Cho `1/a=\frac{x}{y+z};1/b=\frac{y}{z+x};1/c=\frac{z}{x+y}` Chứng minh `\frac{1}{ab}+\frac{1}{bc}+\frac{1}{ca}+\frac{2}{abc}=1`

Đáp án+Giải thích các bước giải:

 `1/a=x/(y+z)`

`=>a=(y+z)/x`

`=>b=(x+z)/y`

`=>c=(x+y)/z`

khi đó ta có:

`a+b+c+2`

`=(y+z)/x+(x+z)/y+(x+y)/z+2`

`=((x+y)xy+(y+z)yz+(x+z)xz+2xyz)/(x+y+z)(1)`

Ta có:

`xy(x+y) + yz(y+z) + xz(x+z) + 2xyz`

`= xy(x+y+z) + yz(x+y+z) + xz(x+z)`

`= (xy+yz)(x+y+z) + xz(x+z)`

`= y(x+z)(x+y+z) + xz(x+z)`

`= (x+z)(xy+y^2+yz) + xz(x+z)`

`= (x+z)(xy+xz+y^2+yz)`

`= (x+z)[x(y+z)+y(y+z)]`

`= (x+z)(y+z)(x+y)`

Do đó `(1)` trở thành:

`((x+y)(y+z)(x+z))/(x+y+z)`

`=(y+z)/x . (x+z)/y . (x+y)/z`

`=abc`

`=a+b+c+2=abc`

`=(a+b+c+2)/(abc)=1`

`=>a/(abc)+b/(abc)+c/(abc)+2/(abc)=1`

`=>1/(bc)+1/(ac)+1/(ab)+2/(abc)=1`

`=>1/(ab)+1/(bc)+1/(ca)+2/(abc)=1(đpcm)`

$\frac{1}{a}$ = $\frac{x}{y+z}$

⇔ a = $\frac{y+z}{x}$ 

⇔ b = $\frac{x+z}{y}$ và c = $\frac{x+y}{z}$ (chứng minh tương tự)

Khi đó:

    a + b + c + 2 

= $\frac{y+z}{x}$ + $\frac{x+z}{y}$ + $\frac{x+y}{z}$ + 2

= $\frac{(x+y)xy + (y+z)yz + (x+z)xz + 2xyz}{xyz}$  (1)

Ta có: xy(x+y) + yz(y+z) + xz(x+z) + 2xyz

= xy(x+y+z) + yz(x+y+z) + xz(x+z)

= (xy+yz)(x+y+z) + xz(x+z)

= y(x+z)(x+y+z) + xz(x+z)

= (x+z)(xy+y²+yz) + xz(x+z)

= (x+z)(xy+xz+y²+yz)

= (x+z)[x(y+z)+y(y+z)]

= (x+z)(y+z)(x+y)

Do đó (1) trở thành: $\frac{(x+y)(y+z)(x+z)}{xyz}$ 

= $\frac{y+z}{x}$ . $\frac{x+z}{y}$ . $\frac{x+y}{z}$

= abc

⇒ a+b+c+2 = abc

⇒ $\frac{a+b+c+2}{abc}$ = 1

⇒ $\frac{a}{abc}$ + $\frac{b}{abc}$ + $\frac{c}{abc}$ + $\frac{2}{abc}$ = 1

⇒ $\frac{1}{bc}$ + $\frac{1}{ac}$ + $\frac{1}{ab}$ + $\frac{2}{abc}$ = 1

⇒ $\frac{1}{ab}$ + $\frac{1}{bc}$ + $\frac{1}{ca}$ + $\frac{2}{abc}$ = 1 (đpcm)