Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
Áp dụng BĐT Cauchy cho 3 số dương, ta được:
\(\frac{1}{x\left(x+1\right)}+\frac{x}{2}+\frac{x+1}{4}\ge\sqrt[3]{\frac{1}{x\left(x+1\right)}.\frac{x}{2}.\frac{x+1}{4}}=3.\sqrt{\frac{1}{4}}=\frac{3}{2}\)
\(\frac{1}{y\left(y+1\right)}+\frac{y}{2}+\frac{y+1}{4}\ge\sqrt[3]{\frac{1}{y\left(y+1\right)}.\frac{y}{2}.\frac{y+1}{4}}=3.\sqrt{\frac{1}{4}}=\frac{3}{2}\)
\(\frac{1}{z\left(z+1\right)}+\frac{z}{2}+\frac{z+1}{4}\ge\sqrt[3]{\frac{1}{z\left(z+1\right)}.\frac{z}{2}.\frac{z+1}{4}}=3.\sqrt{\frac{1}{4}}=\frac{3}{2}\)
\(\Rightarrow\frac{1}{x\left(x+1\right)}+\frac{x}{2}+\frac{x+1}{4}\)\(+\frac{1}{y\left(y+1\right)}+\frac{y}{2}+\frac{y+1}{4}\)
\(+\frac{1}{z\left(z+1\right)}+\frac{z}{2}+\frac{z+1}{4}\ge\frac{3}{2}.3=\frac{9}{2}\)
\(\Leftrightarrow\frac{1}{x^2+x}+\frac{1}{y^2+y}+\frac{1}{z^2+z}+\frac{x+y+z}{2}+\frac{x+y+z+3}{4}\ge\frac{9}{2}\)
\(\Leftrightarrow\frac{1}{x^2+x}+\frac{1}{y^2+y}+\frac{1}{z^2+z}+\frac{3}{2}+\frac{3}{2}\ge\frac{9}{2}\)
\(\Leftrightarrow\frac{1}{x^2+x}+\frac{1}{y^2+y}+\frac{1}{z^2+z}\ge\frac{3}{2}\left(đpcm\right)\)
ta có \(\frac{1}{x^2+x}+\frac{x^2+x}{4}>=2\cdot\sqrt{\frac{1\cdot\left(x^2+x\right)}{\left(x^2+x\right)\cdot4}}=1\)
tương tự => \(\frac{1}{y^2+y}+\frac{y^2+y}{4}>=1;\frac{1}{z^2+z}+\frac{z^2+z}{4}>=1\)
=> VT >= 3-(\(\frac{x^2+x}{4}+\frac{y^2+y}{4}+\frac{z^2+z}{4}\))=3-\(\frac{x^2+y^2+z^2+3}{4}\)
mà \(\frac{x^2}{4}+\frac{y^2}{4}+\frac{z^2}{4}>=\frac{\left(x+y+z\right)^2}{4+4+4}=\frac{3}{4}\)
=> P>= 3-3/4-3/4=3/2
Dấu bằng khi x=y=z=1
Bài bạn Lương Ngọc Anh bị ngược dấu nên sai hoàn toàn. Lời giải:
Ta có:
\(\frac{1}{x^2+x}=\frac{1}{x\left(x+1\right)}=\frac{1}{x}-\frac{1}{x+1}\)
Tương tự, ta được:
\(VT=\left(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}\right)-\left(\frac{1}{x+1}+\frac{1}{y+1}+\frac{1}{z+1}\right)\)
Áp dụng BĐT Schwarz:
\(\frac{1}{x+1}+\frac{1}{y+1}+\frac{1}{z+1}\le\frac{1}{4}\left(3+\frac{1}{x}+\frac{1}{y}+\frac{1}{z}\right)=\frac{3}{4}+\frac{1}{4}\left(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}\right)\)
Do đó:
\(VT\ge\frac{3}{4}\left(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}\right)-\frac{3}{4}\left(1\right)\)
Mặt khác:
\(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}\ge\frac{9}{x+y+z}=3\left(2\right)\)
TỪ (1) VÀ (2) TA CÓ ĐIỀU PHẢI CHỨNG MINH.
