Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
đặt\(A=\dfrac{x^3}{2x+3y+5z}+\dfrac{y^3}{2y+3z+5x}+\dfrac{z^3}{2z+3x+5y}\)
\(=>A=\dfrac{x^4}{2x^2+3xy+5xz}+\dfrac{y^4}{2y^2+3yz+5xy}+\dfrac{z^4}{2z^2+3xz+5yz}\)
BBDT AM-GM
\(=>A\ge\dfrac{\left(x^2+y^2+z^2\right)^2}{2\left(x^2+y^2+z^2\right)+8\left(xy+yz+xz\right)}\)
theo BDT AM -GM ta chứng minh được \(xy+yz+xz\le x^2+y^2+z^2\)
vì \(x^2+y^2\ge2xy\)
\(y^2+z^2\ge2yz\)
\(x^2+z^2\ge2xz\)
\(=>2\left(x^2+y^2+z^2\right)\ge2\left(xy+yz+xz\right)< =>xy+yz+xz\le x^2+y^2+z^2\)
\(=>2\left(x^2+y^2+z^2\right)+8\left(xy+yz+xz\right)\le10\left(x^2+y^2+z^2\right)\)
\(=>A\ge\dfrac{\left(x^2+y^2+z^2\right)^2}{10\left(x^2+y^2+z^2\right)}=\dfrac{x^2+y^2+z^2}{10}=\dfrac{\dfrac{1}{3}}{10}=\dfrac{1}{30}\left(đpcm\right)\)
dấu"=" xảy ra<=>x=y=z=1/3
\(VT\le\dfrac{x}{2x+2y+2}+\dfrac{y}{2yz+2z+2}+\dfrac{z}{2z+2x+2}\)
Nên ta chỉ cần chứng minh: \(\dfrac{x}{x+y+1}+\dfrac{y}{y+z+1}+\dfrac{z}{z+x+1}\le1\)
\(\Leftrightarrow\dfrac{y+1}{x+y+1}+\dfrac{z+1}{y+z+1}+\dfrac{x+1}{z+x+1}\ge2\)
Thật vậy, ta có:
\(VT=\dfrac{\left(x+1\right)^2}{\left(x+1\right)\left(z+x+1\right)}+\dfrac{\left(y+1\right)^2}{\left(y+1\right)\left(x+y+1\right)}+\dfrac{\left(z+1\right)^2}{\left(z+1\right)\left(y+z+1\right)}\)
\(VT\ge\dfrac{\left(x+y+z+3\right)^2}{\left(x^2+y^2+z^2\right)+3\left(x+y+z\right)+xy+yz+zx+3}\)
\(VT\ge\dfrac{6\left(x+y+z\right)+2\left(xy+yz+zx\right)+12}{3\left(x+y+z\right)+xy+yz+zx+6}=2\) (đpcm)
Dấu "=" xảy ra khi \(x=y=z=1\)
\(VT^2\le3\left(\dfrac{1}{2x^2+y^2+3}+\dfrac{1}{2y^2+z^2+3}+\dfrac{1}{2z^2+x^2+3}\right)\)
Mặt khác:
\(\dfrac{1}{2\left(x^2+1\right)+y^2+1}\le\dfrac{1}{4x+2y}=\dfrac{1}{2}\left(\dfrac{1}{x+x+y}\right)\le\dfrac{1}{18}\left(\dfrac{2}{x}+\dfrac{1}{y}\right)\)
\(\Rightarrow VT^2\le\dfrac{1}{6}\left(\dfrac{3}{x}+\dfrac{3}{y}+\dfrac{3}{z}\right)=\dfrac{1}{2}\left(\dfrac{1}{x}+\dfrac{1}{y}+\dfrac{1}{z}\right)=\dfrac{3}{2}\)
\(\Rightarrow VT\le\dfrac{\sqrt{6}}{2}\)
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 ⇔ \(x=y=z=\dfrac{3}{4}\)
Áp dụng BĐT Cauchy-Schwarz:
\(\dfrac{1}{x+y}+\dfrac{1}{x+y}+\dfrac{1}{y+z}+\dfrac{1}{z+x}\ge\dfrac{16}{3x+3y+2z}\\ \Leftrightarrow\dfrac{1}{3x+2y+2z}\le\dfrac{1}{16}\left(\dfrac{2}{x+y}+\dfrac{1}{y+z}+\dfrac{1}{z+x}\right)\\ \Leftrightarrow\sum\dfrac{1}{3x+2y+2z}\le\dfrac{1}{16}\left(\dfrac{4}{x+y}+\dfrac{4}{y+z}+\dfrac{4}{z+x}\right)=\dfrac{4}{16}\cdot6=\dfrac{3}{2}\)
Dấu \("="\Leftrightarrow x=y=z=\dfrac{1}{3}\)
Đề bài:Cho x,y,z dương thỏa mãn \(x\geq y\geq z>0\). CMR
\(\frac{x^2y}{z}+\frac{y^2z}{x}+\frac{z^2x}{y}\geq x^2+y^2+z^2\)
Giải
Áp dụng BĐT Cauchy-Schwarz ta có:
\(\left(\dfrac{x^2y}{z}+\dfrac{y^2z}{x}+\dfrac{z^2x}{y}\right)\left(\dfrac{x^2z}{y}+\dfrac{y^2x}{z}+\dfrac{z^2y}{x}\right)\ge\left(x^2+y^2+z^2\right)^2\)
Vậy ta cần chứng minh \(\dfrac{x^2y}{z}+\dfrac{y^2z}{x}+\dfrac{z^2x}{y}\ge\dfrac{x^2z}{y}+\dfrac{y^2x}{z}+\dfrac{z^2y}{x}\)
Thật vậy ta có: \(\dfrac{x^2y}{z}+\dfrac{y^2z}{x}+\dfrac{z^2x}{y}-\dfrac{x^2z}{y}+\dfrac{y^2x}{z}+\dfrac{z^2y}{x}\ge0\)
\(\Leftrightarrow\dfrac{\left(xy+yz+xz\right)\left(x-y\right)\left(y-z\right)\left(x-z\right)}{xyz}\ge0\) (luôn đúng)
Đề bài chắc chắn là có vấn đề
Thử với \(x=y=z=\dfrac{1}{3}\) thì \(VT=\dfrac{\sqrt{2}}{4}< 2\)
Như bạn sửa điều kiện thành \(x^3+y^3+z^3=1\) thì dấu "=" không xảy ra
Việc chứng minh vế trái lớn hơn 2 (một cách tuyệt đối) khá đơn giản:
\(\dfrac{x^2}{\sqrt{1-x^2}}=\dfrac{x^3}{x\sqrt{1-x^2}}\ge\dfrac{x^3}{\dfrac{x^2+1-x^2}{2}}=2x^3\)
Làm tương tự với 2 số hạng còn lại, sau đó cộng vế
Nhưng đẳng thức không xảy ra.
\(\dfrac{x^3}{y+2z}+\dfrac{y^3}{z+2x}+\dfrac{z^3}{x+2y}=\dfrac{x^4}{xy+2xz}+\dfrac{y^4}{yz+2xy}+\dfrac{z^4}{xz+2yz}\)
\(\ge\dfrac{\left(x^2+y^2+z^2\right)^2}{3\left(xy+yz+zx\right)}\ge\dfrac{\left(x^2+y^2+z^2\right)^2}{3\left(x^2+y^2+z^2\right)}=\dfrac{1}{3}\)
Dấu "=" xảy ra khi \(x=y=z=\dfrac{1}{\sqrt{3}}\)