`(a+b+c)^2=3(ab+bc+ca)`

`<=>a^2+b^2+c^2+2ab+2bc+2ca=3(ab+bc+ca)`

`<=>a^2+b^2+c^2=ab+bc+ca`

`<=>2a^2+2b^2+2c^2=2ab+2bc+2ca`

`<=>(a-b)^2+(b-c)^2+(c-a)^2=0`

`VT>=0`

Dấu "=" xảy ra khi `a=b=c`

`a^3+b^3+c^3=3abc`

`<=>a^3+b^3+c^3-3abc=0`

`<=>(a+b)^3+c^3-3abc-3ab(a+b)=0`

`<=>(a+b)^3+c^3-3ab(a+b+c)=0`

`<=>(a+b+c)(a^2+b^2+c^2-ab-bc-ca)=0`

`**a+b+c=0`

`**a^2+b^2+c^2=ab+bc+ca`

`<=>a=b=c`

Ta có A=\(\left(ab+bc+ca\right)\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)-abc\left(\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2}\right)\)

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

\(A=\left(a+b\right)\left(a^2-ab+b^2\right)+3ab\left[\left(a+b\right)^2-2ab\right]+6a^2b^2=a^2-ab+b^2+3ab\left(1-2ab\right)+6a^2b^2\)

=\(\left(a+b\right)^2-3ab+3ab-6a^2b^2+6a^2b^2=1\)

2) Ta có \(A=\left(a-1\right)\left(b-1\right)\left(c-1\right)=abc-ab-bc-ca+a+b+c-1=0\)

bài 3 : Ta có \(A=\left(x-y\right)\left(x^2+xy+y^2\right)-36xy=12\left(x^2+xy+y^2\right)-36xy=12\left(x^2-2xy+y^2\right)\)

\(=12\left(x-y\right)^2=12.12^2=1728\)

b)

Đề: Cho a, b, c > 0 và abc = ab + bc + ca. Chứng minh rằng: \(\frac{1}{a+2b+3c}+\frac{1}{2a+3b+c}+\frac{1}{3a+b+2c}\le\frac{3}{16}\)

~ ~ ~ ~ ~

\(abc=ab+bc+ca\)

\(\Leftrightarrow1=\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\)

Áp dụng BĐT \(\frac{1}{x+y}\le\frac{1}{4}\left(\frac{1}{x}+\frac{1}{y}\right)\), ta có:

\(\frac{1}{a+2b+3c}+\frac{1}{2a+3b+c}+\frac{1}{3a+b+2c}\)

\(\le\frac{1}{4}\left(\frac{1}{a+c}+\frac{1}{2\left(b+c\right)}+\frac{1}{2\left(a+b\right)}+\frac{1}{b+c}+\frac{1}{2\left(a+c\right)}+\frac{1}{a+b}\right)\)

\(=\frac{1}{4}\left[\frac{3}{2\left(a+c\right)}+\frac{3}{2\left(b+c\right)}+\frac{3}{2\left(a+b\right)}\right]\)

\(=\frac{3}{8}\left(\frac{1}{a+c}+\frac{1}{b+c}+\frac{1}{a+b}\right)\)

\(\le\frac{3}{32}\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}+\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)\)

\(=\frac{3}{16}\) (đpcm)

Dấu "=" xảy ra khi a = b = c 

Bài 2:b) \(9=\left(\frac{1}{a^3}+1+1\right)+\left(\frac{1}{b^3}+1+1\right)+\left(\frac{1}{c^3}+1+1\right)\)

\(\ge3\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)\therefore\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\le3\)

Ta sẽ chứng minh \(P\le\frac{1}{48}\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)^2\)

Ai có cách hay?

1/Đặt a=1/x,b=1/y,c=1/z ->x+y+z=1.