\(\dfrac{x-y}{z^2+1}=\dfrac{x-y}{z^2+xy+yz+zx}=\dfrac{x-y}{z\left(z+y\right)+x\left(z+y\right)}=\dfrac{x-y}{\left(x+z\right)\left(z+y\right)}\)
Tương tự: \(\dfrac{y-z}{x^2+1}=\dfrac{y-z}{\left(x+y\right)\left(x+z\right)}\);\(\dfrac{z-x}{y^2+1}=\dfrac{z-x}{\left(x+y\right)\left(y+z\right)}\)
Cộng vế với vế \(\Rightarrow VT=\dfrac{x-y}{\left(x+z\right)\left(y+z\right)}+\dfrac{y-z}{\left(x+y\right)\left(x+z\right)}+\dfrac{z-x}{\left(x+y\right)\left(y+z\right)}\)
\(=\dfrac{\left(x-y\right)\left(x+y\right)+\left(y-z\right)\left(y+z\right)+\left(z-x\right)\left(z+x\right)}{\left(x+y\right)\left(y+z\right)\left(z+x\right)}\)
\(=\dfrac{x^2-y^2+y^2-z^2+z^2-x^2}{\left(x+y\right)\left(y+z\right)\left(z+x\right)}=0\)(đpcm)
\(taco:\)
\(\frac{1}{x\left(x+1\right)}+\frac{x}{2}+\frac{x+1}{4}\ge3\sqrt[3]{\frac{1}{x\left(x+1\right)}.\frac{x}{2}.\frac{x+1}{4}}=\frac{3}{2}\)
\(\frac{1}{y\left(y+1\right)}+\frac{y}{2}+\frac{y+1}{2}\ge3\sqrt[3]{\frac{1}{y\left(y+1\right)}.\frac{y}{2}.\frac{y+1}{4}}=\frac{3}{2}\)
\(\frac{1}{z\left(z+1\right)}+\frac{z}{2}+\frac{z+1}{4}\ge3\sqrt[3]{\frac{1}{z\left(z+1\right)}.\frac{z}{2}.\frac{z+1}{4}}=\frac{3}{2}\)
\(\frac{1}{x\left(x+1\right)}+\frac{1}{y\left(y+1\right)}+\frac{1}{z\left(z+1\right)}+\frac{x+y+z}{2}+\frac{x+y+z+3}{4}\ge\frac{3}{2}+\frac{3}{2}+\frac{3}{2}\)
\(\Leftrightarrow\frac{1}{x^2+x}+\frac{1}{y^2+y}+\frac{1}{z^2+z}+\frac{3}{2}+\frac{3}{2}\ge\frac{9}{2}\)
\(\Leftrightarrow\frac{1}{x^2+x}+\frac{1}{y^2+y}+\frac{1}{z^2+z}\ge\frac{3}{2}\left(dpcm\right)\)
^^
Mình giải lại bài này cho đầy đủ hơn nhé: (nãy chỉ là hướng dẫn thôi)
Ta sẽ c/m: \(\frac{1}{x^2+x}\ge-\frac{3}{4}x+\frac{5}{4}\) (1).Thật vậy,xét hiệu hai vế,ta có:
\(VT-VP=\frac{\left(3x+4\right)\left(x-1\right)^2}{4\left(x^2+x\right)}\ge0\)
Suy ra \(VT\ge VP\).Vậy (1) đúng.
