chung minh gi va

ghi ra

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

Khi đó:

\(\frac{a+b}{2a-b}+\frac{c+b}{2c-b}=\frac{a+\frac{2ac}{a+c}}{2a-\frac{2ac}{a+c}}+\frac{c+\frac{2ac}{a+c}}{2c-\frac{2ac}{a+c}}\)

\(=\frac{a\left(a+c\right)+2ac}{2a\left(a+c\right)-2ac}+\frac{c\left(a+c\right)+2ac}{2c\left(a+c\right)-2ac}\)

\(=\frac{a^2+3ac}{2a^2}+\frac{c^2+3ac}{2c^2}=\frac{a^2}{2a^2}+\frac{3ac}{2a^2}+\frac{c^2}{2c^2}+\frac{3ac}{2c^2}\)

\(=\frac{1}{2}+\frac{3c}{2a}+\frac{1}{2}+\frac{3a}{2c}=1+\frac{3}{2}\left(\frac{a}{c}+\frac{c}{a}\right)\)

\(\ge1+\frac{3}{2}\cdot2\sqrt{\frac{a}{c}\cdot\frac{c}{a}}=1+3=4\) (Cauchy)

Dấu "=" xảy ra khi: \(a=b=c\)

từ cái đã cho suy ra được \(\frac{2a-b}{ab}=\frac{1}{c}\Rightarrow2a-b=\frac{ab}{c}\)

Chứng minh tương tự =>2c-b=bc/a

Đặt \(M=\frac{a+b}{2a-b}+\frac{c+b}{2c-b}=\frac{c\left(a+b\right)}{ab}+\frac{a\left(b+c\right)}{bc}\)

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

\(=\frac{c}{a}+\frac{c}{b}+\frac{a}{b}+\frac{a}{c}=\left(\frac{a}{b}+\frac{b}{a}\right)+\left(\frac{a}{c}+\frac{c}{a}\right)\ge4\)Cái này tự chứng minh nhé

Dấu = xảy ra khi a=b=c

Theo giả thiết: \(\frac{2}{b}=\frac{1}{a}+\frac{1}{c}\ge\frac{2}{\sqrt{ac}}\Leftrightarrow b^2\le ac\Leftrightarrow\frac{ac}{b^2}\ge1\)

Ta có: \(\frac{1}{a}+\frac{1}{c}=\frac{2}{b}\Leftrightarrow b\left(a+c\right)=2ac\Leftrightarrow2ac-bc=ab\Leftrightarrow2a-b=\frac{ab}{c}\)\(\Rightarrow\frac{a+b}{2a-b}=\frac{a+b}{\frac{ab}{c}}=\frac{ac+bc}{ab}=\frac{c}{b}+\frac{c}{a}\)(1)

Tương tự: \(\frac{b+c}{2c-b}=\frac{a}{c}+\frac{a}{b}\)(2)

Cộng từng vế hai đẳng thức (1), (2) và áp dụng Cô - si, ta được: \(\frac{a+b}{2a-b}+\frac{b+c}{2c-b}\ge\frac{c}{b}+\frac{c}{a}+\frac{a}{c}+\frac{a}{b}\ge4\sqrt[4]{\frac{ca}{b^2}}\ge4\)

Đẳng thức xảy ra khi a = b = c

http://diendantoanhoc.net/topic/71619-cmr-fracab2a-bfraccb2c-bgeq-4/?setlanguage=1&langurlbits=topic/71619-cmr-fracab2a-bfraccb2c-bgeq-4/&langid=1

khó thế

\(\frac{2}{b}=\frac{1}{a}+\frac{1}{c}\ge\frac{4}{a+c}\)

\(\frac{a}{b}+\frac{c}{b}\le2\Leftrightarrow x+y\le2\)

VT=\(\frac{x+1}{2x-1}+\frac{y+1}{2y-1}\)

\(2VT=2+3\left(\frac{1}{2x-1}+\frac{1}{2y-1}\right)\ge2+\frac{3.4}{2\left(x+y\right)-2}\ge8\)

=> dpcm