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NV
14 tháng 1 2021

\(T=\dfrac{\left(xy\right)^2}{zx+zy}+\dfrac{\left(yz\right)^2}{xy+xz}+\dfrac{\left(zx\right)^2}{yx+yz}\ge\dfrac{xy+yz+zx}{2}\ge\dfrac{3}{2}\sqrt[3]{\left(xyz\right)^2}=\dfrac{3}{2}\)

 

NV
17 tháng 8 2021

\(\left(xy+yz+zx\right)^2\ge3xyz\left(x+y+z\right)=9\Rightarrow xy+yz+zx\ge3\)

\(2\left(x^2+y^2\right)-xy\ge\left(x+y\right)^2-\dfrac{1}{4}\left(x+y\right)^2=\dfrac{3}{4}\left(x+y\right)^2\)

Tương tự và nhân vế với vế:

\(VT\ge\dfrac{27}{64}\left[\left(x+y\right)\left(y+z\right)\left(z+x\right)\right]^2\)

Mặt khác ta có:

\(\left(x+y\right)\left(y+z\right)\left(z+x\right)=\left(x+y+z\right)\left(xy+yz+zx\right)-xyz\)

\(\ge\left(x+y+z\right)\left(xy+yz+zx\right)-\sqrt[3]{xyz}.\sqrt[3]{xy.yz.zx}\)

\(\ge\left(x+y+z\right)\left(xy+yz+xz\right)-\dfrac{1}{9}\left(x+y+z\right)\left(xy+yz+zx\right)\)

\(=\dfrac{8}{9}\left(x+y+z\right)\left(xy+yz+zx\right)\ge\dfrac{8}{9}\sqrt{3\left(xy+yz+zx\right)}.\left(xy+yz+zx\right)\)

\(\Rightarrow VT\ge\dfrac{27}{64}.\dfrac{64}{81}.3\left(xy+yz+zx\right)^3\ge3^3=27\) (đpcm)

17 tháng 8 2021

em cảm ơn

 

19 tháng 5 2021

\(gt\Leftrightarrow\dfrac{1}{\sqrt{x}}+\dfrac{1}{\sqrt{y}}+\dfrac{1}{\sqrt{z}}=1\)

\(P=\dfrac{1}{xyz}\left(x\sqrt{2y^2+yz+2z^2}+y\sqrt{2x^2+xz+2z^2}+z\sqrt{2y^2+xy+2x^2}\right)\)

\(=\dfrac{1}{xyz}\left(x\sqrt{\dfrac{5}{4}\left(y+z\right)^2+\dfrac{3}{4}\left(y-z\right)^2}+y\sqrt{\dfrac{5}{4}\left(x+z\right)^2+\dfrac{3}{4}\left(x-z\right)^2}+z\sqrt{\dfrac{5}{4}\left(x+y\right)^2+\dfrac{3}{4}\left(x-y\right)^2}\right)\)

\(\ge\dfrac{1}{xyz}\left[x.\dfrac{\sqrt{5}\left(z+y\right)}{2}+y.\dfrac{\sqrt{5}\left(x+z\right)}{2}+z.\dfrac{\sqrt{5}\left(x+y\right)}{2}\right]\)

\(=\dfrac{\sqrt{5}\left(z+y\right)}{2yz}+\dfrac{\sqrt{5}\left(x+z\right)}{2xz}+\dfrac{\sqrt{5}\left(x+y\right)}{2xy}\)

\(=\dfrac{\sqrt{5}}{3}\left(1+1+1\right)\left(\dfrac{1}{x}+\dfrac{1}{y}+\dfrac{1}{z}\right)\ge\dfrac{\sqrt{5}}{3}\left(\dfrac{1}{\sqrt{x}}+\dfrac{1}{\sqrt{y}}+\dfrac{1}{\sqrt{z}}\right)^2=\dfrac{\sqrt{5}}{3}\) (bunhia)

Dấu = xảy ra khi \(x=y=z=9\)

19 tháng 5 2021

 Thấy : \(\sqrt{2y^2+yz+2z^2}=\sqrt{\dfrac{5}{4}\left(y+z\right)^2+\dfrac{3}{4}\left(y-z\right)^2}\ge\dfrac{\sqrt{5}}{2}\left(y+z\right)>0\) 

CMTT : \(\sqrt{2x^2+xz+2z^2}\ge\dfrac{\sqrt{5}}{2}\left(x+z\right)\)  ; \(\sqrt{2y^2+xy+2x^2}\ge\dfrac{\sqrt{5}}{2}\left(x+y\right)\) 

Suy ra : \(P\ge\dfrac{1}{xyz}.\dfrac{\sqrt{5}}{2}\left[x\left(y+z\right)+y\left(x+z\right)+z\left(x+y\right)\right]\)

\(\Rightarrow P\ge\sqrt{5}\left(\dfrac{1}{x}+\dfrac{1}{y}+\dfrac{1}{z}\right)\) 

