Sửa đề: Chứng minh \(abc\le\dfrac{1}{8}\)

Ta có

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

\(=\dfrac{b}{1+b}+\dfrac{c}{1+c}\ge2\sqrt{\dfrac{bc}{\left(1+b\right)\left(1+c\right)}}\) (1)

Tương tự \(\dfrac{1}{1+b}\ge2\sqrt{\dfrac{ca}{\left(1+c\right)\left(1+a\right)}}\) (2)

và \(\dfrac{1}{1+c}\ge2\sqrt{\dfrac{ab}{\left(1+a\right)\left(1+b\right)}}\) (3)

Nhân (1), (2), (3) với nhau:

\(\dfrac{1}{\left(1+a\right)\left(1+b\right)\left(1+c\right)}\ge\dfrac{8abc}{\left(1+a\right)\left(1+b\right)\left(1+c\right)}\)

\(\Rightarrow abc\le\dfrac{1}{8}\)

Đẳng thức xảy ra \(\Leftrightarrow a=b=c=\dfrac{1}{2}\)

Sửa đề: Chứng minh \(abc\le\dfrac{1}{8}\)

Ta có

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

\(=\dfrac{b}{1+b}+\dfrac{c}{1+c}\ge2\sqrt{\dfrac{bc}{\left(1+b\right)\left(1+c\right)}}\) (1)

Tương tự \(\dfrac{1}{1+b}\ge2\sqrt{\dfrac{ca}{\left(1+c\right)\left(1+a\right)}}\) (2)

và \(\dfrac{1}{1+c}\ge2\sqrt{\dfrac{ab}{\left(1+a\right)\left(1+b\right)}}\) (3)

Nhân (1), (2), (3) với nhau:

\(\dfrac{1}{\left(1+a\right)\left(1+b\right)\left(1+c\right)}\ge\dfrac{8abc}{\left(1+a\right)\left(1+b\right)\left(1+c\right)}\)

\(\Rightarrow abc\le\dfrac{1}{8}\)

Đẳng thức xảy ra \(\Leftrightarrow a=b=c=\dfrac{1}{2}\)

\(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}=\dfrac{1}{a+b+c}\) (ĐKXĐ: \(a\ne0;b\ne0;c\ne0;a+b+c\ne0\))

<=> \(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}-\dfrac{1}{a+b+c}=0\)

<=> \(\dfrac{b}{ab}+\dfrac{a}{ab}+\dfrac{a+b+c}{c\left(a+b+c\right)}-\dfrac{c}{c\left(a+b+c\right)}=0\)

<=> \(\dfrac{a+b}{ab}+\dfrac{a+b}{c\left(a+b+c\right)}=0\)

<=> \(\left(a+b\right)\left(\dfrac{1}{ab}+\dfrac{1}{c\left(a+b+c\right)}\right)=0\)

<=> \(\left(a+b\right)\left[\dfrac{c\left(a+b+c\right)}{abc\left(a+b+c\right)}+\dfrac{ab}{abc\left(a+b+c\right)}\right]=0\)

<=> \(\dfrac{\left(a+b\right)\left(b+c\right)\left(a+c\right)}{abc\left(a+b+c\right)}=0\) [vì c(a + b + c) + ab = ac + bc + c² + ab = a(b + c) + c(b + c) = (a + c)(b + c)]

<=> (a + b)(b + c)(a + c) = 0

câu c : vì nhân hai vế ta được :

(a+b+c)x (ab+bc+ac)=abc

abc+a\(^2\)b+\(a^2\)c + b^2c+ab^2+abc+bc^2+ac^c+abc=abc

abc+a^2b+a^2c+ b^2c+ab^2+abc+bc^2+ac^c=0

(a+c)(a+b)(b+c)=0

Đặt \(\left(a-1\right)\left(b-1\right)\left(c-1\right)>0\) là ( 1)

Ta có : \(\left(a-1\right)\left(b-1\right)\left(c-1\right)>0\)

\(=\left(ab-a-b+1\right)\left(c-1\right)>0\)

\(=a+b+c-ab-bc-ca>0\)

\(=a+b+c-\dfrac{c}{ab}-\dfrac{a}{bc}-\dfrac{b}{ac}>0\)

\(\Leftrightarrow a+b+c>\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}\) ( 2 )

BĐT ( 2 ) đúng . Từ đây ta có thể thấy BĐt ( 1 ) cũng đúng :D

Đặt \(\left(a;b;c\right)\rightarrow\left(\frac{x}{y};\frac{y}{z};\frac{z}{x}\right)\Rightarrow abc=1\left(TMGT\right)\)

Ta có:
\(\frac{1}{a+2}=\frac{1}{\frac{x}{y}+2}=\frac{1}{\frac{x+2y}{y}}=\frac{y}{x+2y}=\frac{y^2}{xy+2y^2}\)

Tương tự:

\(\frac{1}{b+2}=\frac{z^2}{yz+z^2};\frac{1}{c+2}=\frac{x^2}{zx+x^2}\)

Ta có:

\(\frac{x^2}{xz+2x^2}+\frac{y^2}{xy+2y^2}+\frac{z^2}{yz+2z^2}\ge\frac{\left(x+y+z\right)^2}{2\left(x^2+y^2+z^2\right)+xy+yz+zx}\)

