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Ta có BĐT \(x^2+1\ge2x\Leftrightarrow\left(x-1\right)^2\ge0\forall x\in R\)
Tương tự: \(y^2+1\ge2y;z^2+1\ge2z\)
\(\Rightarrow x^2+y^2+z^2+3\ge2\left(x+y+z\right)\left(1\right)\)
Và BĐT \(x^2+y^2+z^2\ge xy+yz+xz\left(2\right)\)
\(\Leftrightarrow\left(x-y\right)^2+\left(y-z\right)^2+\left(z-x\right)^2\ge0\forall x,y,z\in R\)
Cộng theo vế 2 BĐT (1);(2) ta có:
\(2\left(x^2+y^2+z^2\right)+3\ge45\)
\(\Rightarrow2\left(x^2+y^2+z^2\right)\ge42\Rightarrow x^2+y^2+z^2\ge21\)
Khi x=y=z=1
Sửa đề : cho \(CM:x^2+y^2+z^2\ge21\)
Ta có : \(\left(x-y\right)^2+\left(y-z\right)^2+\left(z-x\right)^2\ge0\)
\(\Leftrightarrow2x^2+2y^2+2z^2-2xy-2xy-2xz\ge0\)
\(\Leftrightarrow x^2+y^2+z^2-xy-yz-xz\ge0\)
\(\Rightarrow x^2+y^2+z^2\ge xy+yz+xz\)(1)
Ta lại có : \(\left(x-1\right)^2+\left(y-1\right)^2+\left(z-1\right)^2\ge0\)
\(\Leftrightarrow x^2+y^2+z^2-2x-2y-2z+3\ge0\)
\(\Rightarrow x^2+y^2+z^2\ge2x+2y+2z-3\)(2)
Cộng vế với vế của (1); (2) lại ta được :
\(2\left(x^2+y^2+z^2\right)\ge xy+yz+xy+2x+2y+2z-3\)
\(\Leftrightarrow2\left(x^2+y^2+z^2\right)\ge45-3=42\)
\(\Rightarrow x^2+y^2+z^2\ge\frac{42}{2}=21\)(đpcm)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(\hept{\begin{cases}x^2-2x\sqrt{y}+2y=x\\y^2-2y\sqrt{z}+2z=y\\z^2-2z\sqrt{x}+2x=z\end{cases}}\)
\(\Leftrightarrow x^2-2x\sqrt{y}+2y+y^2-2y\sqrt{z}+2z+z^2-2z\sqrt{x}+2x=x+y+z\)
\(\Leftrightarrow\left(x-\sqrt{y}\right)^2+\left(y-\sqrt{z}\right)^2+\left(z-\sqrt{x}\right)^2=0\)
\(\Rightarrow\hept{\begin{cases}x-\sqrt{y}=0\\y-\sqrt{z}=0\\z-\sqrt{x}=0\end{cases}\Leftrightarrow\hept{\begin{cases}x=\sqrt{y}\\y=\sqrt{z}\\z=\sqrt{x}\end{cases}}}\)
\(\Rightarrow\orbr{\begin{cases}x=y=z=0\\x=y=z=1\end{cases}}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
Ta có:
\(2\left(2x^2+xy+2y^2\right)=3\left(x^2+y^2\right)+\left(x+y\right)^2\ge\dfrac{3}{2}\left(x+y\right)^2+1\left(x+y\right)^2=\dfrac{5}{2}\left(x+y\right)^2\)
\(\Rightarrow\sqrt{2x^2+xy+2y^2}\ge\dfrac{\sqrt{5}}{2}\left(x+y\right)\)
Gợi ý. Dùng cái trên.
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\(gt\Leftrightarrow\dfrac{1}{\sqrt{x}}+\dfrac{1}{\sqrt{y}}+\dfrac{1}{\sqrt{z}}=1\)
\(P=\dfrac{1}{xyz}\left(x\sqrt{2y^2+yz+2z^2}+y\sqrt{2x^2+xz+2z^2}+z\sqrt{2y^2+xy+2x^2}\right)\)
\(=\dfrac{1}{xyz}\left(x\sqrt{\dfrac{5}{4}\left(y+z\right)^2+\dfrac{3}{4}\left(y-z\right)^2}+y\sqrt{\dfrac{5}{4}\left(x+z\right)^2+\dfrac{3}{4}\left(x-z\right)^2}+z\sqrt{\dfrac{5}{4}\left(x+y\right)^2+\dfrac{3}{4}\left(x-y\right)^2}\right)\)
\(\ge\dfrac{1}{xyz}\left[x.\dfrac{\sqrt{5}\left(z+y\right)}{2}+y.\dfrac{\sqrt{5}\left(x+z\right)}{2}+z.\dfrac{\sqrt{5}\left(x+y\right)}{2}\right]\)
\(=\dfrac{\sqrt{5}\left(z+y\right)}{2yz}+\dfrac{\sqrt{5}\left(x+z\right)}{2xz}+\dfrac{\sqrt{5}\left(x+y\right)}{2xy}\)
\(=\dfrac{\sqrt{5}}{3}\left(1+1+1\right)\left(\dfrac{1}{x}+\dfrac{1}{y}+\dfrac{1}{z}\right)\ge\dfrac{\sqrt{5}}{3}\left(\dfrac{1}{\sqrt{x}}+\dfrac{1}{\sqrt{y}}+\dfrac{1}{\sqrt{z}}\right)^2=\dfrac{\sqrt{5}}{3}\) (bunhia)
Dấu = xảy ra khi \(x=y=z=9\)
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 : \(P\ge\dfrac{\sqrt{5}}{3}\)
" = " \(\Leftrightarrow x=y=z=9\)
![](https://rs.olm.vn/images/avt/0.png?1311)
Áp dụng bất đẳng thức Bunhiacopxki:
\(P^2\le\left(1^2+1^2+1^2\right)\left(2x+2y+2z+xy+yz+xz\right)=3\left(4+xy+yz+xz\right)\)
Mặt khác ta có : \(xy+yz+xz\le x^2+y^2+z^2\le\frac{\left(x+y+z\right)^2}{3}=\frac{4}{3}\) (Dấu "=" xảy ra khi x=y=z=2/3)
=> \(P\le\sqrt{3\left(4+\frac{4}{3}\right)}=4\)khi x=y=z=2/3
Vậy Max P = 4 <=> x=y=z=2/3
![](https://rs.olm.vn/images/avt/0.png?1311)
\(b,\)\(ĐK:x\ge\frac{3}{2}\)
\(PT\Leftrightarrow x^2-4+5-\sqrt{6x^2+1}+\sqrt{2x-3}-1=0\)
\(\Leftrightarrow x^2-4+\frac{24-6x^2}{5+\sqrt{6x^2+1}}+\frac{2x-4}{\sqrt{2x-3}+1}=0\)
\(\Leftrightarrow\left(x-2\right)\left(x+2-\frac{12+6x}{5+\sqrt{6x^2+1}}+\frac{2}{\sqrt{2x-3}+1}=0\right)\)
Xét ....
\(\Leftrightarrow x=2\left(tm\right)\)