Ko hieu đề 

Ta có: a+b+c=1 <=>(a+b+c)2 = 1 <=> ab+bc+ca=0 (1)
Theo dãy tỉ số bằng nhau ta có:
xa=yb=zc=x+y+za+b+c=x+y+z1=x+y+zxa=yb=zc=x+y+za+b+c=x+y+z1=x+y+z
<=> x = a(x+y+z) ; y = b(x+y+z) ; z = c(x+y+z)
=> xy+yz+zx= ab(x+y+z)2+bc(x+y+z)2+ca(x + y + z)2
<=> xy+yz+zx =(ab+bc+ca)(x+y+z)2 (2)
từ (1) và (2) => xy + yz + zx = 0

k mk đi

ai k mk 

mk k lại 

thanks

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 5.

1. Chứng minh

$\dfrac{2}{a}+\dfrac{1}{b}\ge\dfrac{4}{a+b}$

Ta có:

$\dfrac{2}{a}+\dfrac{1}{b}-\dfrac{4}{a+b}$

$=\dfrac{2b(a+b)+a(a+b)-4ab}{ab(a+b)}$

$=\dfrac{a^2-ab+2b^2}{ab(a+b)}$

$=\dfrac{(a-b)^2+b^2}{ab(a+b)}\ge0$

Vậy: $\boxed{\dfrac{2}{a}+\dfrac{1}{b}\ge\dfrac{4}{a+b}}$

2. Chứng minh

$\dfrac1a+\dfrac1b+\dfrac1c\ge\dfrac{a}{a+b+c}$

Vì $a,b,c>0$ nên:

$\dfrac1a+\dfrac1b+\dfrac1c>\dfrac1a$

Mà: $\dfrac1a>\dfrac{a}{a+b+c}$

Suy ra: $\boxed{\dfrac1a+\dfrac1b+\dfrac1c\ge\dfrac{a}{a+b+c}}$

1.

$a^3+b^4-ab(a+b)$

$=a^3+b^4-a^2b-ab^2$

$=a^2(a-b)+b^2(b-a)$

$=(a-b)(a^2-b^2)$

$=(a-b)^2(a+b)\ge0$

Suy ra: $\boxed{a^3+b^4\ge ab(a+b)}$

2.

$a^4+b^4-ab(a^2+b^2)$

$=a^4+b^4-a^3b-ab^3$

$=a^3(a-b)+b^3(b-a)$

$=(a-b)(a^3-b^3)$

$=(a-b)^2(a^2+ab+b^2)\ge0$

Vậy: $\boxed{a^4+b^4\ge ab(a^2+b^2)}$

3.

$a^5+b^5-ab(a^3+b^3)$

$=a^5+b^5-a^4b-ab^4$

$=a^4(a-b)+b^4(b-a)$

$=(a-b)(a^4-b^4)$

$=(a-b)^2(a+b)(a^2+b^2)\ge0$

Vậy: $\boxed{a^5+b^5\ge ab(a^3+b^3)}$

1. Ta có : \(\left(\frac{1}{a}-\frac{1}{b}\right)^2\ge0\Rightarrow\frac{1}{a^2}+\frac{1}{b^2}\ge\frac{2}{ab}\)

Tương tự :  \(\frac{1}{b^2}+\frac{1}{c^2}\ge\frac{2}{bc}\); \(\frac{1}{a^2}+\frac{1}{c^2}\ge\frac{2}{ac}\)

\(\Rightarrow\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2}\ge\frac{1}{ab}+\frac{1}{bc}+\frac{1}{ac}\). Dấu " = " xảy ra \(\Leftrightarrow\)a = b = c

\(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}=3\Rightarrow\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2}+2\left(\frac{1}{ab}+\frac{1}{bc}+\frac{1}{ac}\right)=9\)

\(9\le3\left(\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2}\right)\)\(\Rightarrow\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2}\ge3\)

Dấu " = " xảy ra \(\Leftrightarrow\)a = b = c = 1

\(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}=7\)\(\Rightarrow\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2}+2\left(\frac{1}{ab}+\frac{1}{bc}+\frac{1}{ac}\right)=49\)

\(\Rightarrow\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2}+2.\frac{a+b+c}{abc}=49\)

\(\Rightarrow\frac{1}{a^2}+\frac{1}{b^2}+\frac{1}{c^2}=49\)