x,y,z>0 mà x3+y3+z3=\(\frac{3\sqrt{2}}{4}\).Tìm min
\(\frac{x^2}{\sqrt{1-x^2}}+\frac{y^2}{\sqrt{1-y^2}}+\frac{z^2}{\sqrt{1-z^2}}\)
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HA
19 tháng 11 2015
gọi P là cái 1/x+1/y+1/z nha
1) (1/x+1/y+1/z)^2 = 1/x^2 + 1/y^2 + 1/z^2 + 2/(xy) + 2/(yz) + 2/(zx)
---> 3 = P + 2(x+y+z)/(xyz) = P + 2 ---> P = 1
HF
14 tháng 8 2020
\(P=\frac{\sqrt{1+x^2+y^2}}{xy}+\frac{\sqrt{1+y^2+z^2}}{yz}+\frac{\sqrt{1+z^2+x^2}}{zx}\)
\(\ge\text{Σ}\frac{\sqrt{\frac{\left(1+x+y\right)^2}{3}}}{xy}\text{=}\frac{1+x+y}{xy\sqrt{3}}\)
\(=\frac{\sqrt{3}}{3}\left(\frac{1+x+y}{xy}+\frac{1+y+z}{yz}+\frac{1+z+x}{zx}\right)\)
\(=\frac{\sqrt{3}}{3}\left(\frac{1}{xy}+\frac{1}{yz}+\frac{1}{xz}+\frac{1}{x}+\frac{1}{y}+\frac{1}{y}+\frac{1}{z}+\frac{1}{z}+\frac{1}{x}\right)\)
\(=\frac{\sqrt{3}}{3}\left(x+y+z+2xy+2yz+2zx\right)\)\(\ge\frac{\sqrt{3}}{3}\left(3\sqrt[3]{xyz}+2\cdot3\sqrt[3]{x^2y^2z^2}\right)=\frac{\sqrt{3}}{3}\left(3+6\right)=3\sqrt{3}\)
Dấu = xảy ra khi \(x=y=z=1\)
30 tháng 5 2019
Ta có \(\frac{y}{x\sqrt{y^2+1}}=\frac{y\sqrt{xz}}{x\sqrt{y\left(x+y+z\right)+xz}}=\frac{yz}{\sqrt{x\left(y+z\right).z\left(x+y\right)}}\ge\frac{2yz}{2xz+xy+yz}\)
Đặt \(a=xy,b=yz,c=xz\)=> \(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}=1\)
Khi đó
\(P\ge\frac{2b}{2c+a+b}+\frac{2c}{2a+b+c}+\frac{2a}{2b+a+c}\ge\frac{2\left(a+b+c\right)^2}{b^2+c^2+a^2+3\left(ab+bc+ac\right)}\)
Xét \(P\ge\frac{3}{2}\)
=> \(4\left(a+b+c\right)^2\ge3\left(a^2+b^2+c^2\right)+9\left(ab+bc+ac\right)\)
<=> \(a^2+b^2+c^2\ge\left(ab+bc+ac\right)\)(luôn đúng )
Vậy \(MinP=\frac{3}{2}\)khi a=b=c=3=> \(x=y=z=\sqrt{3}\)
KK
8 tháng 4 2021
Áp dụng bđt bunhiacopxki, ta có:
\(\left(x^2+\frac{1}{x^2}\right)\left(1+16\right)\ge\left(x+\frac{4}{x}\right)^2\) => \(x^2+\frac{1}{x^2}\ge\frac{\left(x+\frac{4}{x}\right)^2}{17}\)
=> \(\sqrt{x^2+\frac{1}{x^2}}\ge\frac{x+\frac{4}{x}}{\sqrt{17}}=\frac{x}{\sqrt{17}}+\frac{4}{x\sqrt{17}}\)
CMTT: \(\sqrt{y^2+\frac{1}{y^2}}\ge\frac{y}{\sqrt{17}}+\frac{4}{\sqrt{17}y}\)
\(\sqrt{z^2+\frac{1}{z^2}}\ge\frac{z}{\sqrt{17}}+\frac{4}{\sqrt{17}z}\)
=> A \(\ge\frac{x+y+z}{\sqrt{17}}+\frac{4}{\sqrt{17}}\left(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}\right)\ge\frac{x+y+z}{\sqrt{17}}+\frac{36}{\sqrt{17}\left(x+y+z\right)}\)(bđt: 1/a + 1/b + 1/c > = 9/(a+b+c)
=> A \(\ge\frac{16\left(x+y+z\right)}{\sqrt{17}}+\frac{36}{\sqrt{17}\left(x+y+z\right)}-\frac{15\left(x+y+z\right)}{\sqrt{17}}\)
