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a, Đặt \(\sqrt[4]{a}=x;\sqrt[4]{b}=y.\)Bất đẳng thức ban đầu trở thành: \(\frac{2x^2y^2}{x^2+y^2}\le xy.\)
ta có : \(x^2+y^2\ge2xy\Rightarrow\frac{2x^2y^2}{x^2+y^2}\le\frac{2x^2y^2}{2xy}=xy.\)(đpcm )
dấu " = " xẩy ra khi x = y > 0
vậy bất đăng thức ban đầu đúng. dấu " = " xẩy ra khi a = b >0
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\(VT=\Sigma\frac{xy+yz+zx}{xy}=3+\Sigma\frac{z\left(x+y\right)}{xy}\)
Đến đây để ý \(\frac{1}{2}\left[\frac{z\left(x+y\right)}{xy}+\frac{y\left(z+x\right)}{zx}\right]\ge\sqrt{\frac{\left(z+x\right)\left(x+y\right)}{x^2}}\left(\text{AM - GM}\right)\)
Là xong.
\(\frac{x}{\left(\sqrt{x}-\sqrt{y}\right)\left(\sqrt{x}-\sqrt{z}\right)}+\frac{y}{\left(\sqrt{y}-\sqrt{z}\right)\left(\sqrt{y}-\sqrt{x}\right)}+\)\(\frac{z}{\left(\sqrt{z}-\sqrt{x}\right)\left(\sqrt{z}-\sqrt{y}\right)}\)
\(=-\frac{x}{\left(\sqrt{x}-\sqrt{y}\right)\left(\sqrt{z}-\sqrt{x}\right)}-\frac{y}{\left(\sqrt{y}-\sqrt{z}\right)\left(\sqrt{x}-\sqrt{y}\right)}\)\(-\frac{z}{\left(\sqrt{z}-\sqrt{x}\right)\left(\sqrt{y}-\sqrt{z}\right)}\)
\(=\frac{-x\left(\sqrt{y}-\sqrt{z}\right)-y\left(\sqrt{z}-\sqrt{x}\right)-z\left(\sqrt{x}-\sqrt{y}\right)}{\left(\sqrt{x}-\sqrt{y}\right)\left(\sqrt{y}-\sqrt{z}\right)\left(\sqrt{z}-\sqrt{x}\right)}\)
\(=\frac{-x\sqrt{y}+x\sqrt{z}-y\sqrt{z}+y\sqrt{x}-z\sqrt{x}+z\sqrt{y}}{\left(\sqrt{x}-\sqrt{y}\right)\left(\sqrt{y}-\sqrt{z}\right)\left(\sqrt{z}-\sqrt{x}\right)}\)
\(=\frac{-\sqrt{xy}\left(\sqrt{x}-\sqrt{y}\right)+\sqrt{z}\left(x-y\right)-z\left(\sqrt{x}-y\right)}{\left(\sqrt{x}-\sqrt{y}\right)\left(\sqrt{y}-\sqrt{z}\right)\left(\sqrt{z}-\sqrt{x}\right)}\)
\(=\frac{-\sqrt{xy}+\sqrt{z}\left(\sqrt{x}+\sqrt{y}\right)-z}{\left(\sqrt{y}-\sqrt{z}\right)\left(\sqrt{z}-\sqrt{x}\right)}\)
\(=\frac{-\sqrt{xy}+\sqrt{xz}+\sqrt{yz}-z}{\left(\sqrt{y}-\sqrt{z}\right)\left(\sqrt{z}-\sqrt{x}\right)}\)
\(=\frac{\sqrt{y}\left(\sqrt{z}-\sqrt{x}\right)-\sqrt{z}\left(\sqrt{z}-\sqrt{x}\right)}{\left(\sqrt{y}-\sqrt{z}\right)\left(\sqrt{z}-\sqrt{x}\right)}\)
\(=\frac{\left(\sqrt{z}-\sqrt{x}\right)\left(\sqrt{y}-\sqrt{z}\right)}{\left(\sqrt{y}-\sqrt{z}\right)\left(\sqrt{z}-\sqrt{x}\right)}\)
+ \(\left(\sqrt{x}+\sqrt{y}+\sqrt{z}\right)^2=4\Rightarrow x+y+z+2\left(\sqrt{xy}+\sqrt{yz}+\sqrt{zx}\right)=4\)
\(\Rightarrow\sqrt{xy}+\sqrt{yz}+\sqrt{zx}=1\)
+ \(x+1=x+\sqrt{xy}+\sqrt{yz}+\sqrt{zx}=\sqrt{x}\left(\sqrt{x}+\sqrt{y}\right)+\sqrt{z}\left(\sqrt{x}+\sqrt{y}\right)\)
\(=\left(\sqrt{x}+\sqrt{y}\right)\left(\sqrt{x}+\sqrt{z}\right)\)
