Cho x,y,z>0 và x+y+z=1. Chứng minh \(\dfrac{1+x}{1-x}+\dfrac{1+y}{1-y}+\dfrac{1+z}{1-z}\le\dfrac{2x}{y}+\dfrac{2y}{z}+\dfrac{2z}{x}\)
cho các số dương thỏa x,y,z thỏa mãn \(\dfrac{1}{x}\)+\(\dfrac{1}{y}\)+\(\dfrac{1}{z}\)=4
chứng minh rằng: \(\dfrac{1}{2x+y+z}\)+\(\dfrac{1}{x+2y+z}\)+\(\dfrac{1}{x+y+2z}\)\(\le\)1
Ta cần chứng minh:
\(\dfrac{1}{a+b}\le\dfrac{1}{4}\left(\dfrac{1}{a}+\dfrac{1}{b}\right)\left(1\right)\left(a,b>0\right)\)
\(\Leftrightarrow\dfrac{4}{a+b}\le\dfrac{a+b}{ab}\\ \Leftrightarrow4ab\le\left(a+b\right)^2\\ \Leftrightarrow\left(a-b\right)^2\ge0\left(luôn.đúng\right)\)
\(DBXR\Leftrightarrow a=b\)
Do các phép biến đổi tương đương nên (1) luôn đúng
Áp dụng (1), ta có:
\(\dfrac{1}{2x+y+z}\le\dfrac{1}{4}\left(\dfrac{1}{x+y}+\dfrac{1}{x+z}\right)\le\dfrac{1}{4}\left[\dfrac{1}{4}\left(\dfrac{1}{x}+\dfrac{1}{y}\right)+\dfrac{1}{4}\left(\dfrac{1}{x}+\dfrac{1}{z}\right)\right]=\dfrac{1}{16}\left(\dfrac{2}{x}+\dfrac{1}{y}+\dfrac{1}{z}\right)\)
Chứng minh tương tự, ta được:
\(\dfrac{1}{x+2y+z}\le\dfrac{1}{16}\left(\dfrac{1}{x}+\dfrac{2}{y}+\dfrac{1}{z}\right)\)
\(\dfrac{1}{x+y+2z}\le\dfrac{1}{16}\left(\dfrac{1}{x}+\dfrac{1}{y}+\dfrac{2}{z}\right)\)
Cộng từng vế BĐT, ta được:
\(\dfrac{1}{2x+y+z}+\dfrac{1}{x+2y+z}+\dfrac{1}{x+y+2z}\le\dfrac{1}{16}.\left(\dfrac{4}{x}+\dfrac{4}{y}+\dfrac{4}{z}\right)=\dfrac{1}{4}.\left(\dfrac{1}{x}+\dfrac{1}{y}+\dfrac{1}{z}\right)=\dfrac{1}{4}.4=1\)Hay \(\dfrac{1}{2x+y+z}+\dfrac{1}{x+2y+z}+\dfrac{1}{x+y+2z}\le1\left(đpcm\right)\)
\(DBXR\Leftrightarrow x=y=z=\dfrac{3}{4}\)
Cho x>0, y>0, z>0 và \(\dfrac{1}{x}+\dfrac{1}{y}+\dfrac{1}{z}=4\). CM: \(\dfrac{1}{2x+y+z}+\dfrac{1}{x+2y+z}+\dfrac{1}{x+y+2z}\le1\)
a, Cho x, y, z > 0 \(\in[0,1]\). Chứng minh:
\(\dfrac{x}{yz+1}+\dfrac{y}{xz+1}+\dfrac{z}{xy+1}< 2\)
b, x, y, z > 0 : xyz = 1. Chứng minh:
\(\dfrac{1}{x^2+2y+3}+\dfrac{1}{y^2+2z^2+3}+\dfrac{1}{z^2+2x^2+3}\le2\)
cho 3 số dương x,y,z thỏa mãn \(\dfrac{1}{x}\)+\(\dfrac{1}{y}\)+\(\dfrac{1}{z}\)=4. Chứng minh:
