Cho các số thực dương x,y,z thỏa mãn x+y+z=3. Chứng minh rằng: \(\frac{1}{x^2+x}+\frac{1}{y^2+y}+\frac{1}{z^2+z}>=\frac{3}{2}\)
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23 tháng 2 2019
\(taco:\)
\(\frac{1}{x\left(x+1\right)}+\frac{x}{2}+\frac{x+1}{4}\ge3\sqrt[3]{\frac{1}{x\left(x+1\right)}.\frac{x}{2}.\frac{x+1}{4}}=\frac{3}{2}\)
\(\frac{1}{y\left(y+1\right)}+\frac{y}{2}+\frac{y+1}{2}\ge3\sqrt[3]{\frac{1}{y\left(y+1\right)}.\frac{y}{2}.\frac{y+1}{4}}=\frac{3}{2}\)
\(\frac{1}{z\left(z+1\right)}+\frac{z}{2}+\frac{z+1}{4}\ge3\sqrt[3]{\frac{1}{z\left(z+1\right)}.\frac{z}{2}.\frac{z+1}{4}}=\frac{3}{2}\)
\(\frac{1}{x\left(x+1\right)}+\frac{1}{y\left(y+1\right)}+\frac{1}{z\left(z+1\right)}+\frac{x+y+z}{2}+\frac{x+y+z+3}{4}\ge\frac{3}{2}+\frac{3}{2}+\frac{3}{2}\)
\(\Leftrightarrow\frac{1}{x^2+x}+\frac{1}{y^2+y}+\frac{1}{z^2+z}+\frac{3}{2}+\frac{3}{2}\ge\frac{9}{2}\)
\(\Leftrightarrow\frac{1}{x^2+x}+\frac{1}{y^2+y}+\frac{1}{z^2+z}\ge\frac{3}{2}\left(dpcm\right)\)
^^
T
23 tháng 2 2019
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:
\(VT\ge-\frac{3}{4}\left(x+y+z\right)+\frac{5}{4}.3=\frac{3}{2}^{\left(đpcm\right)}\)
2 tháng 5 2016
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
VD
3 tháng 5 2016
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.
25 tháng 5 2018
\(x^4y+x^2y-x^2y=x^2y\left(x^2+1\right)-x^2y.\)
\(\hept{\begin{cases}\frac{x^2y\left(x^2+1\right)-x^2y}{\left(x^2+1\right)}=x^2y-\frac{x^2y}{\left(x^2+1\right)}\\\frac{y^2z\left(y^2+1\right)-y^2z}{\left(y^2+1\right)}=y^2z-\frac{y^2z}{\left(y^2+1\right)}\\\frac{z^2x\left(z^2+1\right)-z^2x}{\left(z^2+1\right)}=z^2x-\frac{z^2x}{\left(z^2+1\right)}\end{cases}}Vt\ge x^2y+y^2z+z^2x-\left(\frac{x^2y}{x^2+1}+\frac{y^2z}{y^2+1}+\frac{z^2x}{z^2+1}\right)\)
\(\hept{\begin{cases}x^2+1\ge2x\\y^2+1\ge2y\\z^2+1\ge2z\end{cases}\Leftrightarrow\hept{\begin{cases}-\frac{x^2y}{x^2+1}\ge\frac{x^2y}{2x}=\frac{xy}{2}\\\frac{y^2z}{2y}=\frac{yz}{2}\\\frac{z^2x}{2z}=\frac{xz}{2}\end{cases}\Leftrightarrow}VT\ge x^2y+y^2z+z^2x-\left(\frac{xy+yz+zx}{2}\right)}\)
\(x^2y+y^2z+z^2x\ge3\sqrt[3]{x^3y^3z^3}=3\)
\(VT\ge3-\frac{\left(xy+yz+zx\right)}{2}\)
t chỉ làm dc đến đây thôi :))
27 tháng 5 2018
Từ \(VT\ge x^2y+y^2z+z^2x-\left(\frac{xy+yz+zx}{2}\right)\)ta có:
\(x^2y+x^2y+y^2z=x^2y+x^2y+\frac{y}{x}\ge3xy\)(áp dụng BĐT Cauchy)
Tương tự : \(y^2z+y^2z+z^2x\ge3yz\); \(z^2x+z^2x+x^2y\ge3zx\)
Cộng vế theo vế suy ra : \(3\left(x^2y+y^2z+z^2x\right)\ge3\left(xy+yz+zx\right)\)
\(\Leftrightarrow x^2y+y^2z+z^2x\ge xy+yz+zx\)
\(\Leftrightarrow VT\ge\frac{xy+yz+zx}{2}\ge\frac{3\sqrt[3]{x^2y^2z^2}}{2}=\frac{3}{2}\)
Dấu '=' xảy ra khi x = y = z = 1
3 tháng 4 2020
Ta có \(1+x^2=x^2+xy+yz+xz=\left(x+z\right)\left(x+y\right)\)
Khi đó BĐT <=>
