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\(\Leftrightarrow2\sqrt{x-29}+4\sqrt{y-6}+6\sqrt{z-2011}+2032=x+y+z\)
\(\Leftrightarrow x-29-2\sqrt{x-29}+1+y-6-4\sqrt{y-6}+4+z-2011-6\sqrt{z-2011}+9=0\)
\(\Leftrightarrow\left(\sqrt{x-29}-1\right)^2+\left(\sqrt{y-6}-2\right)^2+\left(\sqrt{z-2011}-3\right)^2=0\)
\(\Leftrightarrow\left\{{}\begin{matrix}\sqrt{x-29}-1=0\\\sqrt{y-6}-2=0\\\sqrt{z-2011}-3=0\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}x=30\\y=10\\z=2020\end{matrix}\right.\)
\(a^2+b^2=\left(a+b-c\right)^2=a^2+\left(b-c\right)^2+2a\left(b-c\right)=b^2+\left(a-c\right)^2+2b\left(a-c\right)\)
\(\Rightarrow\left\{{}\begin{matrix}b^2=\left(b-c\right)^2+2a\left(b-c\right)\\a^2=\left(a-c\right)^2+2b\left(a-c\right)\end{matrix}\right.\)
\(\Rightarrow\dfrac{a^2+\left(a-c\right)^2}{b^2+\left(b-c\right)^2}=\dfrac{\left(a-c\right)^2+2b\left(a-c\right)+\left(a-c\right)^2}{\left(b-c\right)^2+2a\left(b-c\right)+\left(b-c\right)^2}\)
\(=\dfrac{\left(a-c\right)\left(a+b-c\right)}{\left(b-c\right)\left(b+a-c\right)}=\dfrac{a-c}{b-c}\) (đpcm)
Đặt \(\left(\sqrt{x};2\sqrt{y};3\sqrt{z}\right)=\left(a;b;c\right)\Rightarrow a;b;c\ge0\)
Ta có:
\(\dfrac{2}{a+b+c}-\dfrac{1}{ab+bc+ca}\le\dfrac{2}{a+b+c}-\dfrac{3}{\left(a+b+c\right)^2}=-3\left(\dfrac{1}{a+b+c}-\dfrac{1}{3}\right)^2+\dfrac{1}{3}\le\dfrac{1}{3}\)
Đẳng thức xảy ra khi và chỉ khi: \(a=b=c=1\Rightarrow\left\{{}\begin{matrix}x=1\\y=\dfrac{1}{4}\\z=\dfrac{1}{9}\end{matrix}\right.\)
Ta có:
\(x^2+1=x^2+xy+yz+zx\)
\(=x\left(x+y\right)+z\left(x+y\right)=\left(x+y\right)\left(x+z\right)\)
Tương tự:
\(\left\{{}\begin{matrix}y^2+1=\left(y+z\right)\left(y+x\right)\\z^2+1=\left(z+y\right)\left(z+x\right)\end{matrix}\right.\)
\(A=x\sqrt{\dfrac{\left(x+y\right)\left(y+z\right)\left(z+x\right)\left(y+z\right)}{\left(x+y\right)\left(z+x\right)}}+y\sqrt{\dfrac{\left(z+x\right)\left(y+z\right)\left(x+y\right)\left(z+x\right)}{\left(x+y\right)\left(y+z\right)}}+z\sqrt{\dfrac{\left(x+y\right)\left(z+x\right)\left(y+z\right)\left(x+y\right)}{\left(z+x\right)\left(y+z\right)}}\)
\(=x\left|y+z\right|+y\left|z+x\right|+z\left|x+y\right|\)
TH1: x,y,z <0
\(A=-x\left(y+z\right)-y\left(z+x\right)-z\left(x+y\right)=-2\)
TH2: x,y,z>0
\(A=x\left(y+z\right)+y\left(z+x\right)+z\left(x+y\right)=2\)
Ta có \(1+z^2=xy+yz+zx+z^2\)
\(=y\left(x+z\right)+z\left(x+z\right)\)
\(=\left(x+z\right)\left(y+z\right)\)
CMTT, \(1+x^2=\left(x+y\right)\left(x+z\right)\) và \(1+y^2=\left(x+y\right)\left(y+z\right)\)
Do đó \(\sqrt{\dfrac{\left(1+y^2\right)\left(1+z^2\right)}{1+x^2}}\) \(=\sqrt{\dfrac{\left(x+y\right)\left(y+z\right)\left(x+z\right)\left(y+z\right)}{\left(x+y\right)\left(x+z\right)}}\)
\(=\sqrt{\left(y+z\right)^2}\) \(=\left|y+z\right|\)
Tương tự như thế, ta được
\(A=x\left|y+z\right|+y\left|z+x\right|+z\left|x+y\right|\)
Cái này không tính ra số cụ thể được nhé bạn. Nó còn phải tùy vào dấu của \(x+y,y+z,z+x\) nữa.
