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Bài 1:Áp dụng C-S dạng engel
\(\frac{3}{xy+yz+xz}+\frac{2}{x^2+y^2+z^2}=\frac{6}{2\left(xy+yz+xz\right)}+\frac{2}{x^2+y^2+z^2}\)
\(\ge\frac{\left(\sqrt{6}+\sqrt{2}\right)^2}{\left(x+y+z\right)^2}=\left(\sqrt{6}+\sqrt{2}\right)^2>14\)
ĐKXĐ: ...
Lấy pt cuối trừ 3 lần pt đầu ta được:
\(\left(\sqrt{x}-\frac{1}{\sqrt{x}}\right)^3+\left(\sqrt{y}-\frac{1}{\sqrt{y}}\right)^3+\left(\sqrt{z}-\frac{1}{\sqrt{z}}\right)^3=\frac{512}{27}\)
Pt (2) tương đương:
\(x+\frac{1}{x}-2+y+\frac{1}{y}-2+z+\frac{1}{z}-2=\frac{64}{9}\)
\(\Leftrightarrow\left(\sqrt{x}-\frac{1}{\sqrt{x}}\right)^2+\left(\sqrt{y}-\frac{1}{\sqrt{y}}\right)^2+\left(\sqrt{z}-\frac{1}{\sqrt{z}}\right)^2=\frac{64}{9}\)
Đặt \(\left(\sqrt{x}-\frac{1}{\sqrt{x}};\sqrt{y}-\frac{1}{\sqrt{y}};\sqrt{z}-\frac{1}{\sqrt{z}}\right)=\left(a;b;c\right)\)
Hệ trở thành:
\(\left\{{}\begin{matrix}a+b+c=\frac{8}{3}\\a^2+b^2+c^2=\frac{64}{9}\\a^3+b^3+c^3=\frac{512}{27}\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}a+b+c=\frac{8}{3}\\ab+bc+ca=0\\a^3+b^3+c^3=\frac{512}{27}\end{matrix}\right.\)
Ta có: \(a^3+b^3+c^3-3abc=\frac{512}{27}-3abc\)
\(\Leftrightarrow\left(a+b+c\right)\left(a^2+b^2+c^2-ab-bc-ca\right)=\frac{512}{27}-3abc\)
\(\Leftrightarrow\frac{8}{3}.\left(\frac{64}{9}-0\right)=\frac{512}{27}-3abc\)
\(\Rightarrow abc=0\)
\(\Rightarrow\left\{{}\begin{matrix}a+b+c=\frac{8}{3}\\ab+bc+ca=0\\abc=0\end{matrix}\right.\) \(\Leftrightarrow\left(a;b;c\right)=\left(0;0;\frac{8}{3}\right)\) và hoán vị
Hay \(\left(x;y;z\right)=\left(1;1;9\right)\) và hoán vị
Bạn xem lại đề câu b và c nhé !
a) \(\sqrt{x^2+2x+4}\ge x-2\) \(\left(ĐK:x\ge2\right)\)
\(\Leftrightarrow x^2+2x+4>x^2-4x+4\)
\(\Leftrightarrow6x>0\Leftrightarrow x>0\) kết hợp với ĐKXĐ
\(\Rightarrow x\ge2\) thỏa mãn đề.
