\(abc=1\) nên tồn tại các số dương x;y;z sao cho \(\left(a;b;c\right)=\left(\dfrac{x}{y};\dfrac{y}{z};\dfrac{z}{x}\right)\)

BĐT cần chứng minh tương đương:

\(\dfrac{y}{x+2y}+\dfrac{z}{y+2z}+\dfrac{x}{z+2x}\le1\)

\(\Leftrightarrow\dfrac{2y}{x+2y}-1+\dfrac{2z}{y+2z}-1+\dfrac{2x}{z+2x}-1\le2-3\)

\(\Leftrightarrow\dfrac{x}{x+2y}+\dfrac{y}{y+2z}+\dfrac{z}{z+2x}\ge1\)

Điều này đúng do:

\(VT=\dfrac{x^2}{x^2+2xy}+\dfrac{y^2}{y^2+2yz}+\dfrac{z^2}{z^2+2xz}\ge\dfrac{\left(x+y+z\right)^2}{x^2+y^2+z^2+2xy+2yz+2zx}=1\)

\(Áp\ dụng\ BĐT\ AM - GM,\ ta\ có: \\\sum\dfrac{1}{a^2+2b^2+3}=\sum\dfrac{1}{(a^2+b^2)+(b^2+1)+2}\le\sum\dfrac{1}{2ab+2b+2} \\=\dfrac{1}{2}\sum\dfrac{1}{ab+b+1}=\dfrac{1}{2}.1=\dfrac{1}{2} \\Đẳng\ thức\ xảy\ ra\ khi\ a=b=c=1.\)

Đặt \(\left(a;b;c\right)=\left(\dfrac{y}{x};\dfrac{z}{y};\dfrac{x}{z}\right)\)

\(\Rightarrow VT=\dfrac{1}{\dfrac{y}{x}\left(\dfrac{z}{y}+1\right)}+\dfrac{1}{\dfrac{z}{y}\left(\dfrac{x}{z}+1\right)}+\dfrac{1}{\dfrac{x}{z}\left(\dfrac{y}{x}+1\right)}\)

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

\(VT\ge\dfrac{\left(x+y+z\right)^2}{2\left(xy+yz+zx\right)}\ge\dfrac{3\left(xy+yz+zx\right)}{2\left(xy+yz+zx\right)}=\dfrac{3}{2}\)

Theo nguyên lý Dirichlet, trong 3 số a;b;c luôn có ít nhất 2 số cùng phía so với 1

Không mất tính tổng quát, giả sử đó là a và b

\(\Rightarrow\left(a-1\right)\left(b-1\right)\ge0\)

\(\Leftrightarrow ab+1\ge a+b\)

\(\Leftrightarrow2\left(ab+1\right)\ge\left(a+1\right)\left(b+1\right)\)

\(\Rightarrow\dfrac{2}{\left(a+1\right)\left(b+1\right)\left(c+1\right)}\ge\dfrac{2}{2\left(ab+1\right)\left(c+1\right)}=\dfrac{1}{\left(ab+1\right)\left(c+1\right)}=\dfrac{1}{\left(\dfrac{1}{c}+1\right)\left(c+1\right)}=\dfrac{c}{\left(c+1\right)^2}\)

Lại có:

\(\dfrac{1}{\left(\sqrt{ab}.\sqrt{\dfrac{a}{b}}+1.1\right)^2}+\dfrac{1}{\left(\sqrt{ab}.\sqrt{\dfrac{b}{a}}+1\right)^2}\ge\dfrac{1}{\left(ab+1\right)\left(\dfrac{a}{b}+1\right)}+\dfrac{1}{\left(ab+1\right)\left(\dfrac{b}{a}+1\right)}=\dfrac{1}{ab+1}\)

\(\Rightarrow P\ge\dfrac{1}{ab+1}+\dfrac{1}{\left(c+1\right)^2}+\dfrac{c}{\left(c+1\right)^2}=\dfrac{1}{\dfrac{1}{c}+1}+\dfrac{1}{\left(c+1\right)^2}+\dfrac{c}{\left(c+1\right)^2}\)

\(\Rightarrow P\ge\dfrac{c}{c+1}+\dfrac{c+1}{\left(c+1\right)^2}=\dfrac{c\left(c+1\right)+c+1}{\left(c+1\right)^2}=\dfrac{\left(c+1\right)^2}{\left(c+1\right)^2}=1\) (đpcm)

Dấu "=" xảy ra khi \(a=b=c=1\)

Chắc là a;b;c hết chứ?

\(VT=\dfrac{a}{a+b+c+b-a}+\dfrac{b}{a+b+c+c-b}+\dfrac{c}{a+b+c+a-c}\)

\(VT=\dfrac{a}{c+2b}+\dfrac{b}{a+2c}+\dfrac{c}{b+2a}=\dfrac{a^2}{ac+2ab}+\dfrac{b^2}{ab+2bc}+\dfrac{c^2}{bc+2ac}\)

\(VT\ge\dfrac{\left(a+b+c\right)^2}{3\left(ab+bc+ca\right)}\ge\dfrac{3\left(ab+bc+ca\right)}{3\left(ab+bc+ca\right)}=1\) (đpcm)

cho x,y,z>0 ,x+y+z=1 chu nhi?

\(\Rightarrow\dfrac{x}{x+y+z+y-x}=\dfrac{x}{2y+z}\)

\(\Rightarrow\dfrac{y}{1+z-y}=\dfrac{y}{x+y+z+z-y}=\dfrac{y}{2z+x}\)

\(\Rightarrow\dfrac{z}{1+x-z}=\dfrac{z}{x+y+z+x-z}=\dfrac{z}{2x+y}\)

\(\Rightarrow A=\dfrac{x}{2y+z}+\dfrac{y}{2z+x}+\dfrac{z}{2x+y}=\dfrac{x^2}{2xy+xz}+\dfrac{y^2}{2zy+xy}+\dfrac{z^2}{2xz+xz}\ge\dfrac{\left(x+y+z\right)^2}{3\left(xy+yz+xz\right)}=1\)

dau"=" xay ra<=>x=y=z=1/3

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

\(\frac{1}{1+a}+\frac{1}{1+b}+\frac{1}{1+c}\ge2\)

\(\Leftrightarrow\frac{1}{1+a}\ge1-\frac{1}{1+b}+1-\frac{1}{1+c}\)

\(\Leftrightarrow\frac{1}{1+a}\ge\frac{b}{1+b}+\frac{c}{1+c}\ge2\sqrt{\frac{bc}{\left(1+b\right)\left(1+c\right)}}\left(1\right)\)

Tương tự:

\(\frac{1}{1+b}\ge2\sqrt{\frac{ac}{\left(1+a\right)\left(1+c\right)}}\left(2\right)\)

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

Nhân (1),(2) và (3) theo vế:

\(\frac{1}{\left(1+a\right)\left(1+b\right)\left(1+c\right)}\ge8\frac{abc}{\left(1+a\right)\left(1+b\right)\left(1+c\right)}\)

\(\Leftrightarrow1\ge8abc\Rightarrow abc\le\frac{1}{8}\)

Dấu "=" xảy ra khi a=b=c=1/2

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