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Question

cho x ; y ; z là các số dương . Chứng minh rằng : x/2x + y + z + y / 2y + z + x + z / 2z + x + y nhỏ hơn hoặc bằng $\frac{3}{4}$

I - Vy Nguyễn · Accepted Answer

Đặt $a=2x+y+z;b=2y+z+x;c=2z+x+y$$	\implies$ $a+b+c=\left(2x+y+z\right)+\left(2y+z+x\right)+\left(2z+x+y\right)$ $	\implies$ $a+b+c=4x+4y+4z$$	\implies$ $x+y+z=\frac{a+b+c}{4}$ +)Ta có : $a=2x+y+z$$\iff$ $a=x+\left(x+y+z\right)$$\iff$ $a-\left(x+y+z\right)=x$$\iff$ $a-\frac{a+b+c}{4}=x$$\iff$ $x=\frac{3a-b-c}{4}$+)Ta có :$b=2y+z+x$$\iff$ $b=y+\left(y+z+x\right)$$\iff$$b-\left(y+z+x\right)=y$$\iff$ $b-\frac{a+b+c}{4}=y$$\iff$$y=\frac{3b-c-a}{4}$+)Ta có :$c=2z+x+y$$\iff$ $c=z+\left(z+x+y\right)$$\iff$ $c-\left(z+x+y\right)=z$$\iff$ $c-\frac{a+b+c}{4}=z$$\iff$$z=\frac{3c-a-b}{4}$​​​$	\implies$​ $\frac{x}{2x+y+z}+\frac{y}{2y+z+x}+\frac{z}{2z+x+y}$  $=\frac{3a-b-c}{4a}+\frac{3b-c-a}{4b}+\frac{3c-a-b}{4c}$ $=\frac{9}{4}-\left(\frac{b}{4a}+\frac{c}{4a}+\frac{c}{4b}+\frac{a}{4b}+\frac{a}{4c}+\frac{b}{4c}\right)$ $=\frac{9}{4}-\frac{1}{4}\left(\frac{b}{a}+\frac{c}{a}+\frac{c}{b}+\frac{a}{b}+\frac{a}{c}+\frac{b}{c}\right)$ $=\frac{9}{4}-\frac{1}{4}\left[\left(\frac{b}{a}+\frac{a}{b}\right)+\left(\frac{c}{a}+\frac{a}{c}\right)+\left(\frac{b}{c}+\frac{c}{b}\right)\right]$Áp dụng bất đẳng thức ( BĐT Cosi ) : $m+n$$	\geq$$2\sqrt{mn}$ $\left(m;n>0\right)$ta được : $\frac{b}{a}+\frac{a}{b}$ $	\geq$ 2 $\sqrt{\frac{b}{a}.\frac{a}{b}}$ = 2 $	\implies$ $\frac{b}{a}+\frac{a}{b}$ $	\geq$ 2 $\frac{c}{a}+\frac{a}{c}$ $	\geq$ 2 $\sqrt{\frac{c}{a}.\frac{a}{c}}$ = 2 $	\implies$ $\frac{c}{a}+\frac{a}{c}$ $	\geq$ 2 $\frac{b}{c}+\frac{c}{b}$ $	\geq$ 2 $\sqrt{\frac{b}{c}.\frac{c}{b}}$ = 2 $	\implies$ $\frac{b}{c}+\frac{c}{b}$ $	\geq$ 2 $	\implies$ $\left(\frac{b}{a}+\frac{a}{b}\right)+\left(\frac{c}{a}+\frac{a}{c}\right)+\left(\frac{b}{c}+\frac{c}{b}\right)$ $	\geq$ 2 + 2 + 2 $	\implies$ ​​​$\left(\frac{b}{a}+\frac{a}{b}\right)+\left(\frac{c}{a}+\frac{a}{c}\right)+\left(\frac{b}{c}+\frac{c}{b}\right)$​ $	\geq$ 6 $	\implies$ $\frac{1}{4}\left[\left(\frac{b}{a}+\frac{a}{b}\right)+\left(\frac{c}{a}+\frac{a}{c}\right)+\left(\frac{b}{c}+\frac{c}{b}\right)\right]$ $	\geq$ $\frac{6}{4}$$	\implies$ $\frac{1}{4}\left[\left(\frac{b}{a}+\frac{a}{b}\right)+\left(\frac{c}{a}+\frac{a}{c}\right)+\left(\frac{b}{c}+\frac{c}{b}\right)\right]$ $	\geq$ $\frac{3}{2}$$	\implies$ $-\frac{1}{4}\left[\left(\frac{b}{a}+\frac{a}{b}\right)+\left(\frac{c}{a}+\frac{a}{c}\right)+\left(\frac{b}{c}+\frac{c}{b}\right)\right]$ $\leq$ $-\frac{3}{2}$$	\implies$ $\frac{9}{4}-\frac{1}{4}\left[\left(\frac{b}{a}+\frac{a}{b}\right)+\left(\frac{c}{a}+\frac{a}{c}\right)+\left(\frac{b}{c}+\frac{c}{b}\right)\right]$ $\leq$ $\frac{9}{4}-\frac{3}{2}$$	\implies$ $\frac{9}{4}-\frac{1}{4}\left[\left(\frac{b}{a}+\frac{a}{b}\right)+\left(\frac{c}{a}+\frac{a}{c}\right)+\left(\frac{b}{c}+\frac{c}{b}\right)\right]$ $\leq$ $\frac{3}{4}$ Câu trả lời: Đặt $a=2x+y+z;b=2y+z+x;c=2z+x+y$$	\implies$ $a+b+c=\left(2x+y+z\right)+\left(2y+z+x\right)+\left(2z+x+y\right)$ $	\implies$ $a+b+c=4x+4y+4z$$	\implies$ $x+y+z=\frac{a+b+c}{4}$ +)Ta có : $a=2x+y+z$$\iff$ $a=x+\left(x+y+z\right)$$\iff$ $a-\left(x+y+z\right)=x$$\iff$ $a-\frac{a+b+c}{4}=x$$\iff$ $x=\frac{3a-b-c}{4}$+)Ta có :$b=2y+z+x$$\iff$ $b=y+\left(y+z+x\right)$$\iff$$b-\left(y+z+x\right)=y$$\iff$ $b-\frac{a+b+c}{4}=y$$\iff$$y=\frac{3b-c-a}{4}$+)Ta có :$c=2z+x+y$$\iff$ $c=z+\left(z+x+y\right)$$\iff$ $c-\left(z+x+y\right)=z$$\iff$ $c-\frac{a+b+c}{4}=z$$\iff$$z=\frac{3c-a-b}{4}$​​​$	\implies$​ $\frac{x}{2x+y+z}+\frac{y}{2y+z+x}+\frac{z}{2z+x+y}$  $=\frac{3a-b-c}{4a}+\frac{3b-c-a}{4b}+\frac{3c-a-b}{4c}$ $=\frac{9}{4}-\left(\frac{b}{4a}+\frac{c}{4a}+\frac{c}{4b}+\frac{a}{4b}+\frac{a}{4c}+\frac{b}{4c}\right)$ $=\frac{9}{4}-\frac{1}{4}\left(\frac{b}{a}+\frac{c}{a}+\frac{c}{b}+\frac{a}{b}+\frac{a}{c}+\frac{b}{c}\right)$ $=\frac{9}{4}-\frac{1}{4}\left[\left(\frac{b}{a}+\frac{a}{b}\right)+\left(\frac{c}{a}+\frac{a}{c}\right)+\left(\frac{b}{c}+\frac{c}{b}\right)\right]$Áp dụng bất đẳng thức ( BĐT Cosi ) : $m+n$$	\geq$$2\sqrt{mn}$ $\left(m;n>0\right)$ta được : $\frac{b}{a}+\frac{a}{b}$ $	\geq$ 2 $\sqrt{\frac{b}{a}.\frac{a}{b}}$ = 2 $	\implies$ $\frac{b}{a}+\frac{a}{b}$ $	\geq$ 2 $\frac{c}{a}+\frac{a}{c}$ $	\geq$ 2 $\sqrt{\frac{c}{a}.\frac{a}{c}}$ = 2 $	\implies$ $\frac{c}{a}+\frac{a}{c}$ $	\geq$ 2 $\frac{b}{c}+\frac{c}{b}$ $	\geq$ 2 $\sqrt{\frac{b}{c}.\frac{c}{b}}$ = 2 $	\implies$ $\frac{b}{c}+\frac{c}{b}$ $	\geq$ 2 $	\implies$ $\left(\frac{b}{a}+\frac{a}{b}\right)+\left(\frac{c}{a}+\frac{a}{c}\right)+\left(\frac{b}{c}+\frac{c}{b}\right)$ $	\geq$ 2 + 2 + 2 $	\implies$ ​​​$\left(\frac{b}{a}+\frac{a}{b}\right)+\left(\frac{c}{a}+\frac{a}{c}\right)+\left(\frac{b}{c}+\frac{c}{b}\right)$​ $	\geq$ 6 $	\implies$ $\frac{1}{4}\left[\left(\frac{b}{a}+\frac{a}{b}\right)+\left(\frac{c}{a}+\frac{a}{c}\right)+\left(\frac{b}{c}+\frac{c}{b}\right)\right]$ $	\geq$ $\frac{6}{4}$$	\implies$ $\frac{1}{4}\left[\left(\frac{b}{a}+\frac{a}{b}\right)+\left(\frac{c}{a}+\frac{a}{c}\right)+\left(\frac{b}{c}+\frac{c}{b}\right)\right]$ $	\geq$ $\frac{3}{2}$$	\implies$ $-\frac{1}{4}\left[\left(\frac{b}{a}+\frac{a}{b}\right)+\left(\frac{c}{a}+\frac{a}{c}\right)+\left(\frac{b}{c}+\frac{c}{b}\right)\right]$ $\leq$ $-\frac{3}{2}$$	\implies$ $\frac{9}{4}-\frac{1}{4}\left[\left(\frac{b}{a}+\frac{a}{b}\right)+\left(\frac{c}{a}+\frac{a}{c}\right)+\left(\frac{b}{c}+\frac{c}{b}\right)\right]$ $\leq$ $\frac{9}{4}-\frac{3}{2}$$	\implies$ $\frac{9}{4}-\frac{1}{4}\left[\left(\frac{b}{a}+\frac{a}{b}\right)+\left(\frac{c}{a}+\frac{a}{c}\right)+\left(\frac{b}{c}+\frac{c}{b}\right)\right]$ $\leq$ $\frac{3}{4}$

I - Vy Nguyễn · Answer

Dấu " = " xảy ra khi a = b = c hay x = y = z Câu trả lời: Dấu " = " xảy ra khi a = b = c hay x = y = z

★Ἣὰἢ ๖ۣۜΝğụŷệť ๖ۣۜĽĩηħ✧ · Answer

https://chat.lazi.vn/upload/images/2020/03/file_xot1584973258.jpgChúc bạn hok tốt!!!Câu trả lời: https://chat.lazi.vn/upload/images/2020/03/file_xot1584973258.jpgChúc bạn hok tốt!!!

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