Question

[Nhok_baobinh]

Cho \(a;b;c\in\left[1;2\right]\)

CMR: \(\left(a+b+c\right)\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)\le10\)

Answer 1

[Phúc]

ta co 

(a+b+c)(\(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\))<=10

<=>\(\frac{a}{b}+\frac{a}{c}+\frac{b}{a}+\frac{b}{c}+\frac{c}{a}+\frac{c}{b}\)(1)\(\le7\)

That vay ta co

Do a,b,c co vai tro nhu nhau nen ta gia su a>=b>=c 

=>(a-b)(b-c)>=0

=> ab+bc>=b²+ac

Do a,b,c khac 0 

=>\(\hept{\begin{cases}1+\frac{c}{a}\ge\frac{b}{a}+\frac{c}{b}\\1+\frac{a}{c}\ge\frac{b}{c}+\frac{a}{b}\end{cases}}\)

=> 2+2(\(\frac{c}{a}+\frac{a}{c}\))>=(1)

Do a,b,c thuoc [1;2]

=> a/c<=2; c/a<=1/2

=>\(\frac{a}{c}+\frac{c}{a}\le\frac{5}{2}\)

=>\(\frac{a}{b}+\frac{a}{c}+\frac{b}{a}+\frac{b}{c}+\frac{c}{a}+\frac{c}{b}\le7\)

=> (a+b+c)(1/a+1/b+1/c)<=10

Answer 2

[ɱ√ρ︵ʙᴇsт➻❥ɴᴀκᴀʀoтн★彡]

Ta có (a+b+c)(1/a+1/b+1/c)=3+a/b + a/c + b/a + b/c + c/a + c/b ≤ 10 

<=> a/b+b/a+b/c+c/a+c/b ≤ 7

Giả sử 1 ≤ c ≤ b ≤ a ≤ 2 thì:

(1 - a/b)(1 - b/c) + (1 - b/a)(1 - c/b) ≥ 0

<=> 2 + a/c + c/a ≥ a/b + b/a + b/c + c/b 

<=> 2+2(a/c+c/a) ≥ a/b + a/c + b/a + b/c + c/a + c/b 

Do 1≤ a,c ≤2

=> 1/2≤ a/c ≤ 2 

=>  (a/c-2)(a/c-1/2) ≤ 0 

=>  a/c+c/a ≤ 5/2 

Mà 2+2(a/c+c/a) ≥ a/b + a/c + b/a + b/c + c/a + c/b  

=> 7 ≥ a/b + a/c + b/a + b/c + c/a + c/b   

=> (a+b+c)(1/a+1/b+1/c) ≤ 10