+ Tương tự : \(y+1=\left(\sqrt{x}+\sqrt{y}\right)\left(\sqrt{y}+\sqrt{z}\right)\); \(z+1=\left(\sqrt{x}+\sqrt{z}\right)\left(\sqrt{y}+\sqrt{z}\right)\)
+ \(P=\sqrt{\left(\sqrt{x}+\sqrt{y}\right)^2\left(\sqrt{y}+\sqrt{z}\right)^2\left(\sqrt{z}+\sqrt{x}\right)^2}\cdot\frac{\sqrt{x}\left(\sqrt{y}+\sqrt{z}\right)+\sqrt{y}\left(\sqrt{x}+\sqrt{z}\right)+\sqrt{z}\left(\sqrt{x}+\sqrt{y}\right)}{\left(\sqrt{x}+\sqrt{y}\right)\left(\sqrt{y}+\sqrt{z}\right)\left(\sqrt{z}+\sqrt{x}\right)}\)
\(=2\left(\sqrt{xy}+\sqrt{yz}+\sqrt{zx}\right)=2\)
\(\sqrt{x}+\sqrt{y}+\sqrt{z}=2\)
\(\Leftrightarrow x+y+z+2\sqrt{xy}+2\sqrt{yz}+2\sqrt{zx}=4\)
\(\Leftrightarrow2+2\left(\sqrt{xy}+\sqrt{yz}+\sqrt{zx}\right)=4\)
\(\Leftrightarrow\sqrt{xy}+\sqrt{yz}+\sqrt{zx}=1\)
Khi đó ta có : \(x+1=x+\sqrt{xy}+\sqrt{yz}+\sqrt{zx}\)
\(\Leftrightarrow x+1=\sqrt{x}\left(\sqrt{x}+\sqrt{y}\right)+\sqrt{z}\left(\sqrt{x}+\sqrt{y}\right)\)
\(\Leftrightarrow x+1=\left(\sqrt{x}+\sqrt{y}\right)\left(\sqrt{z}+\sqrt{x}\right)\)
Tương tự : \(y+1=\left(\sqrt{y}+\sqrt{z}\right)\left(\sqrt{x}+\sqrt{y}\right)\);
\(z+1=\left(\sqrt{z}+\sqrt{x}\right)\left(\sqrt{y}+\sqrt{z}\right)\)
Ta lần lượt xét các biểu thức :
+) \(\sqrt{\left(x+1\right)\left(y+1\right)\left(z+1\right)}\)
\(=\sqrt{\left(\sqrt{x}+\sqrt{y}\right)\left(\sqrt{z}+\sqrt{x}\right)\left(\sqrt{y}+\sqrt{z}\right)\left(\sqrt{x}+\sqrt{y}\right)\left(\sqrt{z}+\sqrt{x}\right)\left(\sqrt{y}+\sqrt{z}\right)}\)
\(=\sqrt{\left[\left(\sqrt{x}+\sqrt{y}\right)\left(\sqrt{z}+\sqrt{x}\right)\left(\sqrt{y}+\sqrt{z}\right)\right]^2}\)
\(=\left(\sqrt{x}+\sqrt{y}\right)\left(\sqrt{z}+\sqrt{x}\right)\left(\sqrt{y}+\sqrt{z}\right)\)
+) \(\frac{\sqrt{x}}{x+1}+\frac{\sqrt{y}}{y+1}+\frac{\sqrt{z}}{z+1}\)
\(=\frac{\sqrt{x}}{\left(\sqrt{y}+\sqrt{x}\right)\left(\sqrt{x}+\sqrt{z}\right)}+\frac{\sqrt{y}}{\left(\sqrt{y}+\sqrt{x}\right)\left(\sqrt{y}+\sqrt{z}\right)}+\frac{\sqrt{z}}{\left(\sqrt{z}+\sqrt{x}\right)\left(\sqrt{y}+\sqrt{z}\right)}\)
\(=\frac{\sqrt{x}\left(\sqrt{y}+\sqrt{z}\right)+\sqrt{y}\left(\sqrt{x}+\sqrt{z}\right)+\sqrt{z}\left(\sqrt{x}+\sqrt{y}\right)}{\left(\sqrt{x}+\sqrt{y}\right)\left(\sqrt{z}+\sqrt{x}\right)\left(\sqrt{y}+\sqrt{z}\right)}\)
\(=\frac{2\left(\sqrt{xy}+\sqrt{yz}+\sqrt{zx}\right)}{\left(\sqrt{z}+\sqrt{x}\right)\left(\sqrt{y}+\sqrt{z}\right)\left(\sqrt{x}+\sqrt{y}\right)}\)
\(=\frac{2}{\left(\sqrt{z}+\sqrt{x}\right)\left(\sqrt{y}+\sqrt{z}\right)\left(\sqrt{x}+\sqrt{y}\right)}\)
Do đó ta có :
\(P=\left(\sqrt{x}+\sqrt{y}\right)\left(\sqrt{z}+\sqrt{x}\right)\left(\sqrt{y}+\sqrt{z}\right)\cdot\frac{2}{\left(\sqrt{z}+\sqrt{x}\right)\left(\sqrt{y}+\sqrt{z}\right)\left(\sqrt{x}+\sqrt{y}\right)}\)
\(P=2\)
Vậy...