d) \(x+y+z+4=2\sqrt{x-2}+4\sqrt{y-3}+6\sqrt{z-5}\)
\(ĐKXĐ:x\ge2,y\ge3,z\ge5\)
Pt tương đương :
\(\left(x-2-2\sqrt{x-2}+1\right)+\left(y-3-4\sqrt{y-3}+4\right)+\left(z-5-6\sqrt{z-5}+9\right)=0\)
\(\Leftrightarrow\left(\sqrt{x-2}-1\right)^2+\left(\sqrt{y-3}-2\right)^2+\left(\sqrt{z-5}-3\right)^2=0\)
Dấu "=" xảy ra \(\Leftrightarrow\hept{\begin{cases}\sqrt{x-2}=1\\\sqrt{y-3}=2\\\sqrt{z-5}=3\end{cases}\Leftrightarrow}\hept{\begin{cases}x=3\\y=7\\z=14\end{cases}}\) ( Thỏa mãn ĐKXĐ )
e) \(\sqrt{x}+\sqrt{y-1}+\sqrt{z-2}=\frac{1}{2}\left(x+y+z\right)\) (1)
\(ĐKXĐ:x\ge0,y\ge1,z\ge2\)
Phương trình (1) tương đương :
\(x+y+z-2\sqrt{x}-2\sqrt{y-1}-2\sqrt{z-2}=0\)
\(\Leftrightarrow\left(x-2\sqrt{x}+1\right)+\left(y-1-2\sqrt{y-1}+1\right)+\left(z-2-2\sqrt{z-2}+1\right)=0\)
\(\Leftrightarrow\left(\sqrt{x}-1\right)^2+\left(\sqrt{y-1}-1\right)^2+\left(\sqrt{z-2}-1\right)^2=0\)
Dấu "=" xảy ra \(\Leftrightarrow\hept{\begin{cases}\sqrt{x}=1\\\sqrt{y-1}=1\\\sqrt{z-2}=1\end{cases}\Leftrightarrow}\hept{\begin{cases}x=1\\y=2\\z=3\end{cases}}\)( Thỏa mãn ĐKXĐ )
Ta có: \(\frac{1}{xy}+\frac{1}{yz}+\frac{1}{zx}=2019\)
\(\Rightarrow\frac{x+y+z}{xyz}=2019\)
\(\Rightarrow x+y+z=2019xyz\)
\(\Rightarrow2019x^2=\frac{x^2+xy+xz}{yz}\)
\(\Rightarrow2019x^2+1=\frac{x^2+xy+xz+yz}{yz}=\frac{\left(x+y\right)\left(x+z\right)}{yz}\)
\(=\left(\frac{x}{y}+1\right)\left(\frac{x}{z}+1\right)\)
\(\Rightarrow\sqrt{2019x^2+1}=\sqrt{\left(\frac{x}{y}+1\right)\left(\frac{x}{z}+1\right)}\)\(\le\frac{1}{2}\left(\frac{x}{y}+\frac{x}{z}+2\right)=1+\frac{x}{2}\left(\frac{1}{y}+\frac{1}{z}\right)\)(cô -si)
\(\Rightarrow\frac{x^2+1+\sqrt{2019x^2+1}}{x}\le\frac{x^2+1+1+\frac{x}{2}\left(\frac{1}{y}+\frac{1}{z}\right)}{x}\)\(=x+\frac{2}{x}+\frac{1}{2}\left(\frac{1}{y}+\frac{1}{z}\right)\)
Tương tự ta có: \(\frac{y^2+1+\sqrt{2019y^2+1}}{y}\le y+\frac{2}{y}+\frac{1}{2}\left(\frac{1}{z}+\frac{1}{x}\right)\)
và \(\frac{z^2+1+\sqrt{2019z^2+1}}{z}\le z+\frac{2}{z}+\frac{1}{2}\left(\frac{1}{x}+\frac{1}{y}\right)\)
Cộng từng vế của các bđt trên, ta được:
\(\text{Σ}_{cyc}\frac{x^2+1+\sqrt{2019x^2+1}}{x}\le x+y+z+3\left(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}\right)\)
Chứng minh được: \(\left(x+y+z\right)^2\ge3\left(xy+yz+zx\right)\)
