt{z+x}};\frac{z}{\sqrt{x+y}}\right)\)

Ta có : \(\left(x+y+z\right)^2\le\left(x+y+z+x+y+z\right)A\)

\(\Rightarrow A\ge\frac{x+y+z}{2}\ge\frac{3\sqrt[3]{xyz}}{2}=\frac{3}{2}\left(Q.E.D\right)\)

Đẳng thức xảy ra khi và chỉ khi \(x=y=z=1\Leftrightarrow a=b=c=1\)

a, \(BĐT\Leftrightarrow\left(a+b\right)\left(a^2-ab+b^2\right)-ab\left(a+b\right)\ge0\Leftrightarrow\left(a+b\right)\left(a^2-ab+b^2-ab\right)\ge0\)

\(\Leftrightarrow\left(a+b\right)\left(a^2-2ab+b^2\right)\ge0\Leftrightarrow\left(a+b\right)\left(a-b\right)^2\ge0\) (luôn đúng vì a,b>0)

Dấu "=" xảy ra <=> a=b

b, Áp dụng bđt câu a ta có: \(a^3+b^3+1\ge ab\left(a+b\right)+abc=ab\left(a+b+c\right)\)

=>\(\frac{1}{a^3+b^3+1}\le\frac{1}{ab\left(a+b+c\right)}\)

Tương tự \(\frac{1}{b^3+c^3+1}\le\frac{1}{bc\left(a+b+c\right)};\frac{1}{c^3+a^3+1}\le\frac{1}{ca\left(a+b+c\right)}\)

Cộng 3 bđt vế theo vế ta được:

\(VT\le\frac{1}{ab\left(a+b+c\right)}+\frac{1}{bc\left(a+b+c\right)}+\frac{1}{ca\left(a+b+c\right)}=\frac{a+b+c}{abc\left(a+b+c\right)}=\frac{1}{abc}=1\left(đpcm\right)\)

Dấu "=" xảy ra <=> a=b=c=1

Cho mk k nhé!

4/1x3x5 = 1/1x3 - 1/3x5
4/3x5x7 = 1/3x5 - 1/5x7
.............
A = 1/1x3 - 1/11x13

1/1x3x5 = 1/4 x (1/1x3 - 1/3x5)
1/3x5x7 = 1/4 x (1/3x5 - 1/5x7)
..........
B = 1/4 x (1/1x3 - 1/11x13)

bài 1 a, hình như có thêm đk là a+b+c=3

Bài 4 nha

Áp dụng BĐT cô si ta có

\(\frac{1}{x^2}+x+x\ge3\sqrt[3]{\frac{1}{x^2}.x.x}=3.\)

Tương tự với y . \(A\ge6\)dấu = xảy ra khi x=y=1

câu 1 mk bị lộn nhưng đáng ra  ca^2 thành c^2a  mới đúng

a)Áp dụng BĐT cosi-schwart:
`A=1/a+1/b+1/c>=9/(a+b+c)`
Mà `a+b+c<=3/2`
`=>A>=9:3/2=6`
Dấu "=" `<=>a=b=c=1/2`
b)Áp dụng BĐT cosi:
`a+1/(4a)>=1`
`b+1/(4b)>=1`
`c+1/(4c)>=1`
`=>a+b+c+1/(4a)+1/(4b)+1/(4c)>=3`
Ta có:
`1/a+1/b+1/c>=6`(Ở câu a)
`=>3/4(1/a+1/b+1/c)>=9/2`
`=>a+b+c+1/(a)+1/(b)+1/(c)>=3+9/2=15/2`
Dấu "=" `<=>a=b=c=1/2`

a)Áp dụng BĐT cosi-schwart:

Mà 

Dấu "=" 
b)Áp dụng BĐT cosi:




Ta có:
(Ở câu a)


Dấu "=" 

Có: \(VT=\frac{abc}{a^2\left(b+c\right)}+\frac{abc}{b^2\left(c+a\right)}+\frac{abc}{c^2\left(a+b\right)}\)

            \(=\frac{bc}{ab+ac}+\frac{ac}{bc+ba}+\frac{ab}{ac+bc}\)

Áp dụng bđt \(\frac{a^2}{x}+\frac{b^2}{y}+\frac{c^2}{z}\ge\frac{\left(a+b+c\right)^2}{x+y+z}\)được

\(VT\ge\frac{\left(\sqrt{ab}+\sqrt{ac}+\sqrt{bc}\right)^2}{2\left(ab+bc+ca\right)}\)

Mà\(\left(\sqrt{ab}+\sqrt{ac}+\sqrt{bc}\right)^2\ge3\left(ab+bc+ca\right)\)(Chuyển vế đưa thành tổng bình phương) 

 \(\Rightarrow VT\ge...\ge\frac{3\left(ab+bc+ca\right)}{2\left(ab+bc+ca\right)}=\frac{3}{2}\)

Dấu "=" khi a=b=c=1