Question

Cho \(x,y,z>2\) thỏa mãn \(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}=1\). Chứng minh rằng :

\(\left(x-2\right)\left(y-2\right)\left(z-2\right)\le1\)

Answer 1

Đặt x-2=a; y-2=b; z-2=c (a,b,c>0)

Ta có: \(\frac{1}{a+2}+\frac{1}{b+2}+\frac{1}{c+2}=1\)

<=>\(\frac{1}{a+2}=1-\frac{1}{b+2}-\frac{1}{c+2}\Leftrightarrow\frac{1}{a+2}=\frac{1}{2}-\frac{1}{b+2}+\frac{1}{2}-\frac{1}{c+2}\)

<=>\(\frac{1}{a+2}=\frac{b}{2\left(b+2\right)}+\frac{c}{2\left(c+2\right)}\ge2\sqrt{\frac{bc}{4\left(b+2\right)\left(c+2\right)}}=\sqrt{\frac{bc}{\left(b+2\right)\left(c+2\right)}}\left(1\right)\)

Tương tự ta cũng có: \(\frac{1}{b+2}\ge\sqrt{\frac{ca}{\left(c+2\right)\left(a+2\right)}}\left(2\right);\frac{1}{c+2}\ge\sqrt{\frac{ab}{\left(a+2\right)\left(b+2\right)}}\left(3\right)\)

Nhân (1),(2),(3) vế theo vế ta được:

\(\frac{1}{\left(a+2\right)\left(b+2\right)\left(c+2\right)}\ge\sqrt{\frac{\left(abc\right)^2}{\left[\left(a+2\right)\left(b+2\right)\left(c+2\right)\right]^2}}\)

<=> \(\frac{1}{\left(a+2\right)\left(b+2\right)\left(c+2\right)}\ge\frac{abc}{\left(a+2\right)\left(b+2\right)\left(c+2\right)}\)

\(\Leftrightarrow abc\le1\Leftrightarrow\left(x-2\right)\left(y-2\right)\left(z-2\right)\le1\) (đpcm)

Dấu "=" xảy ra khi a=b=c=3

Answer 2

Chia hai vế của cho xyz khác 0, ta cần chứng minh:

\(\left(1-\frac{2}{x}\right)\left(1-\frac{2}{y}\right)\left(1-\frac{2}{z}\right)\le\frac{1}{xyz}\)

Đặt \(\left(\frac{1}{x};\frac{1}{y};\frac{1}{z}\right)\rightarrow\left(a;b;c\right)\). Bài toán trở thành:

Cho 0 <a,b,c \(< \frac{1}{2}\) thỏa mãn \(a+b+c=1\). Chứng minh rằng:

\(\left(1-2a\right)\left(1-2b\right)\left(1-2c\right)\le abc\)

\(\Leftrightarrow\left(b+c-a\right)\left(c+a-b\right)\left(a+b-c\right)\le abc\)

BĐT đến đây trở về dạng quen thuộc! Hoặc không thì nó hiển nhiên đúng theo BĐT Schur

Answer 3

\(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}=1\Leftrightarrow\frac{1}{x}=\left(\frac{1}{2}-\frac{1}{y}\right)+\left(\frac{1}{2}-\frac{1}{z}\right)\Leftrightarrow\frac{1}{x}=\frac{1}{2}\left(\frac{y-2}{y}+\frac{z-2}{z}\right)\)

áp dụng BĐT Cauchy ta có \(\frac{1}{x}=\frac{1}{2}\left(\frac{y-2}{y}+\frac{z-2}{y}\right)\ge\sqrt{\frac{\left(y-2\right)\left(z-2\right)}{yz}}\)

Tương tự : \(\frac{1}{y}\ge\sqrt{\frac{\left(x-2\right)\left(z-2\right)}{xz}};\frac{1}{z}\ge\sqrt{\frac{\left(x-2\right)\left(y-2\right)}{xy}}\)

Nhân theo vế ta được \(\frac{1}{xyz}\ge\frac{\left(x-2\right)\left(y-2\right)\left(z-2\right)}{xyz}\Rightarrow\left(x-2\right)\left(y-2\right)\left(z-2\right)\le1\)

Answer 4

Akai Haruma Lightning Farron