Cho x,y,z>0 . Cm : \(\dfrac{1}{x}+\dfrac{1}{y}+\dfrac{1}{z}\ge\dfrac{16}{x+y+z}\)
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Cho x,y,z>0. CM: \(\dfrac{xy}{z^2\left(x+y\right)}+\dfrac{yz}{x^2\left(y+z\right)}+\dfrac{zx}{y^2\left(z+x\right)}\ge\dfrac{1}{2}\left(\dfrac{1}{x}+\dfrac{1}{y}+\dfrac{1}{z}\right)\)
\(VT=\dfrac{\left(\dfrac{1}{z}\right)^2}{\dfrac{1}{x}+\dfrac{1}{y}}+\dfrac{\left(\dfrac{1}{x}\right)^2}{\dfrac{1}{y}+\dfrac{1}{z}}+\dfrac{\left(\dfrac{1}{y}\right)^2}{\dfrac{1}{x}+\dfrac{1}{z}}\ge\dfrac{\left(\dfrac{1}{x}+\dfrac{1}{y}+\dfrac{1}{z}\right)^2}{2\left(\dfrac{1}{x}+\dfrac{1}{y}+\dfrac{1}{z}\right)}=\dfrac{1}{2}\left(\dfrac{1}{x}+\dfrac{1}{y}+\dfrac{1}{z}\right)\)
Dâu "=" xảy ra khi \(x=y=z\)
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CM: \(\dfrac{x}{y^2}+\dfrac{y}{z^2}+\dfrac{z}{x^2}\ge\dfrac{1}{x}+\dfrac{1}{y}+\dfrac{1}{z}\) biêt x,y,z>0
\(\sum\dfrac{1}{x}\cdot\sum\dfrac{x}{y^2}\ge\sum^2\dfrac{1}{x}\)(bunhia)
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với x,y,z>0 và \(x+y+z\ge\dfrac{1}{x}+\dfrac{1}{y}+\dfrac{1}{z}\)
chứng minh đẳng thức \(x+y+z\ge\dfrac{3}{x+y+z}+\dfrac{2}{xyz}\)
\(\Rightarrow\left(x+y+z\right)^2\ge\left(\dfrac{1}{x}+\dfrac{1}{y}+\dfrac{1}{z}\right)^2\ge3\left(\dfrac{1}{xy}+\dfrac{1}{yz}+\dfrac{1}{xz}\right)=\dfrac{3\left(x+y+z\right)}{xyz}\Rightarrow x+y+z\ge\dfrac{3}{xyz}\)
\(x+y+z=\dfrac{x+y+z}{3}+\dfrac{2\left(x+y+z\right)}{3}\ge\dfrac{1}{3}\left(\dfrac{1}{x}+\dfrac{1}{y}+\dfrac{1}{z}\right)+\dfrac{2}{3}.\dfrac{3}{xyz}\ge\dfrac{1}{3}\left(\dfrac{9}{x+y+z}\right)+\dfrac{2}{xyz}=\dfrac{3}{x+y+z}+\dfrac{2}{xyz}\left(đpcm\right)\)
\(dấu"="xảy\) \(ra\Leftrightarrow x=y=z=1\)
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Cho x,y,z là các số dương. CMR:
a) (x+y+z)(dfrac{1}{x+y}+dfrac{1}{y+z}+dfrac{1}{x+z}) ≥dfrac{9}{2}
b) (x+y+z+t)(dfrac{1}{x+y+z}+dfrac{1}{y+z+t}+dfrac{1}{z+t+x}+dfrac{1}{t+x+y}) ≥dfrac{16}{3}
c) dfrac{x^2}{y+z}+dfrac{y^2}{z+x}+dfrac{z^2}{x+y} ≥dfrac{1}{2}left(a+b+cright)
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Cho x,y,z là các số dương. CMR:
a) (x+y+z)(\(\dfrac{1}{x+y}+\dfrac{1}{y+z}+\dfrac{1}{x+z}\)) ≥\(\dfrac{9}{2}\)
b) (x+y+z+t)(\(\dfrac{1}{x+y+z}+\dfrac{1}{y+z+t}+\dfrac{1}{z+t+x}+\dfrac{1}{t+x+y}\)) ≥\(\dfrac{16}{3}\)
c) \(\dfrac{x^2}{y+z}+\dfrac{y^2}{z+x}+\dfrac{z^2}{x+y}\) ≥\(\dfrac{1}{2}\left(a+b+c\right)\)
câu c là \(\dfrac{1}{2}\)(x+y+z) nhé, mih chép nhầm
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dfrac{x^2}{y+z}+dfrac{y^2}{z+x}+dfrac{z^2}{x+y}
do x,y,z≥0 nên x2≥0 , y+z≥0
áp dụng bất đẳng thức cosi cho 2 số dương dfrac{x^2}{y+z} và y+z/4
x^2/y+z +(y+z)/4≥2sqrt{dfrac{x^2}{y+z}.dfrac{left(y+zright)}{4}} x (1)
y^2/x+z+(x+z)/4≥2sqrt{dfrac{y^2}{x+z}.dfrac{x+z}{4}} y (2)
z^2/y+x+(y+x)/4≥2sqrt{dfrac{z^2}{y+x}.dfrac{y+x}{4}} z (3)
từ (1)(2)(3)
➜dfrac{x^2}{y+z}+dfrac{y^2}{z+x}+dfrac{z^2}{x+y}+(y+z/4)+(z+x)/4+(x+y)/4 ≥ x+y+z
⇔dfrac{x^2}{y+z}+dfrac{y^2}{z+x}+dfrac{z^2}{x+y} +(a+b+c)/2 ≥x+...
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\(\dfrac{x^2}{y+z}+\dfrac{y^2}{z+x}+\dfrac{z^2}{x+y}\)
do x,y,z≥0 nên x2≥0 , y+z≥0
áp dụng bất đẳng thức cosi cho 2 số dương \(\dfrac{x^2}{y+z}\) và y+z/4
x^2/y+z +(y+z)/4≥2\(\sqrt{\dfrac{x^2}{y+z}.\dfrac{\left(y+z\right)}{4}}\) =x (1)
y^2/x+z+(x+z)/4≥2\(\sqrt{\dfrac{y^2}{x+z}.\dfrac{x+z}{4}}\) =y (2)
z^2/y+x+(y+x)/4≥2\(\sqrt{\dfrac{z^2}{y+x}.\dfrac{y+x}{4}}\) =z (3)
từ (1)(2)(3)
➜\(\dfrac{x^2}{y+z}+\dfrac{y^2}{z+x}+\dfrac{z^2}{x+y}\)+(y+z/4)+(z+x)/4+(x+y)/4 ≥ x+y+z
⇔\(\dfrac{x^2}{y+z}+\dfrac{y^2}{z+x}+\dfrac{z^2}{x+y}\) +(a+b+c)/2 ≥x+y+z
⇔\(\dfrac{x^2}{y+z}+\dfrac{y^2}{z+x}+\dfrac{z^2}{x+y}\) ≥ (x+y+z)/2
⇔\(\dfrac{x^2}{y+z}+\dfrac{y^2}{z+x}+\dfrac{z^2}{x+y}\) ≥1 (vì x+y+z=2)
vậy giá trị nhỏ nhất của \(\dfrac{x^2}{y+z}+\dfrac{y^2}{z+x}+\dfrac{z^2}{x+y}\) =1
Nham ko phai Nesbit, Cauchy-Schwarz ra luon
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12 tháng 3 2021 lúc 22:40
Cho các số x, y, z dương thỏa mãn: \(\dfrac{1}{x^2}+\dfrac{1}{y^2}+\dfrac{1}{z^2}=3\)
Cmr: \(\dfrac{1}{\left(2x+y+z\right)^2}+\dfrac{1}{\left(2y+z+x\right)^2}+\dfrac{1}{\left(2z+x+y\right)^2}\ge\dfrac{3}{16}\)
Thay $x=\sqrt{\frac{1}{2,5}}; y=z=\sqrt{\frac{1}{0,25}}$ ta thấy đề sai bạn nhé!