\(\frac{1}{\left(x+y\right)\left(x+z\right)}+\frac{1}{\left(y+z\right)\left(x+z\right)}+\frac{1}{\left(x+y\right)\left(y+z\right)}\ge\frac{2}{3}\left(\frac{x}{\sqrt{\left(x+z\right)\left(x+y\right)}}+...\right)\)
<=> \(\frac{x+y+z}{\left(x+y\right)\left(y+z\right)\left(x+z\right)}\ge\frac{1}{3}\left(\frac{x\sqrt{y+z}+y\sqrt{x+z}+z\sqrt{x+y}}{\sqrt{\left(x+y\right)\left(y+z\right)\left(x+z\right)}}\right)^3\)
<=>\(\left(x+y+z\right)\sqrt{\left(x+y\right)\left(x+z\right)\left(y+z\right)}\ge\frac{1}{3}\left(x\sqrt{y+z}+y\sqrt{x+z}+z\sqrt{x+y}\right)^3\)
<=> \(\left(x+y+z\right)\sqrt{\left(x+y\right)\left(y+z\right)\left(x+z\right)}\ge\frac{1}{3}\left(\sqrt{x\left(1-yz\right)}+\sqrt{y\left(1-xz\right)}+\sqrt{z\left(1-xy\right)}\right)^3\)(1)
Xét \(\left(x+y\right)\left(y+z\right)\left(x+z\right)\ge\frac{8}{9}\left(x+y+z\right)\left(xy+yz+xz\right)\)
<=> \(9\left[xy\left(x+y\right)+yz\left(y+z\right)+xz\left(x+z\right)+2xyz\right]\ge8\left(xy\left(x+y\right)+xz\left(x+z\right)+yz\left(y+z\right)+3xyz\right)\)
<=> \(xy\left(y+x\right)+yz\left(y+z\right)+xz\left(x+z\right)\ge6xyz\)
<=> \(x\left(y-z\right)^2+z\left(x-y\right)^2+y\left(x-z\right)^2\ge0\)luôn đúng
Khi đó (1) <=>
\(\left(x+y+z\right).\frac{2\sqrt{2}}{3}.\sqrt{x+y+z}\ge\frac{1}{3}\left(\sqrt{x\left(1-yz\right)}+....\right)^3\)
<=> \(\sqrt{2\left(x+y+z\right)}\ge\sqrt{x\left(1-yz\right)}+\sqrt{y\left(1-xz\right)}+\sqrt{z\left(1-xy\right)}\)
Áp dụng buniacopxki cho vế phải ta có
\(\sqrt{x\left(1-yz\right)}+\sqrt{y\left(1-xz\right)}+\sqrt{z\left(1-xy\right)}\le\sqrt{\left(x+y+z\right)\left(3-xy-yz-xz\right)}\)
\(=\sqrt{2\left(x+y+z\right)}\)
=> BĐT được CM
Dấu bằng xảy ra khi \(x=y=z=\frac{1}{\sqrt{3}}\)
27 tháng 10 2020
Nếu x; y; z là các số nguyên dương mà x y z = 1 => x = y = z = 1
=> bất đẳng thức luôn xảy ra dấu bằng
Sửa đề 1 chút cho z; y; x là các số dương
Ta có: \(\frac{x^2}{y+1}+\frac{y+1}{4}\ge2\sqrt{\frac{x^2}{y+1}.\frac{y+1}{4}}=x\)
=> \(\frac{x^2}{y+1}\ge x-\frac{y+1}{4}\)
Tương tự:
\(\frac{x^2}{y+1}+\frac{y^2}{z+1}+\frac{z^2}{z+1}\ge x+y+z-\frac{y+1}{4}-\frac{z+1}{4}-\frac{x+1}{4}\)
\(=\frac{3}{4}\left(x+y+z\right)-\frac{3}{4}\ge\frac{3}{4}.3\sqrt[3]{xyz}-\frac{3}{4}=\frac{3}{2}\)
Dấu "=" xảy ra <=> x = y = z = 1
22 tháng 6 2020
Theo AM - GM và Bunhiacopski ta có được
\(x^2+y^2\ge\frac{\left(x+y\right)^2}{2};\frac{1}{x^2}+\frac{1}{y^2}\ge\frac{2}{xy}\ge\frac{8}{\left(x+y\right)^2}\)
Khi đó \(LHS\ge\left[\frac{\left(x+y\right)^2}{2}+z^2\right]\left[\frac{8}{\left(x+y\right)^2}+\frac{1}{z^2}\right]\)
\(\)\(=\left[\frac{1}{2}+\left(\frac{z}{x+y}\right)^2\right]\left[8+\left(\frac{x+y}{z}\right)^2\right]\)
Đặt \(t=\frac{z}{x+y}\ge1\)
Khi đó:\(LHS\ge\left(\frac{1}{2}+t^2\right)\left(8+\frac{1}{t^2}\right)=8t^2+\frac{1}{2t^2}+5\)