2a) \(VT=\frac{\left(\frac{1}{a^3}+\frac{1}{b^3}\right)\left(\frac{1}{a}+\frac{1}{b}\right)}{\frac{1}{a}+\frac{1}{b}}\ge\frac{\left(\frac{1}{a^2}+\frac{1}{b^2}\right)^2}{\frac{1}{a}+\frac{1}{b}}\)

\(=\frac{\left[\frac{\left(a^2+b^2\right)^2}{a^4b^4}\right]}{\frac{a+b}{ab}}=\frac{\left(a^2+b^2\right)^2}{a^3b^3\left(a+b\right)}\ge\frac{\left(a+b\right)^3}{4\left(ab\right)^3}\)

\(\ge\frac{\left(a+b\right)^3}{4\left[\frac{\left(a+b\right)^2}{4}\right]^3}=\frac{16}{\left(a+b\right)^3}\)

Thôi đành dồn về bậc dễ chịu hơn vậy :))
\(9=\frac{1}{a^3}+1+\frac{1}{a^3}+\frac{1}{b^3}+1+\frac{1}{b^3}+\frac{1}{c^3}+1+\frac{1}{c^3}\)

\(\ge\frac{3}{a^2}+\frac{3}{b^2}+\frac{3}{c^2}\Rightarrow\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2}\le3\)

Đến đây ta có đánh giá bằng 2 cách như sau:

Cách 1:

Theo Bunhiacopski ta dễ có:

\(\left[2a+\left(b+c\right)\right]^2\ge4\cdot2a\left(b+c\right)\Rightarrow\frac{1}{\left(2a+b+c\right)^2}\le\frac{1}{8a\left(b+c\right)}\)

\(\le\frac{1}{8}\left[\frac{1}{4a^2}+\frac{1}{\left(b+c\right)^2}\right]\le\frac{1}{8}\left[\frac{1}{4a^2}+\frac{1}{4bc}\right]\le\frac{1}{8}\left[\frac{1}{4a^2}+\frac{1}{8}\left(\frac{1}{b^2}+\frac{1}{c^2}\right)\right]\)

Khi đó:

\(P\le\frac{1}{8}\left[\frac{1}{4a^2}+\frac{1}{8b^2}+\frac{1}{8c^2}+\frac{1}{4b^2}+\frac{1}{8a^2}+\frac{1}{8c^2}+\frac{1}{4c^2}+\frac{1}{8a^2}+\frac{1}{8b^2}\right]=\frac{3}{16}\)

Cách 2:

Áp dụng liên tiếp BĐT phụ dạng \(\frac{1}{x+y}\le\frac{1}{4}\left(\frac{1}{x}+\frac{1}{y}\right)\) ta dễ có rằng:

\(\frac{1}{\left(2a+b+c\right)^2}=\left(\frac{1}{2a+b+c}\right)^2=\frac{1}{16}\left(\frac{1}{a+b}+\frac{1}{a+c}\right)^2=\frac{1}{16}\left[\frac{1}{\left(a+b\right)^2}+\frac{1}{\left(a+c\right)^2}+\frac{2}{\left(a+b\right)\left(a+c\right)}\right]\)

\(\Rightarrow16P\le\frac{2}{\left(a+b\right)^2}+\frac{2}{\left(b+c\right)^2}+\frac{2}{\left(c+a\right)^2}+\frac{2}{\left(a+b\right)\left(b+c\right)}+\frac{2}{\left(b+c\right)\left(c+a\right)}+\frac{2}{\left(c+a\right)\left(a+b\right)}\)

\(\le\frac{4}{\left(a+b\right)^2}+\frac{4}{\left(b+c\right)^2}+\frac{4}{\left(c+a\right)^2}\)

\(\le4\cdot\frac{1}{16}\left[\left(\frac{1}{a}+\frac{1}{b}\right)^2+\left(\frac{1}{b}+\frac{1}{c}\right)^2+\left(\frac{1}{c}+\frac{1}{a}\right)^2\right]\)

\(=\frac{1}{2}\left(\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2}+\frac{1}{ab}+\frac{1}{bc}+\frac{1}{ca}\right)\)

\(\le\frac{1}{2}\cdot\left(3+\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2}\right)\le3\)

\(\Rightarrow P\le\frac{3}{16}\)

Đẳng thức xảy ra tại a=b=c=1