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Cho x>0, y>0, z>0 và \(\dfrac{1}{x}+\dfrac{1}{y}+\dfrac{1}{z}=4\). CM: \(\dfrac{1}{2x+y+z}+\dfrac{1}{x+2y+z}+\dfrac{1}{x+y+2z}\le1\)
Cho x, y, z > 0 và xyz=1. CMR :
\(\dfrac{x^2}{1+y}+\dfrac{y^2}{1+z}+\dfrac{z^2}{1+z}\ge\dfrac{3}{2}\)
Đề sai nhé, \(\dfrac{z^2}{x+1}\) mới đúng nha
\(\dfrac{x^2}{y+1}+\dfrac{y^2}{z+1}+\dfrac{z^2}{x+1}\ge\dfrac{\left(x+y+z\right)^2}{x+y+z+3}\left(\text{Svácxơ}\right)\)
\(\ge\dfrac{\left(x+y+z\right)^2}{2\left(x+y+z\right)}=\dfrac{x+y+z}{2}\ge\dfrac{3}{2}\)
Dấu "=" xảy ra \(\Leftrightarrow x=y=z=1\)
Ta có: \(x+y+z\ge3\sqrt[3]{xyz}=3\)
\(\Rightarrow x+y+z+3\le2\left(x+y+z\right)\)
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Cho x,y,z > 0 có xy+yz+xz = 3xyz CMR : \(\dfrac{x^3}{x^2+z}+\dfrac{y^3}{y^2+x}+\dfrac{z^3}{z^2+y}\ge\dfrac{1}{2}\left(\dfrac{1}{x}+\dfrac{1}{y}+\dfrac{1}{z}\right)\)
Lời giải:
Ta có: \(xy+yz+xz=3xyz\Rightarrow \frac{1}{x}+\frac{1}{y}+\frac{1}{z}=3\)
Mà theo BĐT Cauchy-Schwarz: \(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}\geq \frac{9}{x+y+z}\)
Do đó: \(3\geq \frac{9}{x+y+z}\Rightarrow x+y+z\geq 3\)
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Ta có: \(\text{VT}=x-\frac{xz}{x^2+z}+y-\frac{xy}{y^2+x}+z-\frac{yz}{z^2+y}\)
\(=(x+y+z)-\left(\frac{xy}{y^2+x}+\frac{yz}{z^2+y}+\frac{xz}{x^2+z}\right)\)
\(\geq x+y+z-\frac{1}{2}\left(\frac{xy}{\sqrt{xy^2}}+\frac{yz}{\sqrt{z^2y}}+\frac{xz}{\sqrt{x^2z}}\right)\) (AM-GM)
\(=x+y+z-\frac{1}{2}(\sqrt{x}+\sqrt{y}+\sqrt{z})\)
Tiếp tục AM-GM: \(\sqrt{x}+\sqrt{y}+\sqrt{z}\leq \frac{x+1}{2}+\frac{y+1}{2}+\frac{z+1}{2}=\frac{x+y+z+3}{2}\)
Suy ra:
\(\text{VT}\geq x+y+z-\frac{1}{2}.\frac{x+y+z+3}{2}=\frac{3}{4}(x+y+z)-\frac{3}{4}\)
\(\geq \frac{9}{4}-\frac{3}{4}=\frac{3}{2}=\frac{1}{2}\left(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}\right)\)
Ta có đpcm
Dấu bằng xảy ra khi $x=y=z=1$
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