A \(\ge2\sqrt{\frac{16\left(x+y+z\right)}{\sqrt{17}}\cdot\frac{36}{\sqrt{17}\left(x+y+z\right)}}-\frac{15\cdot\frac{3}{2}}{\sqrt{17}}\)(Bđt cosi + bđt: x + y + z < = 3/2)
A \(\ge\frac{48}{\sqrt{17}}-\frac{45}{2\sqrt{17}}=\frac{3\sqrt{17}}{2}\)
Dấu "=" xảy ra <=> x = y= z = 1/2
Vậy MinA = \(\frac{3\sqrt{17}}{2}\) <=> x = y = z = 1/2
31 tháng 7 2019
Ta có \(\frac{\sqrt{x^2+2y^2}}{xy}=\sqrt{\frac{1}{y^2}+\frac{2}{x^2}}\)
Áp dụng BĐT Buniacoxki ta có
\(\sqrt{\left(\frac{1}{y^2}+\frac{2}{x^2}\right)\left(1+2\right)}\ge\sqrt{\left(\frac{1}{y}+\frac{2}{x}\right)^2}=\frac{1}{y}+\frac{2}{x}\)
=> \(\sqrt{3}A\ge3\left(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}\right)=3\)
=> \(A\ge\sqrt{3}\)
\(MinA=\sqrt{3}\)khi x=y=z=3
3 tháng 10 2023
Ta có với x,y,z >0 thì:\(\dfrac{x^2}{\sqrt{1-x^2}}=\dfrac{x^3}{x\sqrt{1-x^2}}\)
Bất đẳng thức Cô si ta có:
\(x\sqrt{1-x^2}\le\dfrac{x^2+1-x^2}{2}=\dfrac{1}{2}\\ \Rightarrow\dfrac{1}{x\sqrt{1-x^2}}\ge2\\ \Rightarrow\dfrac{x^3}{x\sqrt{1-x^2}}\ge2x^3\Leftrightarrow\dfrac{x^2}{\sqrt{1-x^2}}\ge2x^3\)
Tương tự: \(\dfrac{y^2}{\sqrt{1-y^2}}\ge2y^3;\dfrac{z^2}{\sqrt{1-z^2}}\ge2z^3\)
Từ đó ta có:\(\dfrac{x^2}{\sqrt{1-x^2}}+\dfrac{y^2}{\sqrt{1-y^2}}+\dfrac{z^2}{\sqrt{1-z^2}}\ge2\left(x^3+y^3+z^3\right)=2\left(dpcm\right)\)
BH
1
DT
18 tháng 5 2019
áp dụng bất đẳng thức Cauchy ngược dấu cho 2 số không âm ta có
\(\sqrt{\left(x-1\right).1}\le\frac{x-1+1}{2}=\frac{x}{2}\Rightarrow\frac{x}{\sqrt{x-1}}\ge2.\)
\(\sqrt{\left(\frac{y}{\sqrt{2}}-\sqrt{2}\right).\sqrt{2}}\le\frac{\frac{y}{\sqrt{2}}-\sqrt{2}+\sqrt{2}}{2}=\frac{y}{2\sqrt{2}}\Rightarrow\frac{y}{\sqrt{y-2}}\ge2\sqrt{2}.\)
\(\sqrt{\left(\frac{z}{\sqrt{3}}-\sqrt{3}\right).\sqrt{3}}\le\frac{\frac{z}{\sqrt{3}}-\sqrt{3}+\sqrt{3}}{2}=\frac{z}{2\sqrt{3}}\Rightarrow\frac{z}{\sqrt{z-3}}\ge2\sqrt{3}\)
\(\Rightarrow A\ge2+2\sqrt{2}+2\sqrt{3}\)
Vậy Min \(A=2+2\sqrt{2}+2\sqrt{3}\)
\(\Leftrightarrow\hept{\begin{cases}x-1=1\\\frac{y}{\sqrt{2}}-\sqrt{2}=\sqrt{2}\\\frac{z}{\sqrt{3}}-\sqrt{3}=\sqrt{3}\end{cases}\Leftrightarrow}\hept{\begin{cases}x=2\\y=4\\z=6\end{cases}\left(tmđk\right)}\)
Ta có: \(\frac{x^2}{\sqrt{1-x^2}}+\frac{y^2}{\sqrt{1-y^2}}+\frac{z^2}{\sqrt{1-z^2}}\)
\(=\frac{x^3}{x\sqrt{1-x^2}}+\frac{y^3}{y\sqrt{1-y^2}}+\frac{z^3}{z\sqrt{1-z^2}}\)
\(\ge\frac{x^3}{\frac{x^2+1-x^2}{2}}+\frac{y^3}{\frac{y^2+1-y^2}{2}}+\frac{z^3}{\frac{z^2+1-z^2}{2}}\)
\(=2x^3+2y^3+2z^3=2\left(x^3+y^3+z^3\right)=2\cdot\frac{3\sqrt{2}}{4}=\frac{3\sqrt{2}}{2}\)
Dấu "=" xảy ra khi: \(x=y=z=\frac{1}{\sqrt{2}}\)