Thiết lập hai BĐT còn lại tương tự và cộng theo vế,ta có:
\(VT\ge-\frac{3}{4}\left(x+y+z\right)+\frac{5}{4}.3=\frac{3}{2}^{\left(đpcm\right)}\)
Bài này áp dụng BĐT này nhé , với x,y > 0 ta có :
\(\frac{1}{x}+\frac{1}{y}\ge\frac{4}{x+y}\) ( Cách chứng minh thì chuyển vế quy đồng nhé )
Áp dụng vào bài toán ta có :
\(\frac{1}{2x+y+z}=\frac{1}{4}\left(\frac{4}{\left(x+y\right)+\left(z+x\right)}\right)\le\frac{1}{4}\left(\frac{1}{x+y}+\frac{1}{z+x}\right)=\frac{1}{16}\left(\frac{4}{x+y}+\frac{4}{z+x}\right)\)
\(\le\frac{1}{16}\left(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}+\frac{1}{x}\right)\)
\(\Rightarrow\frac{1}{2x+y+z}\le\frac{1}{16}\left(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}+\frac{1}{x}\right)\)
Tương tự ta có :
\(\frac{1}{x+2y+z}\le\frac{1}{16}\left(\frac{1}{x}+\frac{1}{y}+\frac{1}{y}+\frac{1}{z}\right)\)
\(\frac{1}{x+y+2z}\le\frac{1}{16}\left(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}+\frac{1}{z}\right)\)
Do đó : \(\frac{1}{2x+y+z}+\frac{1}{x+2y+z}+\frac{1}{x+y+2z}\le\frac{1}{16}\left(\frac{4}{x}+\frac{4}{y}+\frac{4}{z}\right)=\frac{1}{4}\left(x+y+z\right)=1\)
Dấu "=" xảy ra \(\Leftrightarrow x=y=z=\frac{3}{4}\) (đpcm)
Ta có: \(\frac{1}{2x+y+z}\le\frac{1}{4}\left(\frac{1}{x+y}+\frac{1}{x+z}\right)\le\frac{1}{16}\left(\frac{2}{x}+\frac{1}{y}+\frac{1}{z}\right)\)
Tương tự: \(\frac{1}{x+2y+z}\le\frac{1}{16}\left(\frac{1}{x}+\frac{2}{y}+\frac{1}{z}\right)\)
\(\frac{1}{x+y+2z}\le\frac{1}{16}\left(\frac{1}{x}+\frac{1}{y}+\frac{2}{z}\right)\)
Cộng vế theo vế có: \(VT\le\frac{1}{16}\left(\frac{4}{x}+\frac{4}{y}+\frac{4}{z}\right)=1\)
...
=>\(\left(x+y+z\right)\left(\frac{x}{y+z}+\frac{y}{z+x}+\frac{z}{x+y}\right)=1\)
=>\(\frac{x^2}{y+z}+\frac{xy}{y+z}+\frac{xz}{y+z}+\frac{xy}{z+x}+\frac{y^2}{z+x}+\frac{yz}{z+x}+\frac{xz}{x+y}+\frac{yz}{x+y}+\frac{z^2}{x+y}=1\)
=>\(\left(\frac{x^2}{y+z}+\frac{y^2}{z+x}+\frac{z^2}{x+y}\right)+\left(\frac{xy}{y+z}+\frac{xz}{y+z}+\frac{xy}{z+x}+\frac{yz}{z+x}+\frac{xz}{x+y}+\frac{yz}{x+y}\right)=1\)
=>\(\left(\frac{x^2}{y+z}+\frac{y^2}{z+x}+\frac{z^2}{x+y}\right)+\left(\frac{xy+xz}{y+z}+\frac{xy+yz}{z+x}+\frac{xz+yz}{x+y}\right)=1\)
=>\(\left(\frac{x^2}{y+z}+\frac{y^2}{z+x}+\frac{z^2}{x+y}\right)+\left(x+y+z\right)=1\)
=>\(\left(\frac{x^2}{y+z}+\frac{y^2}{z+x}+\frac{z^2}{x+y}\right)+1=1\)
=>\(\frac{x^2}{y+z}+\frac{y^2}{z+x}+\frac{z^2}{x+y}=0\)
