Question

Cho các số dương x;y;z ; CMR:

\(\dfrac{1}{x+3y}+\dfrac{1}{y+3z}+\dfrac{1}{z+3x}\ge\dfrac{1}{x+2y+z}+\dfrac{1}{y+2z+x}+\dfrac{1}{z+2x+y};.\)

Answer 1

Haha không giỡn nữa :v

Áp dụng BĐT Cauchy-Schwarz ta có:

\(L.H.S=Σ\dfrac{1}{2x+y+z}=7Σ\dfrac{1}{2\left(x+3y\right)+\left(y+3z\right)+4\left(z+3x\right)}\)

\(=\dfrac{1}{7}Σ\dfrac{\left(2+1+4\right)^2}{2\left(x+3y\right)+\left(y+3z\right)+4\left(z+3x\right)}\)

\(\le\dfrac{1}{7}Σ\left(\dfrac{2^2}{2\left(x+3y\right)}+\dfrac{1^2}{y+3z}+\dfrac{4^2}{4\left(z+3x\right)}\right)\)

\(=\dfrac{1}{7}Σ\left(\dfrac{2}{x+3y}+\dfrac{1}{y+3z}+\dfrac{4}{z+3x}\right)\)

\(=\dfrac{1}{7}Σ\dfrac{7}{x+3y}=Σ\dfrac{1}{x+3y}=R.H.S\)

Answer 2

Áp dụng bất đẳng thức \(\dfrac{1}{x}+\dfrac{1}{y}\le\dfrac{4}{x+y}\) \(\forall x,y>0\)

\(\Rightarrow\left\{{}\begin{matrix}\dfrac{1}{x+3y}+\dfrac{1}{y+2z+x}\le\dfrac{4}{2x+4y+2z}=\dfrac{2}{x+2y+z}\\\dfrac{1}{y+3z}+\dfrac{1}{z+2x+y}\le\dfrac{4}{2x+2y+4z}=\dfrac{2}{x+y+2z}\\\dfrac{1}{z+3x}+\dfrac{1}{x+2y+z}\le\dfrac{4}{4x+2y+2z}=\dfrac{2}{2x+y+z}\end{matrix}\right.\)

\(\Rightarrow\dfrac{1}{x+3y}+\dfrac{1}{y+3z}+\dfrac{1}{z+3x}+\dfrac{1}{y+2z+x}+\dfrac{1}{z+2x+y}+\dfrac{1}{x+2y+z}\le\dfrac{2}{x+2y+z}+\dfrac{2}{x+y+2z}+\dfrac{2}{2x+y+z}\)

\(\Rightarrow VT\le\left(\dfrac{2}{x+2y+z}-\dfrac{1}{x+2y+z}\right)+\left(\dfrac{2}{x+y+2z}-\dfrac{1}{y+x+2z}\right)+\left(\dfrac{2}{2x+y+z}-\dfrac{1}{z+2x+y}\right)\)

\(\Rightarrow VT\le\dfrac{1}{x+2y+z}+\dfrac{1}{x+y+2z}+\dfrac{1}{2x+y+z}\)

\(\Leftrightarrow\dfrac{1}{x+3y}+\dfrac{1}{y+3z}+\dfrac{1}{z+3x}\le\dfrac{1}{x+2y+z}+\dfrac{1}{x+y+2z}+\dfrac{1}{2x+y+z}\) ( đpcm )

Answer 3

cau nay cau de y mot y la ra

chi lam the nay thoi cac cai sau cau dua vao ma lam tuong tu\(\dfrac{1}{x+3y}+\dfrac{1}{x+y+2z}\ge\dfrac{4}{2x+4y+2z}=\dfrac{2}{x+2y+z}\)

Answer 4

có cách nhưng mà xài Tích Phân, ko bt you học chưa :)

Answer 5

bỏ tích phân đi, chắc chưa học r`

\(BDT\LeftrightarrowΣ\left(3x^2+16x^4y+x^4z+20x^3y^2+2x^3z^2-8x^3yz-34x^2y^2z\right)\ge0\)

Đúng theo BĐT Muirhead, AM-GM và Rearrangement

Answer 6

Đặt \(a;b;c\) là các số dương. Theo AM-GM có:

\(\sum_{cyc}ab^3=\frac{1}{7}\sum_{cyc}(4ca^3+bc^3+2ab^3)\geq\frac{1}{7}\sum_{cyc}7\sqrt[7]{a^{14}b^7c^7}=\sum_{cyc}a^2bc.\)

Đặt \(a=t^x;b=t^y;c=t^z\left(t>0\right)\)

Do đó \(\sum_{cyc}\left(t^{x+3y}-t^{2x+y+z}\right)\geq0\)

\(\Leftrightarrow\sum_{cyc}\left(t^{x+3y-1}-t^{2x+y+z-1}\right)\geq0\)

Vì vậy \(\int\limits_{0}^1\sum_{cyc}\left(t^{x+3y-1}-t^{2x+y+z-1}\right)dt\geq0\)

Hay \(\sum_{cyc}\left(\frac{1}{x+3y}-\frac{1}{2x+y+z}\right)\geq0\)

Vui tí, ko bt bn hiểu ko :))

Answer 7

su dung hang dang thuc 1/a+1/b<=4/a+b

vd 1/(3x+y )+1/(x+y++2z)<=4/(4x+2y+2z)=2/(2x+y+z)

tuong tu =>

Answer 8