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cj tìm ra nghiệm r e! x=y=z=1 . Nhưg mà vẫn chưa giải ra đc
Vì \(x,y,z\in\left[0;1\right]\) nên \(\left(x-1\right)\left(z-1\right)\ge0\Leftrightarrow xz+1\ge x+z\)
\(\Rightarrow xz+1+y\ge x+y+z\) \(\Rightarrow\dfrac{x}{1+y+zx}\le\dfrac{x}{x+y+z}\)
Tương tự ta có:
\(\dfrac{x}{1+y+xz}+\dfrac{y}{1+z+xy}+\dfrac{z}{1+x+yz}\le\dfrac{x}{x+y+z}+\dfrac{y}{x+y+z}+\dfrac{z}{z+y+z}=1\)
hay \(\dfrac{3}{x+y+z}\le1\Leftrightarrow x+y+z\ge3\)
Mà \(x;y;z\in\left[0;1\right]\Rightarrow x+y+z\le3\)
\(\Rightarrow x+y+z=3\) và \(x=y=z=1\)
a/ \(\frac{1}{1+x}+\frac{1}{1+y}\le\frac{2}{1+\sqrt{xy}}\)
\(\Leftrightarrow\left(1+x\right)\left(1+\sqrt{xy}\right)+\left(1+y\right)\left(1+\sqrt{xy}\right)-2\left(1+x\right)\left(1+y\right)\le0\)
\(\Leftrightarrow x\sqrt{xy}+2\sqrt{xy}+y\sqrt{xy}-x-y-2xy\le0\)
\(\Leftrightarrow\sqrt{xy}\left(x-2\sqrt{xy}+y\right)-\left(x-2\sqrt{xy}+y\right)\le0\)
\(\Leftrightarrow\left(\sqrt{x}-\sqrt{y}\right)^2\left(\sqrt{xy}-1\right)\le0\) đúng vì \(x,y\le1\)
b/ Vì \(\hept{\begin{cases}0\le x\le y\le z\le t\\yt\le1\end{cases}}\)
\(\Rightarrow\hept{\begin{cases}xz\le1\\yt\le1\end{cases}}\)
Áp dụng câu a ta được
\(\frac{1}{1+x}+\frac{1}{1+y}+\frac{1}{1+z}+\frac{1}{1+t}\le\frac{2}{1+\sqrt{xz}}+\frac{2}{1+\sqrt{yt}}\le\frac{4}{1+\sqrt[4]{xyzt}}\)
4) Áp dụng bất đẳng thức Bunyakovsky
\(\Rightarrow\left(x^4+yz\right)\left(1+1\right)\ge\left(x^2+\sqrt{yz}\right)^2\)
\(\Rightarrow\dfrac{x^2}{x^4+yz}\le\dfrac{2x^2}{\left(x^2+\sqrt{yz}\right)^2}\)
Tượng tự ta có \(\left\{{}\begin{matrix}\dfrac{y^2}{y^4+xz}\le\dfrac{2y^2}{\left(y^2+\sqrt{xz}\right)^2}\\\dfrac{z^2}{z^4+xy}\le\dfrac{2z^2}{\left(z^2+\sqrt{xy}\right)^2}\end{matrix}\right.\)
\(\Rightarrow VT\le2\left[\dfrac{x^2}{\left(x^2+\sqrt{yz}\right)^2}+\dfrac{y^2}{\left(y^2+\sqrt{xz}\right)^2}+\dfrac{z^2}{\left(z^2+\sqrt{xy}\right)}\right]\)
Chứng minh rằng \(2\left[\dfrac{x^2}{\left(x^2+\sqrt{yz}\right)^2}+\dfrac{y^2}{\left(y^2+\sqrt{xz}\right)^2}+\dfrac{z^2}{\left(z^2+\sqrt{xy}\right)}\right]\le\dfrac{3}{2}\)
\(\Leftrightarrow\dfrac{x^2}{\left(x^2+\sqrt{yz}\right)^2}+\dfrac{y^2}{\left(y^2+\sqrt{xz}\right)^2}+\dfrac{z^2}{\left(z^2+\sqrt{xy}\right)^2}\le\dfrac{3}{4}\)
