cho x,y,z là các số dương. CMR: D=x/2x+y+z + y/2y+z+x +z/2z+y+x < hoặc = 3/4 30/07/2021 Bởi Brielle cho x,y,z là các số dương. CMR: D=x/2x+y+z + y/2y+z+x +z/2z+y+x < hoặc = 3/4
Giải thích các bước giải: Đặt $2x+y+z=a, 2y+z+x=b, 2z+x+y=c$ $\rightarrow \begin{cases}a+b+c=4x+4y+4z\\ x=a-\dfrac{a+b+c}{4}=\dfrac{3a-b-c}{4}\\ y=b-\dfrac{a+b+c}{4}=\dfrac{3b-a-c}{4}\\ z=c-\dfrac{a+b+c}{4}=\dfrac{3c-a-b}{4}\end{cases}$ $\rightarrow D=\dfrac{3a-b-c}{4a}+\dfrac{3b-c-a}{4b}+\dfrac{3c-a-b}{4c}$ $\rightarrow D=\dfrac{9}{4}-\dfrac{1}{4}(\dfrac{a}{b}+\dfrac{b}{c}+\dfrac{c}{a}+\dfrac{a}{c}+\dfrac{c}{b}+\dfrac{b}{a})$ $\rightarrow D\le \dfrac{9}{4}-\dfrac{1}{4}(6\sqrt[6]{\dfrac{a}{b}.\dfrac{b}{c}.\dfrac{c}{a}.\dfrac{a}{c}.\dfrac{c}{b}.\dfrac{b}{a})}$ $\rightarrow D\le \dfrac{9}{4}-\dfrac{3}{2}$ $\rightarrow D\le \dfrac{3}{4}$ Bình luận
Ta chứng minh với a,b,c >0 thì (a+b+c)( $\frac{1}{a}$ + $\frac{1}{b}$ +$\frac{1}{c}$ )≥ 9 (*) Thật vậy, (*) ⇔ 3 + $\frac{a}{b}$ +$\frac{a}{c}$ +$\frac{b}{a}$ +$\frac{b}{c}$ +$\frac{c}{a}$ +$\frac{c}{b}$ ≥ 9 ⇔ ($\frac{a}{b}$ +$\frac{b}{a}$-2) + ($\frac{a}{c}$ +$\frac{c}{a}$-2) + ($\frac{b}{c}$ +$\frac{c}{b}$-2) ≥ 0 ⇔ $\frac{a^2 -2ab+b^2}{ab}$ +$\frac{a^2-2ac+c^2}{ac}$ +$\frac{b^2-2bc+c^2}{bc}$ ≥ 0 ⇔ $\frac{(a-b)^2}{ab}$ +$\frac{(a-c)^2}{ac}$ +$\frac{(b-c)^2}{bc}$ ≥0 (luôn đúng với a,b,c >0) Dấu ”=” xảy ra ⇔ a=b=c Có: 3-D= (1-$\frac{x}{2x+y+z}$) + (1-$\frac{y}{x+2y+z}$)+ (1-$\frac{z}{x+y+2z}$) = $\frac{x+y+z}{2x+y+z}$ +$\frac{x+y+z}{x+2y+z}$+$\frac{x+y+z}{x+y+2z}$ = (x+y+z) ($\frac{1}{2x+y+z}$ +$\frac{1}{x+2y+z}$+$\frac{1}{x+y+2z}$) = $\frac{1}{4}$ . [(2x+y+z)+ (x+2y+z) +(x+y+2z) ]($\frac{1}{2x+y+z}$ +$\frac{1}{x+2y+z}$+$\frac{1}{x+y+2z}$) ≥ $\frac{1}{4}$ . 9 (theo cmt) ⇒ 3-D≥ 9/4 ⇔ D≤ 3/4 Dấu ”=” xảy ra ⇔ 2x+y+z=x+2y+z=x+y+2z⇔x=y=z Bình luận
Giải thích các bước giải:
Đặt $2x+y+z=a, 2y+z+x=b, 2z+x+y=c$
$\rightarrow \begin{cases}a+b+c=4x+4y+4z\\ x=a-\dfrac{a+b+c}{4}=\dfrac{3a-b-c}{4}\\ y=b-\dfrac{a+b+c}{4}=\dfrac{3b-a-c}{4}\\ z=c-\dfrac{a+b+c}{4}=\dfrac{3c-a-b}{4}\end{cases}$
$\rightarrow D=\dfrac{3a-b-c}{4a}+\dfrac{3b-c-a}{4b}+\dfrac{3c-a-b}{4c}$
$\rightarrow D=\dfrac{9}{4}-\dfrac{1}{4}(\dfrac{a}{b}+\dfrac{b}{c}+\dfrac{c}{a}+\dfrac{a}{c}+\dfrac{c}{b}+\dfrac{b}{a})$
$\rightarrow D\le \dfrac{9}{4}-\dfrac{1}{4}(6\sqrt[6]{\dfrac{a}{b}.\dfrac{b}{c}.\dfrac{c}{a}.\dfrac{a}{c}.\dfrac{c}{b}.\dfrac{b}{a})}$
$\rightarrow D\le \dfrac{9}{4}-\dfrac{3}{2}$
$\rightarrow D\le \dfrac{3}{4}$
Ta chứng minh với a,b,c >0 thì (a+b+c)( $\frac{1}{a}$ + $\frac{1}{b}$ +$\frac{1}{c}$ )≥ 9 (*)
Thật vậy, (*) ⇔ 3 + $\frac{a}{b}$ +$\frac{a}{c}$ +$\frac{b}{a}$ +$\frac{b}{c}$ +$\frac{c}{a}$ +$\frac{c}{b}$ ≥ 9
⇔ ($\frac{a}{b}$ +$\frac{b}{a}$-2) + ($\frac{a}{c}$ +$\frac{c}{a}$-2) + ($\frac{b}{c}$ +$\frac{c}{b}$-2) ≥ 0
⇔ $\frac{a^2 -2ab+b^2}{ab}$ +$\frac{a^2-2ac+c^2}{ac}$ +$\frac{b^2-2bc+c^2}{bc}$ ≥ 0
⇔ $\frac{(a-b)^2}{ab}$ +$\frac{(a-c)^2}{ac}$ +$\frac{(b-c)^2}{bc}$ ≥0 (luôn đúng với a,b,c >0)
Dấu ”=” xảy ra ⇔ a=b=c
Có: 3-D= (1-$\frac{x}{2x+y+z}$) + (1-$\frac{y}{x+2y+z}$)+ (1-$\frac{z}{x+y+2z}$)
= $\frac{x+y+z}{2x+y+z}$ +$\frac{x+y+z}{x+2y+z}$+$\frac{x+y+z}{x+y+2z}$
= (x+y+z) ($\frac{1}{2x+y+z}$ +$\frac{1}{x+2y+z}$+$\frac{1}{x+y+2z}$)
= $\frac{1}{4}$ . [(2x+y+z)+ (x+2y+z) +(x+y+2z) ]($\frac{1}{2x+y+z}$ +$\frac{1}{x+2y+z}$+$\frac{1}{x+y+2z}$) ≥ $\frac{1}{4}$ . 9 (theo cmt)
⇒ 3-D≥ 9/4 ⇔ D≤ 3/4
Dấu ”=” xảy ra ⇔ 2x+y+z=x+2y+z=x+y+2z⇔x=y=z