Ta có : \(\sqrt{xy}+\sqrt{yz}+\sqrt{xz}=\sqrt{xyz}\Leftrightarrow\dfrac{1}{\sqrt{x}}+\dfrac{1}{\sqrt{y}}+\dfrac{1}{\sqrt{z}}=1\) 

Mặt khác :   \(\dfrac{1}{x}+\dfrac{1}{y}+\dfrac{1}{z}\ge\dfrac{\left(\dfrac{1}{\sqrt{x}}+\dfrac{1}{\sqrt{y}}+\dfrac{1}{\sqrt{z}}\right)^2}{3}=\dfrac{1}{3}\)

Suy ra : \(P\ge\dfrac{\sqrt{5}}{3}\)

" = " \(\Leftrightarrow x=y=z=9\)

21 tháng 2 2019

C/m: \(\sqrt{2x^2+xy+2y^2}\ge\dfrac{\sqrt{5}}{2}\left(x+y\right)\)

\(\Rightarrow2x^2+xy+2y^2\ge\dfrac{5}{4}\left(x^2+2xy+y^2\right)\)

\(\Leftrightarrow8x^2+4xy+8y^2\ge5x^2+10xy+5y^2\)

\(\Leftrightarrow3\left(x-y\right)^2\ge0\left(LĐ\right)\)

Vậy \(\sqrt{2x^2+xy+2y^2}\ge\dfrac{\sqrt{5}}{2}\left(x+y\right)\)

CMTT: \(\sqrt{2y^2+yz+2z^2}\ge\dfrac{\sqrt{5}}{2}\left(y+z\right)\);

\(\sqrt{2z^2+zx+2x^2}\ge\dfrac{\sqrt{5}}{2}\left(x+z\right)\)

Vậy H=\(\sqrt{2x^2+xy+2y^2}+\sqrt{2y^2+yz+2z^2}+\sqrt{2z^2+xz+2z^2}\ge\sqrt{5}\left(x+y+z\right)=2019\)Hmin=2019\(\Leftrightarrow x=y=z=\dfrac{\dfrac{2019}{\sqrt{5}}}{3}\)

21 tháng 2 2019

Khos quas

16 tháng 5 2019

Có \(xy+yz+zx=xyz\)\(\Leftrightarrow\)\(\frac{xy+yz+zx}{xyz}=1\)\(\Leftrightarrow\)\(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}=1\)

\(\frac{x^2y}{y+2x}+\frac{y^2z}{z+2y}+\frac{z^2x}{x+2z}=\frac{1}{\frac{1}{x^2}+\frac{2}{xy}}+\frac{1}{\frac{1}{y^2}+\frac{2}{yz}}+\frac{1}{\frac{1}{z^2}+\frac{2}{zx}}\ge\frac{9}{\frac{1}{x^2}+\frac{1}{y^2}+\frac{1}{z^2}+2\left(\frac{1}{xy}+\frac{1}{yz}+\frac{1}{zx}\right)}\)

\(=\frac{9}{\left(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}\right)^2}=\frac{9}{1^2}=9\)

Dấu "=" ko xảy ra \(\Rightarrow\)\(\frac{x^2y}{y+2x}+\frac{y^2z}{z+2y}+\frac{z^2x}{x+2z}>9\)

14 tháng 5 2021

Ta có x2-xy+y2=\(\left(\dfrac{x+y}{2}\right)^2+3\left(\dfrac{x-y}{2}\right)^2\)\(\ge\)\(\left(\dfrac{x+y}{2}\right)^2\)

=>\(\dfrac{\sqrt{x^2-xy+y^2}}{x+y+2z}\ge\dfrac{x+y}{2\left(x+y+2z\right)}\)(1) . Tương tự ...

Đặt \(\left\{{}\begin{matrix}y+z=a\\x+z=b\\x+y=c\end{matrix}\right.\)(a,b,c>0). Khi đó ta có :

S=\(\dfrac{1}{2}\left(\dfrac{c}{a+b}+\dfrac{b}{a+c}+\dfrac{a}{b+c}\right)\ge\dfrac{3}{4}\)  (Netbit)

NV
24 tháng 8 2021

\(\dfrac{x^3}{2y+1}+\dfrac{2y+1}{9}+\dfrac{1}{3}\ge3\sqrt[3]{\dfrac{x^3\left(2y+1\right)}{27\left(2y+1\right)}}=x\)

Tương tự: \(\dfrac{y^3}{2z+1}+\dfrac{2z+1}{9}+\dfrac{1}{3}\ge y\) ; \(\dfrac{z^3}{2x+1}+\dfrac{2x+1}{9}+\dfrac{1}{3}\ge z\)

Cộng vế:

\(VT+\dfrac{2\left(x+y+z\right)+3}{9}+1\ge x+y+z\)

\(\Rightarrow VT\ge\dfrac{7}{9}\left(x+y+z\right)-\dfrac{4}{3}\ge\dfrac{7}{9}.3\sqrt[3]{xyz}-\dfrac{4}{3}=1\) (đpcm)

Dấu "=" xảy ra khi \(x=y=z=1\)