CMR
Nếu x,y,z > 0 thỏa mãn \(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}=4\) thì \(\frac{1}{2x+y+z}+\frac{1}{x+2y+z}+\frac{1}{x+y+2z}< =1\)
Cm:
Nếu x,y,z >0 thỏa mãn
\(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}=4\)
thì \(\frac{1}{2x+y+z}+\frac{1}{x+2y+z}+\frac{1}{x+y+2z}\le1\)
Giải:
Ta có: x, y, z >0
Áp dụng BĐT Cô si ta có:
\(\left(x+y\right)\ge2\sqrt{xy}\) và \(\left(\frac{1}{x}+\frac{1}{y}\right)\ge2\sqrt{\frac{1}{xy}}\)
=> \(\left(x+y\right)\left(\frac{1}{x}+\frac{1}{y}\right)\ge2\sqrt{xy}.2\sqrt{\frac{1}{xy}}=4\)
<=> \(\frac{1}{x}+\frac{1}{y}\ge\frac{4}{x+y}\Leftrightarrow\frac{1}{x+y}\le4\left(\frac{1}{x}+\frac{1}{y}\right)\) (*)
Áp dụng (*) ta có:
\(\frac{1}{2x+y+z}=\frac{1}{x+y+x+z}=\frac{1}{\left(x+y\right)+\left(x+z\right)}\le\frac{1}{4}\left(\frac{1}{x+y}+\frac{1}{x+z}\right)\le\frac{1}{16}\left(\frac{1}{x}+\frac{1}{y}+\frac{1}{x}+\frac{1}{z}\right)\) (1)
\(\frac{1}{x+2y+z}=\frac{1}{x+y+y+z}=\frac{1}{\left(x+y\right)+\left(y+z\right)}\le\frac{1}{4}\left(\frac{1}{x+y}+\frac{1}{y+z}\right)\le\frac{1}{16}\left(\frac{1}{x}+\frac{1}{y}+\frac{1}{y}+\frac{1}{z}\right)\) (2)
\(\frac{1}{x+y+2z}=\frac{1}{x+z+y+z}=\frac{1}{\left(x+z\right)+\left(y+z\right)}\le\frac{1}{4}\left(\frac{1}{x+z}+\frac{1}{y+z}\right)\le\frac{1}{16}\left(\frac{1}{x}+\frac{1}{z}+\frac{1}{y}+\frac{1}{z}\right)\) (3)
Cộng 2 vế của (1), (2), (3) ta có
\(\frac{1}{2x+y+z}+\frac{1}{x+2y+z}+\frac{1}{x+y+2z}\le1\) (đpcm)
Cho các số thực dương x,y,z thỏa mãn: \(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}=4\)
CMR: \(\frac{1}{2x+y+z}+\frac{1}{2y+x+z}+\frac{1}{2z+x+y}\le1\)
Với 2 số dương bất kì: ( 1 )
\(\frac{1}{x}+\frac{1}{y}\ge\frac{4}{x+y}\)Vì x và y dương nên \(\frac{1}{x}+\frac{1}{y}\ge\frac{4}{x+y}\Leftrightarrow\left(x+y\right)^2\ge4xy\)
\(\Leftrightarrow\left(x-y\right)^2\ge0\forall x;y\)
Áp dụng ( 1 ): \(\frac{4}{2x+y+z}=\frac{4}{\left(x+y\right)+\left(x+z\right)}\le\frac{1}{x+y}+\frac{1}{x+z}\)
Mà: \(\frac{1}{x+y}+\frac{1}{x+z}\le\frac{1}{4}\left(\frac{1}{x}+\frac{1}{y}+\frac{1}{x}+\frac{1}{z}\right)=\frac{1}{4}\)\(=\frac{1}{4}\left(\frac{2}{x}+\frac{1}{y}+\frac{1}{z}\right)\)
Nên: \(\frac{1}{2x+y+z}\le\frac{1}{16}\left(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}\right)\)
Tương tự ta có: \(\frac{1}{x+2y+z}\le\frac{1}{16}\left(\frac{1}{x}+\frac{2}{y}+\frac{1}{z}\right)\)
Và \(\frac{1}{x+y+2z}\le\frac{1}{16}\left(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}\right)\)
