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Ta có \(1+x^2=x^2+xy+yz+xz=\left(x+y\right)\left(x+z\right)\)
Tương tự \(1+y^2=\left(x+y\right)\left(y+z\right)\)
\(1+z^2=\left(x+z\right)\left(y+z\right)\)
Thay vào A ta được
\(P=x\sqrt{\left(y+z\right)^2}+y\sqrt{\left(x+z\right)^2}+z\sqrt{\left(x+y\right)^2}\)
=2(xy+xz+yz)=2
\(b,VT=VP\)
\(\Leftrightarrow\frac{x}{xy+yz+zx+x^2}+\frac{y}{xy+yz+zx+y^2}+\frac{z}{xy+yz+zx+z^2}\)
\(=\frac{2xyz}{\sqrt{\left(xy+yz+zx+x^2\right)\left(xy+yz+zx+y^2\right)\left(xy+yz+zx+z^2\right)}}\)
\(\Leftrightarrow\frac{x}{\left(x+y\right)\left(x+z\right)}+\frac{y}{\left(x+y\right)\left(y+z\right)}+\frac{z}{\left(x+z\right)\left(y+z\right)}\)
\(=\frac{2xyz}{\sqrt{\left(x+y\right)\left(x+z\right)\left(y+z\right)\left(y+x\right)\left(z+x\right)\left(y+z\right)}}\)
\(\Leftrightarrow\frac{x\left(y+z\right)+y\left(x+z\right)+z\left(x+y\right)}{\left(x+y\right)\left(y+z\right)\left(z+x\right)}=\frac{2xyz}{\left(x+y\right)\left(y+z\right)\left(z+x\right)}\)
\(\Leftrightarrow xy+xz+xy+yz+xz+yz=2xyz\)
\(\Leftrightarrow2=2xyz\)
\(\Leftrightarrow xyz=1\)
Đù =)))
Ta đặt \(\hept{\begin{cases}x+z=a\\y+z=b\end{cases}\Rightarrow ab=1}\)
\(BĐT\Leftrightarrow\frac{1}{\left(a-b\right)^2}+\frac{1}{a^2}+\frac{1}{b^2}\ge4\)
Ta có
\(\frac{1}{\left(a-b\right)^2}+\frac{1}{a^2}+\frac{1}{b^2}=\frac{1}{\left(a-\frac{1}{a}\right)^2}+a^2+\frac{1}{a^2}\)
\(=\frac{1}{\left(a-\frac{1}{a}\right)^2}+\left(a-\frac{1}{a}\right)^2+2\)
\(\ge2+2=4\)
\(\frac{x^4}{y^2\left(x+z\right)}+\frac{x+z}{4}\ge2\sqrt{\frac{x^4}{y^2\left(x+z\right)}.\frac{x+z}{4}}=\frac{x^2}{y}\)
ttu ta sẽ có vt \(\ge\frac{x^2}{y}+\frac{y^2}{z}+\frac{z^2}{x}-\frac{x+y+z}{2}\ge\frac{\left(x+y+z\right)^2}{x+y+z}-\frac{x+y+z}{2}=\frac{x+y+z}{2}\)
alo kb ko
Ta có: \(\frac{x+y}{\left(x-y\right)^2}+\frac{y+z}{\left(y-z\right)^2}+\frac{z+x}{\left(z-x\right)^2}\ge\frac{9}{x+y+z}\)
\(\Leftrightarrow\left(x+y+z\right)\left(\frac{x+y}{\left(x-y\right)^2}+\frac{y+z}{\left(y-z\right)^2}+\frac{z+x}{\left(z-x\right)^2}\right)\ge9\)
giả sử \(x>y>z\ge0\)
Ta có các bđt sau:
+) \(x+y+z\ge x+y\)
+) \(\frac{y+z}{\left(y-z\right)^2}\ge\frac{1}{y}\Leftrightarrow y\left(y+z\right)\ge\left(y-z\right)^2\Leftrightarrow z\left(3y-z\right)\ge0\) (luôn đúng)
+) \(\frac{z+x}{\left(z-x\right)^2}\ge\frac{1}{x}\)
Đặt \(A=\left(x+y+z\right)\left(\frac{x+y}{\left(x-y\right)^2}+\frac{y+z}{\left(y-z\right)^2}+\frac{z+x}{\left(z-x\right)^2}\right)\)
\(\Rightarrow A\ge\left(x+y\right)\left(\frac{x+y}{\left(x-y\right)^2}+\frac{1}{y}+\frac{1}{x}\right)\)
Ta có bđt cơ bản : \(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\ge\frac{9}{a+b+c}\left(\forall a,b,c>0\right)\)
Áp dụng ta được:
\(\frac{x+y}{\left(x-y\right)^2}+\frac{1}{y}+\frac{1}{x}=\left(x+y\right)\left(\frac{1}{\left(x+y\right)^2-4xy}+\frac{1}{xy}\right)\)
\(=\left(x+y\right)\left(\frac{1}{\left(x+y\right)^2-4xy}+\frac{1}{2xy}+\frac{1}{2xy}\right)\ge\frac{9\left(x+y\right)}{\left(x+y\right)^2}=\frac{9}{x+y}\)
Vậy \(\left(x+y\right)\left(\frac{x+y}{\left(x-y\right)^2}+\frac{1}{y}+\frac{1}{x}\right)\ge\left(x+y\right)\frac{9}{x+y}=9\Rightarrow A\ge9\)
Đẳng thức xảy ra <=> \(\hept{\begin{cases}z=0\\\left(x+y\right)^2-4xy=2xy\end{cases}\Leftrightarrow\hept{\begin{cases}z=0\\x=\left(2\pm\sqrt{3}\right)y\end{cases}}}\)