Question

cho a,b,c, là ba số dương. CM rằng :

a, (a+b)(1/a+1/b) lớn hơn hoặc bằng 4

b, (a+b+c)(1/a+1/b+1/c) lớn hon hoặc bằng 9

Answer 1

a.

Xét hiệu:

\(\left(a+b\right)\left(\dfrac{1}{a}+\dfrac{1}{b}\right)-4\)

\(=1+\dfrac{a}{b}+\dfrac{b}{a}+1-4\)

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

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

\(=\dfrac{\left(a-b\right)^2}{ab}\ge0\) ( luôn đúng)

Suy ra:

\(\left(a+b\right)\left(\dfrac{1}{a}+\dfrac{1}{b}\right)\ge4\)

Answer 2

b.

Đặt:

\(A=\)\(\left(a+b+c\right)\left(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}\right)\)

\(=1+\dfrac{a}{b}+\dfrac{a}{c}+\dfrac{b}{a}+1+\dfrac{b}{c}+\dfrac{c}{a}+\dfrac{c}{b}+1\)

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

Áp dụng BĐT Cauchy cho 2 số không âm, ta có:

\(\dfrac{a}{b}+\dfrac{b}{a}\ge2\sqrt{\dfrac{a}{b}.\dfrac{b}{a}}=2\) (2)

\(\dfrac{a}{c}+\dfrac{c}{a}\ge2\sqrt{\dfrac{a}{c}.\dfrac{c}{a}}=2\) (3)

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

Từ (1)(2)(3)(4) cộng vế theo vế, ta được:

\(A\ge3+2+2+2=9\)

=> BĐT luôn đúng

=> \(\left(a+b+c\right)\left(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}\right)\ge9\)

Answer 3

b)Đặt \(A=\left(a+b+c\right)\left(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}\right)\)

\(A=1+\dfrac{a}{b}+\dfrac{a}{c}+\dfrac{b}{a}+1+\dfrac{b}{c}+\dfrac{c}{a}+\dfrac{c}{b}+1\)

\(A=3+\left(\dfrac{a}{b}+\dfrac{b}{a}\right)+\left(\dfrac{a}{c}+\dfrac{c}{a}\right)+\left(\dfrac{b}{c}+\dfrac{c}{b}\right)\)

Ta chứng minh bđt sau:\(\dfrac{x}{y}+\dfrac{y}{x}\ge2\)

\(\Leftrightarrow\dfrac{x^2+y^2}{xy}\ge2\)

\(\Leftrightarrow x^2+y^2\ge2xy\)

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

Áp dụng\(\Rightarrow P\ge3+2+2+2=9\left(đpcm\right)\)

Answer 4

a) c2 : ( a + b)( \(\dfrac{1}{a}+\dfrac{1}{b}\)) ≥ 4

⇔ 1 + \(\dfrac{a}{b}+\dfrac{b}{a}+1\) ≥ 4

Áp dụng BĐT : \(\dfrac{y}{x}+\dfrac{x}{y}\) ≥ 2 ( x > 0 , y > 0 )

⇒ 2 + \(\dfrac{a}{b}+\dfrac{b}{a}\) ≥ 4

⇔ \(\dfrac{a}{b}+\dfrac{b}{a}\) ≥ 2 ( đpcm)

Answer 5

b) \(\left(a+b+c\right)\left(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}\right)\) ≥ 9

⇔ 1 + \(\dfrac{a}{b}+\dfrac{b}{a}+\dfrac{b}{c}+\dfrac{c}{b}+\dfrac{a}{c}+\dfrac{c}{a}\) + 1 + 1 ≥ 9

⇔ 3 + \(\dfrac{a}{b}+\dfrac{b}{a}+\dfrac{b}{c}+\dfrac{c}{b}+\dfrac{a}{c}+\dfrac{c}{a}\) ≥ 9 ( *)

Áp dụng BĐT : \(\dfrac{y}{x}+\dfrac{x}{y}\) ≥ 2 ( x > 0 , y > 0 )

Ta có :

*) \(\dfrac{a}{b}+\dfrac{b}{a}\) ≥ 2 ( a > 0 , b > 0 ) ( 1)

*) \(\dfrac{b}{c}+\dfrac{c}{b}\) ≥ 2 ( b > 0 ; c > 0 ) ( 2)

*) \(\dfrac{a}{c}+\dfrac{c}{a}\) ≥ 2 ( a > 0 ; c > 0) ( 3)

Từ ( * ; 1 ; 2 ; 3 ) ⇒ đpcm