Cộng vế với vế các bất đẳng thức kết hợp với điều kiện \(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}=4\) nên ta có đpcm
cho x,y,z là các số dương thỏa mãn: \(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}=1\)
CMR: \(\frac{1}{2x+y+z}+\frac{1}{x+2y+z}+\frac{1}{x+y+2z}\le\frac{1}{4}\)
Cho x, y, z > 0 thỏa mãn: \(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}=4\)
Chứng minh: \(\frac{1}{2x+y+z}+\frac{1}{x+2y+z}+\frac{1}{x+y+2z}\le1\)
\(\hept{\begin{cases}\frac{1}{2x+y+z}=\frac{1}{x+y+x+z}\\\frac{1}{2z+y+x}=\frac{1}{z+y+x+z}\\\frac{1}{2y+x+z}=\frac{1}{x+y+y+z}\end{cases}}\)
Áp dụng BĐT \(\frac{1}{a}+\frac{1}{b}\ge\frac{4}{a+b}\)
\(\hept{\begin{cases}\frac{1}{x+y+x+z}\le\frac{1}{4}\left(\frac{1}{x+y}+\frac{1}{x+z}\right)\\\frac{1}{z+y+x+z}\le\frac{1}{4}\left(\frac{1}{x+z}+\frac{1}{y+z}\right)\\\frac{1}{x+y+y+z}\le\frac{1}{4}\left(\frac{1}{x+y}+\frac{1}{y+z}\right)\end{cases}}\)
\(\Rightarrow\frac{1}{2x+y+z}+\frac{1}{2y+z+x}+\frac{1}{2z+x+y}\le\frac{1}{2}\left(\frac{1}{x+y}+\frac{1}{y+z}+\frac{1}{z+x}\right)\)
\(\hept{\begin{cases}\frac{1}{x+y}\le\frac{1}{4}\left(\frac{1}{x}+\frac{1}{y}\right)\\\frac{1}{x+z}\le\frac{1}{4}\left(\frac{1}{x}+\frac{1}{z}\right)\\\frac{1}{z+y}\le\frac{1}{4}\left(\frac{1}{z}+\frac{1}{y}\right)\end{cases}}\Rightarrow\frac{1}{x+y}+\frac{1}{y+z}+\frac{1}{x+z}\le\frac{1}{2}\left(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}\right)\)
\(\Rightarrow\frac{1}{2x+y+z}+\frac{1}{2z+x+y}+\frac{1}{2y+z+x}\le\frac{1}{2}\cdot\frac{1}{2}\cdot4=1\)
\("="\Leftrightarrow x=y=z=0,75\)
bùi huyền ơi làm sao để k cho bạn được
Cho x, y, z>0 thỏa mãn \(\frac{1}{\sqrt{2x-1}}+\frac{1}{\sqrt{2y-1}}+\frac{1}{\sqrt{2z-1}}=3\). Tìm GTLN của:
\(A=\frac{2x+y}{x\left(x+2y\right)}+\frac{2y+z}{y\left(y+2z\right)}+\frac{2z+x}{z\left(z+2x\right)}\)
\(ĐK:x,y,z>\frac{1}{2}\)
Ta có: \(\left(x+2y\right)^2=\left(\frac{3y}{2}+\frac{y+2x}{2}\right)^2\ge4.\frac{3y}{2}.\frac{y+2x}{2}=3y\left(2x+y\right)\)\(\Rightarrow\frac{2x+y}{x+2y}\le\frac{x+2y}{3y}\Rightarrow\frac{2x+y}{x\left(x+2y\right)}\le\frac{x+2y}{3xy}=\frac{1}{3}\left(\frac{2}{x}+\frac{1}{y}\right)\)
Tương tự: \(\frac{2y+z}{y\left(y+2z\right)}\le\frac{1}{3}\left(\frac{2}{y}+\frac{1}{z}\right)\); \(\frac{2z+x}{z\left(z+2x\right)}\le\frac{1}{3}\left(\frac{2}{z}+\frac{1}{x}\right)\)
Cộng theo vế ba bất đẳng thức trên, ta được: \(VT\le\frac{1}{x}+\frac{1}{y}+\frac{1}{z}\le\frac{1}{\sqrt{2x-1}}+\frac{1}{\sqrt{2y-1}}+\frac{1}{\sqrt{2z-1}}=3\)
Đẳng thức xảy ra khi x = y = z = 1
Từ giả thiết \(\Rightarrow x,y,z>\frac{1}{2}\)
Áp dụng \(\left(a+b\right)^2\ge4ab\) taoi có:
\(\left(x+2y\right)^2=\left(\frac{2x+y}{2}+\frac{3y}{2}\right)^2\ge4\left(\frac{2x+y}{2}\right)\frac{3y}{2}\)
\(\Rightarrow\left(x+2y\right)^2\ge3y\left(2x+y\right)\)
Dấu '=' xảy ra <=> x=y
\(\Rightarrow\frac{2x+y}{x+2y}\le\frac{x+2y}{3y}\Rightarrow\frac{2x+y}{x\left(x+2y\right)}\le\frac{1}{3}\left(\frac{2}{x}+\frac{1}{y}\right)\)
Tương tự: \(\frac{2y+z}{y\left(y+2z\right)}\le\frac{1}{3}\left(\frac{2}{y}+\frac{1}{z}\right),\frac{2z+x}{z\left(z+2x\right)}\le\frac{1}{3}\left(\frac{2}{z}+\frac{1}{x}\right)\)
\(\Rightarrow A\le\frac{1}{x}+\frac{1}{y}+\frac{1}{z}\)
Dấu '=' xảy ra <=>x=y=z
Lại có: \(\sqrt{\left(2x-1\right)1}\le\frac{\left(2x-1\right)+1}{2}\Rightarrow\sqrt{\left(2x-1\right)}\le x\Rightarrow\frac{1}{x}\le\frac{1}{\sqrt{2x-1}}\)
Tương tự: \(\frac{1}{y}\le\frac{1}{\sqrt{2y-1}},\frac{1}{z}\le\frac{1}{\sqrt{2z-1}}\)
\(\Rightarrow A\le\frac{1}{\sqrt{2x-1}}+\frac{1}{\sqrt{2y-1}}+\frac{1}{\sqrt{2z-1}}=3\)
Dấu '=' xảy ra <=> x=y=z=1
Vậy GTLN của A=3 khi x=y=z=1
Cho x, y, z > 0 thỏa mãn \(x^2+y^2+z^2=1\) . CMR: \(\frac{x^3}{y+2z}+\frac{y^3}{z+2x}+\frac{z^3}{x+2y}\ge\frac{1}{3}\)
\(A=\frac{x^3}{y+2z}+\frac{y^3}{z+2x}+\frac{z^3}{x+2y}\)
\(=\frac{x^4}{xy+2zx}+\frac{y^4}{yz+2xy}+\frac{z^4}{zx+2yz}\)
\(\ge\frac{\left(x^2+y^2+z^2\right)^2}{3\left(xy+yz+zx\right)}\ge\frac{x^2+y^2+z^2}{3}=\frac{1}{3}\)
Cho x,y,z thuộc Z thỏa mãn \(\frac{1}{\sqrt{2x-1}}+\frac{1}{\sqrt{2y-1}}+\frac{1}{\sqrt{2z-1}}=3\).
Tìm GTLN của A=\(\frac{2x+y}{x\left(x+2y\right)}+\frac{2y+z}{y\left(y+2z\right)}+\frac{2z+x}{z\left(z+2x\right)}\)
\(ĐKXĐ:x,y,z\ge1\left(x,y,z\inℤ\right)\)
Ta có: \(\left(x+2y\right)^2=\left(\frac{2x+y}{2}+\frac{3y}{2}\right)^2\ge4.\frac{2x+y}{2}.\frac{3y}{2}=3y\left(2x+y\right)\)
\(\Rightarrow\frac{2x+y}{x+2y}\le\frac{x+2y}{3y}\Rightarrow\frac{2x+y}{x\left(x+2y\right)}\le\frac{1}{3}\left(\frac{2}{x}+\frac{1}{y}\right)\)
Tương tự: \(\frac{2y+z}{y\left(y+2x\right)}\le\frac{1}{3}\left(\frac{2}{y}+\frac{1}{z}\right)\);\(\frac{2z+x}{z\left(z+2x\right)}\le\frac{1}{3}\left(\frac{2}{z}+\frac{1}{x}\right)\)
\(\Rightarrow A\le\frac{1}{3}.3\left(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}\right)=\frac{1}{x}+\frac{1}{y}+\frac{1}{z}\)(*)
Ta có: \(\sqrt{2x-1}=\sqrt{\left(2x-1\right).1}\le\frac{2x-1+1}{2}=x\)(BĐT Cô - si)
\(\Rightarrow\frac{1}{x}\le\frac{1}{\sqrt{2x-1}}\)
Tương tự: \(\frac{1}{y}\le\frac{1}{\sqrt{2y-1}}\);\(\frac{1}{z}\le\frac{1}{\sqrt{2z-1}}\)
\(\Rightarrow\frac{1}{x}+\frac{1}{y}+\frac{1}{z}\le\frac{1}{\sqrt{2x-1}}+\frac{1}{\sqrt{2y-1}}+\frac{1}{\sqrt{2z-1}}=3\)(**)