Áp dụng bất đẳng thức Cauchy
\(\Rightarrow x^2+\sqrt{yz}\ge2\sqrt{x^2\sqrt{yz}}=2x\sqrt{\sqrt{yz}}\)
\(\Rightarrow\left(x^2+\sqrt{yz}\right)^2\ge4x^2\sqrt{yz}\)
\(\Rightarrow\dfrac{x^2}{\left(x^2+\sqrt{yz}\right)^2}\le\dfrac{x^2}{4x^2\sqrt{yz}}=\dfrac{1}{4\sqrt{yz}}\)
Tượng tự ta có \(\left\{{}\begin{matrix}\dfrac{y^2}{\left(y^2+\sqrt{xz}\right)^2}\le\dfrac{1}{4\sqrt{xz}}\\\dfrac{z^2}{\left(z^2+\sqrt{xy}\right)^2}\le\dfrac{1}{4\sqrt{xy}}\end{matrix}\right.\)
\(\Leftrightarrow\dfrac{x^2}{\left(x^2+\sqrt{yz}\right)^2}+\dfrac{y^2}{\left(y^2+\sqrt{xz}\right)^2}+\dfrac{z^2}{\left(z^2+\sqrt{xy}\right)^2}\le\dfrac{1}{4}\left(\dfrac{1}{\sqrt{xy}}+\dfrac{1}{\sqrt{yz}}+\dfrac{1}{\sqrt{xz}}\right)\)
Chứng minh rằng \(\dfrac{1}{4}\left(\dfrac{1}{\sqrt{xy}}+\dfrac{1}{\sqrt{yz}}+\dfrac{1}{\sqrt{xz}}\right)\le\dfrac{3}{4}\)
\(\Leftrightarrow\dfrac{1}{\sqrt{xy}}+\dfrac{1}{\sqrt{yz}}+\dfrac{1}{\sqrt{xz}}\le3\)
Theo đề bài ta có \(x^2+y^2+z^2=3xyz\)
\(\Rightarrow\dfrac{x}{yz}+\dfrac{y}{xz}+\dfrac{z}{xy}=3\)
\(\Rightarrow\dfrac{1}{\sqrt{xy}}+\dfrac{1}{\sqrt{yz}}+\dfrac{1}{\sqrt{xz}}\le3\)
\(\Leftrightarrow\dfrac{1}{\sqrt{xy}}+\dfrac{1}{\sqrt{yz}}+\dfrac{1}{\sqrt{xz}}\le\dfrac{x}{yz}+\dfrac{y}{xz}+\dfrac{z}{xy}\)
Áp dụng bất đẳng thức Cauchy
\(\Rightarrow\dfrac{1}{\sqrt{xy}}\le\dfrac{\dfrac{1}{x}+\dfrac{1}{y}}{2}\)
Tượng tự ta có \(\left\{{}\begin{matrix}\dfrac{1}{\sqrt{xz}}\le\dfrac{\dfrac{1}{x}+\dfrac{1}{z}}{2}\\\dfrac{1}{\sqrt{yz}}\le\dfrac{\dfrac{1}{z}+\dfrac{1}{y}}{2}\end{matrix}\right.\)
\(\Rightarrow\dfrac{1}{\sqrt{xy}}+\dfrac{1}{\sqrt{yz}}+\dfrac{1}{\sqrt{xz}}\le\dfrac{1}{x}+\dfrac{1}{y}+\dfrac{1}{z}\) (1)
Áp dụng bất đẳng thức Cauchy
\(\Rightarrow\dfrac{x}{yz}+\dfrac{y}{xz}\ge2\sqrt{\dfrac{1}{z^2}}=\dfrac{2}{z}\)
Tượng tự ta có \(\left\{{}\begin{matrix}\dfrac{y}{xz}+\dfrac{z}{xy}\ge\dfrac{2}{x}\\\dfrac{x}{zy}+\dfrac{z}{xy}\ge\dfrac{2}{y}\end{matrix}\right.\)
\(\Rightarrow2\left(\dfrac{x}{yz}+\dfrac{y}{xz}+\dfrac{z}{xy}\right)\ge2\left(\dfrac{1}{x}+\dfrac{1}{y}+\dfrac{1}{z}\right)\)
\(\Leftrightarrow\dfrac{x}{yz}+\dfrac{y}{xz}+\dfrac{z}{xy}\ge\dfrac{1}{x}+\dfrac{1}{y}+\dfrac{1}{z}\) (2)
Từ (1) và (2)