Mặt khác \(xy+yz+zx\le x^2+y^2+z^2\)

\(\Rightarrow2\left(x^2+y^2+z^2\right)+xy+yz+zx\le3\left(x^2+y^2+z^2\right)\)

Rồi OK.Đến đây tịt r:( GOD nào vào thông não hộ ạ:(

Sửa đề thành \(\le1\).Bài này cứ quy đồng full nha! Em có làm ở đây r: Câu hỏi của Nguyễn Linh Chi - Toán lớp 0 - Học toán với OnlineMath

Còn một kiểu trâu bò đó là:

Đặt \(\left(a;b;c\right)=\left(\frac{x^2}{yz};\frac{y^2}{zx};\frac{z^2}{xy}\right)\). BĐT quy về:

\(\Sigma_{cyc}\frac{2yz}{x^2+2yz}\le2\Leftrightarrow3-\left(\Sigma_{cyc}\frac{x^2}{x^2+2yz}\right)\le2\)

Ta có: \(VT\le3-\frac{\left(x+y+z\right)^2}{\left(x^2+y^2+z^2+2xy+2yz+2zx\right)}=3-1=2\)(Q.E.D)

ĐẲng thức xảy ra khi ...

Is that true?

Đặt \(\left\{{}\begin{matrix}x=\dfrac{1}{a}\\y=\dfrac{1}{b}\\z=\dfrac{1}{c}\end{matrix}\right.\)\(\Rightarrow\left\{{}\begin{matrix}x,y,z>0\\xyz=1\end{matrix}\right.\) và BĐT cần chứng minh là:

\(\dfrac{x^2}{y+z}+\dfrac{y^2}{x+z}+\dfrac{z^2}{x+y}\ge\dfrac{3}{2}\)

Áp dụng BĐT Cauchy-Schwarz dạng Engel và AM-GM ta có:

\(VT=\dfrac{x^2}{y+z}+\dfrac{y^2}{x+z}+\dfrac{z^2}{x+y}\)

\(\ge\dfrac{\left(x+y+z\right)^2}{2\left(x+y+z\right)}=\dfrac{x+y+z}{2}\ge\dfrac{3\sqrt[3]{xyz}}{2}=\dfrac{3}{2}=VP\)

Xảy ra khi \(x=y=z=1 \Rightarrow a=b=c=1\)

ai tick cho mik , mik tick lại cho !^__<nhớ giải câu hỏi nhé ! thanks

Từ (a-1)(b-1)(c-1)>0 (*)

<=>(ab-b-a+1)(c-1)>0

<=> abc-ab-bc+b-ac+a+c-1>0

<=> a+b+c-ab-ac-bc>0

<=> a+b+c-\(\dfrac{abc}{c}-\dfrac{abc}{b}-\dfrac{abc}{a}\)>0

<=> a+b+c - \(\dfrac{1}{c}-\dfrac{1}{b}-\dfrac{1}{a}>0\)

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

(1) đúng => (*) đúng

2: Ta có: \(\dfrac{a^2}{b+c}+\dfrac{b^2}{c+a}+\dfrac{c^2}{a+b}=\dfrac{a\left(a+b+c\right)}{b+c}+\dfrac{b\left(a+b+c\right)}{c+a}+\dfrac{c\left(a+b+c\right)}{a+b}-a-b-c=\left(a+b+c\right)\left(\dfrac{a}{b+c}+\dfrac{b}{c+a}+\dfrac{c}{a+b}\right)=a+b+c-a-b-c=0\)

1: Sửa đề: Cho \(x,y,z\ne0\) và \(\dfrac{1}{x}+\dfrac{2}{y}+\dfrac{1}{z}=\dfrac{2}{2x+y+2z}\).

CM:....

Đặt 2x = x', 2z = z'.

Ta có: \(\dfrac{2}{x'}+\dfrac{2}{y}+\dfrac{2}{z'}=\dfrac{2}{x'+y+z'}\)

\(\Leftrightarrow\dfrac{1}{x'}+\dfrac{1}{y}+\dfrac{1}{z'}=\dfrac{1}{x'+y+z'}\)

\(\Leftrightarrow\dfrac{1}{x'}-\dfrac{1}{x'+y+z'}+\dfrac{1}{y}+\dfrac{1}{z'}=0\)

\(\Leftrightarrow\dfrac{y+z'}{x'\left(x'+y+z'\right)}+\dfrac{y+z'}{yz'}=0\)

\(\Leftrightarrow\dfrac{\left(y+z'\right)\left(yz'+x'^2+x'y+x'z'\right)}{x'yz'\left(x'+y+z'\right)}=0\)

\(\Leftrightarrow\dfrac{\left(x'+y\right)\left(y+z'\right)\left(z'+x'\right)}{x'yz'\left(x'+y+z'\right)}=0\Leftrightarrow\left(2x+y\right)\left(y+2z\right)\left(2z+2x\right)=0\Leftrightarrow\left(2x+y\right)\left(y+2z\right)\left(z+x\right)=0\left(đpcm\right)\)