Ta có BĐT:
\(\frac{1}{xy}+\frac{1}{yz}+\frac{1}{xz}\le\frac{1}{x^2}+\frac{1}{y^2}+\frac{1}{z^2}\)
\(\Leftrightarrow6\left(\frac{1}{xy}+\frac{1}{yz}+\frac{1}{xz}\right)+2016\le6\left(\frac{1}{x^2}+\frac{1}{y^2}+\frac{1}{z^2}\right)+2016\)
\(\Leftrightarrow7.\left(\frac{1}{x^2}+\frac{1}{y^2}+\frac{1}{z^2}\right)\le6\left(\frac{1}{x^2}+\frac{1}{y^2}+\frac{1}{z^2}\right)+2016\)
\(\Leftrightarrow\frac{1}{x^2}+\frac{1}{y^2}+\frac{1}{z^2}\le2016\)
Xét \(P=\frac{1}{\sqrt{3\left(2x^2+y^2\right)}}+\frac{1}{\sqrt{3\left(2y^2+z^2\right)}}+\frac{1}{\sqrt{3\left(2z^2+x^2\right)}}\)
\(P^2=\left(\frac{1}{\sqrt{3}}.\frac{1}{\sqrt{2x^2+y^2}}+\frac{1}{\sqrt{3}}.\frac{1}{\sqrt{2y^2+z^2}}+\frac{1}{\sqrt{3}}.\frac{1}{\sqrt{2z^2+x^2}}\right)^2\)
Áp dụng BĐT Bunhiacopxki ta có:
\(P^2\le\left(\left(\frac{1}{\sqrt{3}}\right)^2+\left(\frac{1}{\sqrt{3}}\right)^2+\left(\frac{1}{\sqrt{3}}\right)^2\right)\left(\left(\frac{1}{\sqrt{2x^2+y^2}}\right)^2+\left(\frac{1}{\sqrt{2y^2+z^2}}\right)^2+\left(\frac{1}{\sqrt{2z^2+x^2}}\right)^2\right)\)
\(\Leftrightarrow P^2\le\frac{1}{2x^2+y^2}+\frac{1}{2y^2+z^2}+\frac{1}{2z^2+x^2}\)
Mặt khác ta có:
\(\frac{1}{2x^2+y^2}=\frac{1}{x^2+x^2+y^2}\le\frac{1}{9}\left(\frac{1}{x^2}+\frac{1}{x^2}+\frac{1}{y^2}\right)\)
\(\frac{1}{2y^2+z^2}\le\frac{1}{9}\left(\frac{1}{y^2}+\frac{1}{y^2}+\frac{1}{z^2}\right)\)
\(\frac{1}{2z^2+x^2}\le\frac{1}{9}\left(\frac{1}{z^2}+\frac{1}{z^2}+\frac{1}{x^2}\right)\)
\(\Rightarrow P^2\le\frac{1}{3}\left(\frac{1}{x^2}+\frac{1}{y^2}+\frac{1}{z^2}\right)\le\frac{1}{3}.2016=672\)
\(\Rightarrow P\le4\sqrt{42}\)
Dấu '=' xảy ra khi \(x=y=z=\sqrt{\frac{1}{672}}\)
\(\sqrt{x-29}+2\sqrt{y-6}+3\sqrt{z-2011}+1016=\dfrac{1}{2}\left(x+y+z\right)\)\(\Leftrightarrow2\sqrt{x-29}+4\sqrt{y-6}+6\sqrt{z-2011}+2032=x+y+z\)\(\Leftrightarrow-2\sqrt{x-29}-4\sqrt{y-6}-6\sqrt{z-2011}-2032=-x-y-z\)\(\Leftrightarrow(x-29-2\sqrt{x-29}+1)+(y-6-2\cdot2\sqrt{y-6}+2^2)+(z-2011-2\cdot3\sqrt{z-2011}+3^2)=0\)\(\Leftrightarrow\left(\sqrt{x-29}-1\right)^2+\left(\sqrt{y-6}-2\right)^2+\left(\sqrt{z-2011}-3\right)^2=0\)\(\Leftrightarrow\left\{{}\begin{matrix}\sqrt{x-29}-1=0\\\sqrt{y-6}-2=0\\\sqrt{z-2011}-3=0\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}\sqrt{x-29}=1\\\sqrt{y-6}=2\\\sqrt{z-2011}=3\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x-29=1\\y-6=4\\z-2011=9\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}x=30\\y=10\\z=2020\end{matrix}\right.\)
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