\(\Rightarrow\dfrac{1}{\sqrt{xy}}+\dfrac{1}{\sqrt{yz}}+\dfrac{1}{\sqrt{xz}}\le3\) ( đpcm )
Vậy \(\dfrac{1}{4}\left(\dfrac{1}{\sqrt{xy}}+\dfrac{1}{\sqrt{yz}}+\dfrac{1}{\sqrt{xz}}\right)\le\dfrac{3}{4}\)
\(\Rightarrow2\left[\dfrac{x^2}{\left(x^2+\sqrt{yz}\right)^2}+\dfrac{y^2}{\left(y^2+\sqrt{xz}\right)^2}+\dfrac{z^2}{\left(z^2+\sqrt{xy}\right)}\right]\le\dfrac{3}{2}\)
Mà \(VT\le2\left[\dfrac{x^2}{\left(x^2+\sqrt{yz}\right)^2}+\dfrac{y^2}{\left(y^2+\sqrt{xz}\right)^2}+\dfrac{z^2}{\left(z^2+\sqrt{xy}\right)}\right]\)
\(\Rightarrow VT\le\dfrac{3}{2}\) ( đpcm )
Dấu " = " xảy ra khi \(x=y=z=1\)
3. Ta có :\(x^2\left(1-2x\right)=x.x.\left(1-2x\right)\le\dfrac{\left(x+x+1-2x\right)^3}{27}=\dfrac{1}{27}\)(bđt cô si)
Dấu "=" xảy ra khi :x=1-2x\(\Leftrightarrow x=\dfrac{1}{3}\)
Vậy max của Qlaf 1/27 khi x=1/3
Từ đề bài ta có:
\(\left\{{}\begin{matrix}\left(x+1\right)\left(y+1\right)\left(z+1\right)\ge0\\\left(x-3\right)\left(y-3\right)\left(3-z\right)\ge0\end{matrix}\right.\)
\(\Leftrightarrow\left\{{}\begin{matrix}xyz+\left(xy+yz+zx\right)+\left(x+y+z\right)+1\ge0\\-xyz+3\left(xy+yz+zx\right)-9\left(x+y+z\right)+27\ge0\end{matrix}\right.\)
Lấy trên + dưới ta được
\(4\left(xy+yz+zx\right)-8\left(x+y+z\right)+28\ge0\)
\(\Leftrightarrow4\left(xy+yz+zx\right)+20\ge0\)
\(\Leftrightarrow2\left(x+y+z\right)^2+20\ge2x^2+2y^2+2z^2\)
\(\Leftrightarrow x^2+y^2+z^2\le11\)
Bài này Karamata là vừa :D
Giả sử \(a\ge b\ge c\)
Khi \(f\left(x\right)=x^2\) là hàm lồi trên \(\left[-1,3\right]\) và \((-1,-1,3)\succ(a,b,c)\)
Theo Karamata's inequality ta có:
\(11=\left(-1\right)^2+\left(-1\right)^2+3^2\ge a^2+b^2+c^2\)
Ta có \(\sqrt{1+x^2}+\sqrt{2x}\le\sqrt{2}\left(x+1\right)\)
\(\sqrt{1+y^2}+\sqrt{2y}\le\sqrt{2}\left(y+1\right)\)
\(\sqrt{1+z^2}+\sqrt{2z}\le\sqrt{2}\left(z+1\right)\)
\(\Rightarrow\sqrt{1+x^2}+\sqrt{1+y^2}+\sqrt{1+z^2}+\sqrt{2x}+\sqrt{2y}+\sqrt{2z}\le\sqrt{2}\left(x+y+z+3\right)\le6\sqrt{2}\)
Ta lại có \(\sqrt{x}+\sqrt{y}+\sqrt{z}\le\sqrt{3\left(x+y+z\right)}\le3\)
Theo đề bài ta có
\(\sqrt{1+x^2}+\sqrt{1+y^2}+\sqrt{1+z^2}+3\left(\sqrt{x}+\sqrt{y}+\sqrt{z}\right)\)
\(\le6\sqrt{2}+\left(3-\sqrt{2}\right)\left(\sqrt{x}+\sqrt{y}+\sqrt{z}\right)\le3\sqrt{2}+9\)
Dấu = xảy ra khi x = y = z = 1