\(\dfrac{1}{2x+y+z}\)+\(\dfrac{1}{x+2y+z}\)+\(\dfrac{1}{x+y+2z}\)≤1
Bổ đề:\(\dfrac{1}{x}+\dfrac{1}{y}\ge\dfrac{4}{x+y}\Leftrightarrow\dfrac{1}{x+y}\le\dfrac{1}{4}\left(\dfrac{1}{x}+\dfrac{1}{y}\right)\)
Ta có:\(\dfrac{1}{2x+y+z}\le\dfrac{1}{4}\left(\dfrac{1}{x+y}+\dfrac{1}{x+z}\right)\le\dfrac{1}{4}.\dfrac{1}{4}\left(\dfrac{1}{x}+\dfrac{1}{y}+\dfrac{1}{x}+\dfrac{1}{z}\right)\)
Tương tự ta có:\(\dfrac{1}{2y+z+x}\le\dfrac{1}{4}.\dfrac{1}{4}\left(\dfrac{1}{y}+\dfrac{1}{z}+\dfrac{1}{y}+\dfrac{1}{x}\right)\)
\(\dfrac{1}{2z+x+y}\le\dfrac{1}{4}.\dfrac{1}{4}\left(\dfrac{1}{z}+\dfrac{1}{x}+\dfrac{1}{z}+\dfrac{1}{y}\right)\)
Cộng vế với vế ta có:
\(\dfrac{1}{2x+y+z}+\dfrac{1}{2y+z+x}+\dfrac{1}{2z+x+y}\le\dfrac{1}{16}\left[4\left(\dfrac{1}{x}+\dfrac{1}{y}+\dfrac{1}{z}\right)\right]=\dfrac{1}{16}.4.4=1\)
Dấu "=" xảy ra ⇔ \(x=y=z=\dfrac{3}{4}\)
Cho x, y, z > 0 và \(\dfrac{1}{x}+\dfrac{1}{y}+\dfrac{1}{z}=4\) .
CMR : \(\dfrac{1}{2x+y+z}+\dfrac{1}{x+2y+z}+\dfrac{1}{x+y+2z}\le1\)
Ta có bất đẳng thức AM-GM dạng phân thức sau:
\(\dfrac{1}{a}+\dfrac{1}{b}\ge\dfrac{4}{a+b}\Rightarrow \dfrac{1}{a+b}\le\dfrac{1}{4}(\dfrac{1}{a}+\dfrac{1}{b})\)
Dấu ''='' xảy ra khi và chỉ khi a=b
Quay lại bài toán: Áp dụng bđt trên, ta có:
\(\dfrac{1}{2x+y+z}=\dfrac{1}{(x+y)+(x+z)}\le\dfrac{1}{4}(\dfrac{1}{x+y}+\dfrac{1}{x+z})\\ \le\dfrac{1}{16}(\dfrac{1}{x}+\dfrac{1}{y}+\dfrac{1}{x}+\dfrac{1}{z})=\dfrac{1}{16}(\dfrac{2}{x}+\dfrac{1}{y}+\dfrac{1}{z})\)
Tương tự:
\(\dfrac{1}{x+2y+z}\le\dfrac{1}{16}(\dfrac{1}{x}+\dfrac{2}{y}+\dfrac{1}{z})\); \(\dfrac{1}{x+y+2z}\le\dfrac{1}{16}(\dfrac{1}{x}+\dfrac{1}{y}+\dfrac{2}{z})\)
Cộng 3 phân thức lại, ta có:
\(\dfrac{1}{2x+y+z}+\dfrac{1}{x+2y+z}+\dfrac{1}{x+y+2z}\le\dfrac{1}{4}(\dfrac{1}{x}+\dfrac{1}{y}+\dfrac{1}{z})=\dfrac{1}{4}.4=1\)
Dấu ''='' xảy ra khi và chỉ khi: \(x=y=z=\dfrac{3}{4}\)
Cho \(\left\{{}\begin{matrix}x,y,z>0\\xyz=1\end{matrix}\right.\). Chứng minh rằng \(\dfrac{1}{\sqrt{x+2y}}+\dfrac{1}{\sqrt{y+2z}}+\dfrac{1}{\sqrt{z+2x}}\le\sqrt{3}\).