A
Áp dụng BĐT cosi ta có
\(\sqrt{\left(2x-1\right).1}\le\frac{2x-1+1}{2}=x\)
\(x\sqrt{5-4x^2}\le\frac{x^2+5-4x^2}{2}=\frac{-3x^2+5}{2}\)
Khi đó
\(A\le3x+\frac{-3x^2+5}{2}=\frac{-3x^2+6x+5}{2}=\frac{-3\left(x-1\right)^2}{2}+4\le4\)
MaxA=4 khi \(\hept{\begin{cases}2x-1=1\\x^2=5-4x^2\\x=1\end{cases}\Rightarrow}x=1\)
B
Áp dụng BĐT cosi ta có :
\(x^2+y^2+z^2\ge\frac{1}{3}\left(x+y+z\right)^2\)
=> \(x+y+z\le\sqrt{3\left(x^2+y^2+z^2\right)}\)
=> \(B\le\frac{xyz.\left(\sqrt{3\left(x^2+y^2+z^2\right)}+\sqrt{x^2+y^2+z^2}\right)}{\left(x^2+y^2+z^2\right)\left(xy+yz+xz\right)}=\frac{xyz.\left(\sqrt{3}+1\right)}{\left(xy+yz+xz\right)\sqrt{x^2+y^2+z^2}}\)
Lại có \(x^2+y^2+z^2\ge3\sqrt[3]{x^2y^2z^2}\); \(xy+yz+xz\ge3\sqrt[3]{x^2y^2z^2}\)
=> \(\sqrt{x^2+y^2+z^2}\left(xy+yz+xz\right)\ge3\sqrt[3]{x^2y^2z^2}.\sqrt{3\sqrt[3]{x^2y^2z^2}}=3\sqrt{3}.xyz\)
=> \(B\le\frac{\sqrt{3}+1}{3\sqrt{3}}=\frac{3+\sqrt{3}}{9}\)
\(MaxB=\frac{3+\sqrt{3}}{9}\)khi x=y=z
Đặt \(\sqrt{x}=x;\sqrt{y}=y;\sqrt{z}=z\) cho dễ nhìn.
\(\Rightarrow\hept{\begin{cases}x+y+z=2\\x^2+y^2+z^2=2\end{cases}}\)
\(\Rightarrow x^2+y^2+z^2+2\left(xy+yz+zx\right)=4\)
\(\Leftrightarrow xy+yz+zx=1\)
Ta có:
\(x\left(1+y^2\right)\left(1+z^2\right)+y\left(1+z^2\right)\left(1+x^2\right)+z\left(1+x^2\right)\left(1+y^2\right)\)
\(=x^2y^2z+y^2z^2x+z^2x^2y+x^2y+x^2z+y^2x+y^2z+z^2x+z^2y+x+y+z\)
\(=xyz\left(xy+yz+zx\right)+x^2\left(2-x\right)+y^2\left(2-y\right)+z^2\left(2-z\right)+2\)
\(=-2xyz+2\left(x^2+y^2+z^2\right)-\left(x^3+y^3+z^3-3xyz\right)+2\)
\(=-2xyz+6-\left(x+y+z\right)\left(x^2+y^2+z^2-xy-yz-zx\right)\)
\(=-2xyz+6-2=-2xyz+4\)
Ta lại có:
\(\left(1+x^2\right)\left(1+y^2\right)\left(1+z^2\right)=x^2y^2z^2+x^2y^2+y^2z^2+z^2x^2+x^2+y^2+z^2+1\)
\(=x^2y^2z^2+\left(xy+yz+zx\right)^2-2xyz\left(xy+yz+zx\right)+3\)
\(=x^2y^2z^2-2xyz+4=\left(xyz-2\right)^2\)
\(\Rightarrow A=\sqrt{\left(xyz-2\right)^2}.\frac{4-2xyz}{\left(xyz-2\right)^2}\)
Tới đây bí :((
chứng minh $\sqrt{x(y+1)}+\sqrt{y(z+1)}+\sqrt{z(x+1)}\leq \frac{3}{2}\sqrt{(x+1)(y+1)(z+1)}$ - Bất đẳng thức và cực trị - Diễn đàn Toán học