Từ (*) và (**) suy ra \(A=\frac{2x+y}{x\left(x+2y\right)}+\frac{2y+z}{y\left(y+2z\right)}+\frac{2z+x}{z\left(z+2x\right)}\le3\)
Đẳng thức xảy ra khi x = y = z = 1
Từ đẳng thức đã cho suy ra \(x>\frac{1}{2};y>\frac{1}{2};z>\frac{1}{2}\)
Áp dụng\(\left(a+b\right)^2\ge4ab\)ta có \(\left(x+2y\right)^2=\left(\frac{2x+y}{2}+\frac{3y}{2}\right)^2\ge4\cdot\frac{2x+y}{2}\cdot\frac{3y}{2}\)
\(\Rightarrow\left(x+2y\right)^2\ge3y\left(2x+y\right)\)(Dấu "=" xảy ra <=> x=y)
=> \(\frac{2x+y}{x+2y}\le\frac{x+2y}{3y}\Rightarrow\frac{2x+y}{x\left(x+2y\right)}\le\frac{1}{3}\left(\frac{2}{x}+\frac{1}{y}\right)\)
Tương tự \(\hept{\begin{cases}\frac{2y+z}{y\left(y+2z\right)}\le\frac{1}{3}\left(\frac{2}{y}+\frac{1}{z}\right)\\\frac{2z+x}{z\left(z+2x\right)}\le\frac{1}{3}\left(\frac{2}{z}+\frac{1}{x}\right)\end{cases}}\)
=> \(A\le\frac{1}{x}+\frac{1}{y}+\frac{1}{z}\)(Dấu "=" xảy ra <=> x=y=z)
Ta có \(\sqrt{\left(2x-1\right)\cdot1}\le\frac{\left(2x-1\right)+1}{2}\Rightarrow\sqrt{2x-1}\le x\Rightarrow\frac{1}{x}\le\frac{1}{\sqrt{2x-1}}\)
Tương tự \(\hept{\begin{cases}\frac{1}{y}\le\frac{1}{\sqrt{2y-1}}\\\frac{1}{z}\le\frac{1}{\sqrt{2z-1}}\end{cases}}\)
Do đó \(A\le\frac{1}{\sqrt{2x-1}}+\frac{1}{\sqrt{2y-1}}+\frac{1}{\sqrt{2z-1}}=3\)(dấu "=" xảy ra <=> x=y=z=1)
Vậy MaxA=3 đạt được khi x=y=z=1
Chứng minh rằng:
a, nếu x+y=1 thì \(\frac{x}{y^3-1}+\frac{y}{x^3-1}+\frac{2\left(xy-2\right)}{x^2y^2+3}=0\)
b, nếu x,y,z khác -1 thì\(\frac{xy+2x+1}{xy+x+y+1}+\frac{yz+2y+1}{yz+z+y+1}+\frac{zx+2z+1}{zx+z+x+1}=3\)
c, Cho x,y,z đôi một khác nhau thỏa mãn\(\frac{x}{y-z}+\frac{y}{z-x}+\frac{z}{x-y}=0\) thì\(\frac{x}{\left(y-z\right)^2}+\frac{y}{\left(z-x\right)^2}+\frac{z}{\left(x-y\right)^2}=0\)
cho x,y,z > 0. Cmr: \(\frac{2x}{x^6+y^4}+\frac{2y}{y^6+z^4}+\frac{2z}{z^6+x^4}\le\frac{1}{x^4}+\frac{1}{y^4}+\frac{1}{z^4}\)
\(\frac{1}{x^4}+\frac{1}{y^4}=\frac{x^2}{x^6}+\frac{1}{y^4}\ge\frac{\left(x+1\right)^2}{x^6+y^4}\ge\frac{4x}{x^6+y^4}\)
tương tự
\(\frac{1}{y^4}+\frac{1}{z^4}\ge\frac{4y}{y^6+z^4}\);
\(\frac{1}{z^4}+\frac{1}{x^4}\ge\frac{4z}{z^6+x^4}\);
cộng vế với vế => đpcm
Dấu "=" xảy ra <=> x=y=z=1
Cách khác:
\(x^6+y^4\ge2\sqrt{x^6y^4}=2x^3y^2\)
\(\Rightarrow\frac{2}{x^6+y^4}\le\frac{1}{x^2y^2}\)
CMTT , ta có VT \(\le\frac{1}{x^2y^2}+\frac{1}{y^2z^2}+\frac{1}{z^2x^2}\)
Bổ đề: \(a^2+b^2+c^2\ge ab+bc+ca\) ( luôn đúng)
\(\Rightarrow\frac{1}{x^2y^2}+\frac{1}{y^2z^2}+\frac{1}{z^2x^2}\le\frac{1}{x^4}+\frac{1}{y^4}+\frac{1}{z^4}\)
ĐPCM
Dấu " =" xảy ra khi x=y=z=1