Đề bài sai, phản ví dụ: \(x=y=\dfrac{1}{16};z=256\)
Nói chung, chỉ cần 2 biến đủ nhỏ là BĐT này đều sai
Cho x, y, z>0. Chứng minh rằng:
\(\dfrac{x}{x+2y+3z}+\dfrac{y}{y+2z+3x}+\dfrac{z}{z+2x+3y}\ge\dfrac{1}{2}\)
\(VT=\dfrac{x^2}{x^2+2xy+3zx}+\dfrac{y^2}{y^2+2yz+3xy}+\dfrac{z^2}{z^2+2zx+3yz}\)
\(VT\ge\dfrac{\left(x+y+z\right)^2}{x^2+y^2+z^2+5xy+5yz+5zx}=\dfrac{\left(x+y+z\right)^2}{\left(x+y+z\right)^2+3\left(xy+yz+zx\right)}\ge\dfrac{\left(x+y+z\right)^2}{\left(x+y+z\right)^2+\left(x+y+z\right)^2}=\dfrac{1}{2}\)
Cho các số dương x,y,z và \(x^2+y^2+z^2=1\).Chứng minh rằng:\(\dfrac{x^3}{y+2z}+\dfrac{y^3}{z+2x}+\dfrac{z^3}{x+2y}\ge\dfrac{1}{3}\)
\(\dfrac{x^3}{y+2z}+\dfrac{y^3}{z+2x}+\dfrac{z^3}{x+2y}=\dfrac{x^4}{xy+2xz}+\dfrac{y^4}{yz+2xy}+\dfrac{z^4}{xz+2yz}\)
\(\ge\dfrac{\left(x^2+y^2+z^2\right)^2}{3\left(xy+yz+zx\right)}\ge\dfrac{\left(x^2+y^2+z^2\right)^2}{3\left(x^2+y^2+z^2\right)}=\dfrac{1}{3}\)
Dấu "=" xảy ra khi \(x=y=z=\dfrac{1}{\sqrt{3}}\)
cho x,y,z là các số dương thỏa mãn :\(\dfrac{1}{x+y}+\dfrac{1}{y+z}+\dfrac{1}{z+x}=6\)
chứng minh \(\dfrac{1}{3x+3y+2z}+\dfrac{1}{3x+2y+3z}+\dfrac{1}{2x+2y+3z}\le\dfrac{3}{2}\)
Sửa đề nhé\(\dfrac{1}{3x+3y+2z}=\dfrac{1}{\left(z+x\right)+\left(z+y\right)+\left(x+y\right)+\left(x+y\right)}\)
\(\le\dfrac{1}{16}\left(\dfrac{1}{x+z}+\dfrac{1}{z+y}+\dfrac{1}{x+y}+\dfrac{1}{x+y}\right)\)
CMTT và cộng theo vế:
\(VT\le\dfrac{1}{16}\left(\dfrac{1}{x+z}+\dfrac{1}{z+y}+\dfrac{1}{x+y}+\dfrac{1}{x+y}+\dfrac{1}{x+y}+\dfrac{1}{y+z}+\dfrac{1}{x+z}+\dfrac{1}{x+z}+\dfrac{1}{x+z}+\dfrac{1}{x+y}+\dfrac{1}{y+z}+\dfrac{1}{y+z}\right)\)
\(=\dfrac{1}{16}.24=\dfrac{3}{2}\)
\("="\Leftrightarrow x=y=z=\dfrac{1}{4}\)