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cái này giống này - Here. Mỗi tội bài này Min=22 khi x=y=1/2
a: \(=\dfrac{3}{2}\sqrt{6}+\dfrac{2}{3}\sqrt{6}-2\sqrt{3}=\dfrac{13}{6}\sqrt{6}-2\sqrt{3}\)
b: \(VT=\dfrac{\sqrt{xy}\left(\sqrt{x}+\sqrt{y}\right)}{\sqrt{xy}}\cdot\left(\sqrt{x}+\sqrt{y}\right)=\left(\sqrt{x}+\sqrt{y}\right)^2\)
c: \(VT=\dfrac{\sqrt{y}}{\sqrt{x}\left(\sqrt{x}-\sqrt{y}\right)}+\dfrac{\sqrt{x}}{\sqrt{y}\left(\sqrt{y}-\sqrt{x}\right)}\)
\(=\dfrac{y-x}{\sqrt{xy}\left(\sqrt{x}-\sqrt{y}\right)}=\dfrac{-\left(\sqrt{x}+\sqrt{y}\right)}{\sqrt{xy}}\)
Áp dụng BĐT svacxơ, ta có
\(\frac{1}{x^2+xy}+\frac{1}{y^2+xy}\ge\frac{4}{x^2+y^2+2xy}=\frac{4}{\left(x+y\right)^2}\ge4\)
Dấu = xảy ra <=>x=y=1/2
^_^
\(VT=\dfrac{1}{x^2+xy}+\dfrac{1}{y^2+xy}\)
\(\ge\dfrac{4}{x^2+2xy+y^2}\)
\(=\dfrac{4}{\left(x+y\right)^2}>4\)
Cách khác.
Ta có: \(A=\dfrac{1}{x\left(x+y\right)}+\dfrac{1}{y\left(x+y\right)}=\dfrac{1}{x+y}\left(\dfrac{1}{x}+\dfrac{1}{y}\right)\)
\(=\dfrac{1}{x+y}.\dfrac{x+y}{xy}=\dfrac{1}{xy}\)
Áp dụng BĐT cho các số x,y >0 , ta có:
\(x+y\ge2\sqrt{xy}\Rightarrow\left(x+y\right)^2\ge4xy\Rightarrow\dfrac{\left(x+y\right)^2}{4}\ge xy\)
Và x+y \(\le\)1 \(\Rightarrow xy\le\dfrac{1}{4}\) \(\Rightarrow A\ge\dfrac{1}{\dfrac{1}{4}}=4\)
Dấu ''='' xảy ra khi x = y =0,5
Lời giải:
Ta có \(B=\frac{x}{y}+\frac{y}{x}+\frac{xy}{x^2+xy+y^2}=\frac{8}{9}\left(\frac{x}{y}+\frac{y}{x}\right)+\frac{1}{9}\left(\frac{x}{y}+\frac{y}{x}\right)+\frac{xy}{x^2+xy+y^2}\)
\(=\frac{8}{9}\left(\frac{x}{y}+\frac{y}{x}\right)+\frac{x^2+xy+y^2}{9xy}+\frac{xy}{x^2+xy+y^2}-\frac{1}{9}\)
Áp dụng BĐT AM-GM:
\(\frac{x}{y}+\frac{y}{x}\geq 2\)
\(\frac{x^2+xy+y^2}{9xy}+\frac{xy}{x^2+xy+y^2}\geq 2\sqrt{\frac{1}{9}}=\frac{2}{3}\)
Do đó: \(B\geq \frac{8}{9}.2+\frac{2}{3}-\frac{1}{9}=\frac{7}{3}\Leftrightarrow B_{\min}=\frac{7}{3}\)
Dấu bằng xảy ra khi $x=y$
Ta chứng minh BĐT: \(x^2+y^2+z^2\ge xy+yz+xz\)
\(\Leftrightarrow2\left(x^2+y^2+z^2\right)\ge2\left(xy+yz+xz\right)\)
\(\Leftrightarrow\left(x-y\right)^2+\left(y-z\right)^2+\left(z-x\right)^2\ge0\) ( luôn đúng)
Áp dụng BĐT Cauchy - Schwarz dạng Engel ta có:
\(\dfrac{1}{1+xy}+\dfrac{1}{1+yz}+\dfrac{1}{1+xz}\ge\dfrac{\left(1+1+1\right)^2}{1+xy+1+yz+1+xz}=\dfrac{9}{3+xy+yz+xz}\ge\dfrac{9}{3+3}=\dfrac{9}{6}=\dfrac{3}{2}\)\(\RightarrowĐPCM\)
\("="\Leftrightarrow x=y=z=1\)
Cách 2:
Áp dụng BĐT AM - GM, ta có:
\(\dfrac{1}{1+xy}+\dfrac{1+xy}{4}\ge2\sqrt{\dfrac{1}{1+xy}.\dfrac{1+xy}{4}}=1\)
\(\dfrac{1}{1+yz}+\dfrac{1+yz}{4}\ge1\)
\(\dfrac{1}{1+zx}+\dfrac{1+zx}{4}\ge1\)
Cộng vế theo vế BĐT, ta được:
\(\dfrac{1}{1+xy}+\dfrac{1}{1+yz}+\dfrac{1}{1+zx}+\dfrac{1+1+1+xy+yz+zx}{4}\ge1+1+1\)
\(\Leftrightarrow\dfrac{1}{1+xy}+\dfrac{1}{1+yz}+\dfrac{1}{1+zx}+\dfrac{3+xy+yz+zx}{4}\ge3\)
\(\Leftrightarrow\dfrac{1}{1+xy}+\dfrac{1}{1+yz}+\dfrac{1}{1+xz}\ge3-\dfrac{3+xy+yz+zx}{4}\ge3-\dfrac{3+\left(x^2+y^2+z^2\right)}{4}=3-\dfrac{3+3}{4}=\dfrac{3}{2}\)\("="\Leftrightarrow x=y=z=1\)
\(VT\ge\dfrac{4}{x^2+y^2+2xy}=\dfrac{4}{\left(x+y\right)^2}\ge4\) (vì \(x+y\le1\) )
Dấu "=" xảy ra \(\Leftrightarrow x=y=\dfrac{1}{2}\)
Ta có đpcm