\(\Rightarrow3\left(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}\right)=\frac{3\left(xy+yz+zx\right)}{xyz}=\frac{2019.3\left(xy+yz+zx\right)}{2019xyz}\)
\(\le\frac{2019\left(x+y+z\right)^2}{x+y+z}=2019\left(x+y+z\right)\)
\(\Rightarrow VT\le2020\left(x+y+z\right)=2020.2019xyz\)
Vậy \(\text{Σ}_{cyc}\frac{x^2+1+\sqrt{2019x^2+1}}{x}\le2019.2020xyz\left(đpcm\right)\)
Theo bài ra ta có:
\(\frac{1}{xy}+\frac{1}{yz}+\frac{1}{zx}=\frac{z}{xyz}+\frac{x}{xyz}+\frac{y}{xyz}=\frac{x+y+z}{xyz}=2019\)
\(\Rightarrow x+y+z=2019xyz\)
\(\Rightarrow2019x^2=\frac{x^2+xy+xz}{yz}\)
\(\Rightarrow2019x^2+1=\frac{x^2+xy+xz+yz}{yz}=\frac{\left(x+y\right)\left(x+z\right)}{yz}=\left(\frac{x}{y}+1\right)\left(\frac{x}{z}+1\right)\)
\(\Rightarrow\sqrt{2019x^2+1}=\sqrt{\left(\frac{x}{y}+1\right)\left(\frac{x}{z}+1\right)}\le\frac{1}{2}\left(\frac{x}{y}+\frac{x}{z}+2\right)=1+\frac{x}{2}\left(\frac{1}{y}+\frac{1}{z}\right)\)(Theo BĐT Cosi)
\(\Rightarrow\frac{x^2+1+\sqrt{2019^2+1}}{x}\le\frac{x+1+1+\frac{x}{2}\left(\frac{1}{y}+\frac{1}{z}\right)}{x}=x+\frac{2}{x}+\frac{1}{2}\left(\frac{1}{y}+\frac{1}{z}\right)\)
Tương tự:
\(\frac{y^2+1+\sqrt{2019y^2+1}}{y}\le y+\frac{2}{y}+\frac{1}{2}\left(\frac{1}{z}+\frac{1}{x}\right)\)
\(\frac{z^2+1+\sqrt{2019z^2+1}}{z}\le z+\frac{2}{z}+\frac{1}{2}\left(\frac{1}{x}+\frac{1}{y}\right)\)
\(\Rightarrow VT\le x+y+z+3\left(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}\right)\)
Chứng minh được: \(\left(x+y+z\right)^2\ge3\left(xy+yz+zx\right)\)
\(\Rightarrow3\left(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}\right)=\frac{3\left(xy+yz+zx\right)}{xyz}=\frac{2019\cdot3\left(xy+yz+zx\right)}{2019xyz}\le\frac{2019\left(x+y+z\right)^2}{x+y+z}\)\(=2019\left(x+y+z\right)\)
\(\Rightarrow VT\le2020\left(x+y+z\right)=2020\cdot2019xyz=VP\)
=> ĐPCM
b, Ta có
\(\frac{\sqrt{x}+1}{y+1}=\frac{\left(\sqrt{x}+1\right)\left(y+1\right)-y-y\sqrt{x}}{y+1}=\sqrt{x}+1-\frac{y\left(\sqrt{x}+1\right)}{y+1}\)
Mà \(y+1\ge2\sqrt{y}\)
=> \(\frac{\sqrt{x}+1}{y+1}\ge\sqrt{x}+1-\frac{1}{2}\sqrt{y}\left(\sqrt{x}+1\right)\)
Khi đó
\(P\ge\frac{1}{2}\left(\sqrt{x}+\sqrt{y}+\sqrt{z}\right)+3-\frac{1}{2}\left(\sqrt{xy}+\sqrt{yz}+\sqrt{xz}\right)\)
Mà \(\sqrt{xy}+\sqrt{yz}+\sqrt{xz}\le\frac{\left(\sqrt{x}+\sqrt{y}+\sqrt{z}\right)^2}{3}=3\)
=> \(P\ge\frac{1}{2}.3+3-\frac{3}{2}=3\)
Vậy MinP=3 khi x=y=z=1