\(=\left(\frac{1}{2t^2}+\frac{t^2}{2}\right)+\frac{15t^2}{2}+5\ge\frac{27}{2}\)
Vậy ta có đpcm
T
23 tháng 6 2020
Ta có:
\(VT-VP=\frac{\left(x^2+y^2\right)\left(\Sigma xy\right)\left(\Sigma x\right)\left[z\left(x+y\right)-xy\right]\left(z-x-y\right)}{x^2y^2z^2\left(x+y\right)^2}+\frac{\left(x-y\right)^2\left(2x+y\right)^2\left(x+2y\right)^2}{2x^2y^2\left(x+y\right)^2}\ge0\)
Vì \(z\left(x+y\right)-xy\ge\left(x+y\right)^2-xy\ge4xy-xy>0\)
KN
9 tháng 12 2020
Ta có: \(x+y+z=xyz\Rightarrow x=\frac{x+y+z}{yz}\Rightarrow x^2=\frac{x^2+xy+xz}{yz}\Rightarrow x^2+1=\frac{\left(x+y\right)\left(x+z\right)}{yz}\)\(\Rightarrow\sqrt{x^2+1}=\sqrt{\frac{\left(x+y\right)\left(x+z\right)}{yz}}\le\frac{\frac{x+y}{y}+\frac{x+z}{z}}{2}=1+\frac{x}{2}\left(\frac{1}{y}+\frac{1}{z}\right)\)\(\Rightarrow\frac{1+\sqrt{1+x^2}}{x}\le\frac{2+\frac{x}{2}\left(\frac{1}{y}+\frac{1}{z}\right)}{x}=\frac{2}{x}+\frac{1}{2}\left(\frac{1}{y}+\frac{1}{z}\right)\)
Tương tự: \(\frac{1+\sqrt{1+y^2}}{y}\le\frac{2}{y}+\frac{1}{2}\left(\frac{1}{z}+\frac{1}{x}\right)\); \(\frac{1+\sqrt{1+z^2}}{z}\le\frac{2}{z}+\frac{1}{2}\left(\frac{1}{x}+\frac{1}{y}\right)\)
Cộng theo vế ba bất đẳng thức trên, ta được: \(\frac{1+\sqrt{1+x^2}}{x}+\frac{1+\sqrt{1+y^2}}{y}+\frac{1+\sqrt{1+z^2}}{z}\le3\left(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}\right)=3.\frac{xy+yz+zx}{xyz}\)\(\le3.\frac{\frac{\left(x+y+z\right)^2}{3}}{xyz}=\frac{\left(x+y+z\right)^2}{xyz}=\frac{\left(xyz\right)^2}{xyz}=xyz\)
Đẳng thức xảy ra khi \(x=y=z=\sqrt{3}\)
25 tháng 4 2021
\(\frac{x^2}{y+1}+\frac{y+1}{4}\ge x;\frac{y^2}{z+1}+\frac{z+1}{4}\ge y;\frac{z^2}{x+1}+\frac{x+1}{4}\ge z\)
\(\Rightarrow VT\ge\frac{3}{4}\left(x+y+z\right)-\frac{3}{4}\ge\frac{3}{4}.2=\frac{3}{2}\)
Áp dụng BĐT Cauchy cho 3 số dương, ta được:
\(\frac{1}{x\left(x+1\right)}+\frac{x}{2}+\frac{x+1}{4}\ge\sqrt[3]{\frac{1}{x\left(x+1\right)}.\frac{x}{2}.\frac{x+1}{4}}=3.\sqrt{\frac{1}{4}}=\frac{3}{2}\)
\(\frac{1}{y\left(y+1\right)}+\frac{y}{2}+\frac{y+1}{4}\ge\sqrt[3]{\frac{1}{y\left(y+1\right)}.\frac{y}{2}.\frac{y+1}{4}}=3.\sqrt{\frac{1}{4}}=\frac{3}{2}\)
\(\frac{1}{z\left(z+1\right)}+\frac{z}{2}+\frac{z+1}{4}\ge\sqrt[3]{\frac{1}{z\left(z+1\right)}.\frac{z}{2}.\frac{z+1}{4}}=3.\sqrt{\frac{1}{4}}=\frac{3}{2}\)
\(\Rightarrow\frac{1}{x\left(x+1\right)}+\frac{x}{2}+\frac{x+1}{4}\)\(+\frac{1}{y\left(y+1\right)}+\frac{y}{2}+\frac{y+1}{4}\)
\(+\frac{1}{z\left(z+1\right)}+\frac{z}{2}+\frac{z+1}{4}\ge\frac{3}{2}.3=\frac{9}{2}\)
\(\Leftrightarrow\frac{1}{x^2+x}+\frac{1}{y^2+y}+\frac{1}{z^2+z}+\frac{x+y+z}{2}+\frac{x+y+z+3}{4}\ge\frac{9}{2}\)
\(\Leftrightarrow\frac{1}{x^2+x}+\frac{1}{y^2+y}+\frac{1}{z^2+z}+\frac{3}{2}+\frac{3}{2}\ge\frac{9}{2}\)
\(\Leftrightarrow\frac{1}{x^2+x}+\frac{1}{y^2+y}+\frac{1}{z^2+z}\ge\frac{3}{2}\left(đpcm\right)\)