Lời giải:

Ta có: \(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}=1\Leftrightarrow xy+yz+xz=xyz\)

\(\Rightarrow x^2+xy+yz+xz=x^2+xyz=x(x+yz)\)

\(\Leftrightarrow x+yz=\frac{x^2+xy+yz+xz}{x}=\frac{(x+y)(x+z)}{x}\)

\(\Rightarrow \sqrt{x+yz}=\sqrt{\frac{(x+y)(x+z)}{x}}\)

Áp dụng BĐT Bunhiacopxky:\((x+y)(x+z)\geq (x+\sqrt{yz})^2\)

\(\Rightarrow \sqrt{x+yz}=\sqrt{\frac{(x+y)(x+z)}{x}}\geq \frac{x+\sqrt{yz}}{\sqrt{x}}\)

Hoàn toàn tương tự:

\(\sqrt{y+xz}\geq \frac{y+\sqrt{xz}}{\sqrt{y}}\); \(\sqrt{z+xy}\geq \frac{z+\sqrt{xy}}{\sqrt{z}}\)

Cộng theo vế các BĐT đã thu được ta có:

\(\text{VT}\geq \frac{x+\sqrt{yz}}{\sqrt{x}}+\frac{y+\sqrt{xz}}{\sqrt{y}}+\frac{z+\sqrt{xy}}{\sqrt{z}}=\sqrt{x}+\sqrt{y}+\sqrt{z}+\frac{xy+yz+xz}{\sqrt{xyz}}\)

\(\Leftrightarrow \text{VT}\geq \sqrt{x}+\sqrt{y}+\sqrt{z}+\frac{xyz}{\sqrt{xyz}}=\sqrt{x}+\sqrt{y}+\sqrt{z}+\sqrt{xyz}=\text{VP}\)

Do đó ta có đpcm.

Dấu bằng xảy ra khi \(x=y=z=3\)

\(\dfrac{\sqrt{1\left(x-1\right)}}{x}\le\dfrac{1+x-1}{2x}=\dfrac{1}{2}\) ( cauchy )

TT,\(\dfrac{\sqrt{y-2}}{y}\le\dfrac{1}{2\sqrt{2}};\dfrac{\sqrt{z-3}}{z}\le\dfrac{1}{2\sqrt{3}}\)

cộng vế theo vế => đpcm

Thì biết pass facebook thôi chứ cũng không biết có hack không

Bạn ấy đăng nhập bằng FACEBOOK mà

Ta có \(\sqrt{xy}+\sqrt{yz}+\sqrt{zx}=\sqrt{xyz}\left(x,y,z>0\right)\).

\(\Leftrightarrow\frac{1}{\sqrt{x}}+\frac{1}{\sqrt{y}}+\frac{1}{\sqrt{z}}=1\).

\(P=\frac{1}{xyz}\left(x\sqrt{2y^2+yz+2z^2}+y\sqrt{2z^2+xz+2x^2}+z\sqrt{2x^2+xy+y^2}\right)\)\(\left(x,y,z>0\right)\).

Ta có: 

\(\sqrt{2y^2+2yz+2z^2}=\sqrt{\frac{5}{4}\left(y^2+2yz+z^2\right)+\frac{3}{4}\left(y^2-2yz+z^2\right)}\)

\(=\sqrt{\frac{5}{4}\left(y+z\right)^2+\frac{3}{4}\left(y-z\right)^2}\).

Ta có:

\(\frac{3}{4}\left(y-z\right)^2\ge0\forall y;z>0\).

\(\Leftrightarrow\frac{3}{4}\left(y-z\right)^2+\frac{5}{4}\left(y+z\right)^2\ge\frac{5}{4}\left(y+z\right)^2\forall y;z>0\).

\(\Rightarrow\sqrt{\frac{3}{4}\left(y-z\right)^2+\frac{5}{4}\left(y+z\right)^2}\ge\frac{\sqrt{5}}{2}\left(y+z\right)\forall y,z>0\).

\(\Leftrightarrow\sqrt{2y^2+yz+2z^2}\ge\frac{\sqrt{5}}{2}\left(y+z\right)\forall y;z>0\).

\(\Leftrightarrow x\sqrt{2y^2+yz+2z^2}\ge\frac{\sqrt{5}}{2}x\left(y+z\right)\forall x;y;z>0\left(1\right)\).

Chứng minh tương tự, ta được:

\(y\sqrt{2x^2+xz+2z^2}\ge\frac{\sqrt{5}}{2}y\left(x+z\right)\forall x;y;z>0\left(2\right)\).

Chứng minh tương tự, ta được:

\(z\sqrt{2x^2+xy+2y^2}\ge\frac{\sqrt{5}}{2}z\left(x+y\right)\forall x;y;z>0\left(3\right)\).

Từ \(\left(1\right),\left(2\right),\left(3\right)\), ta được:

\(x\sqrt{2y^2+yz+2z^2}+y\sqrt{2z^2+xz+2x^2}+z\sqrt{2x^2+xy+2y^2}\)\(\ge\)\(\frac{\sqrt{5}}{2}\left[x\left(y+z\right)+y\left(x+z\right)+z\left(x+y\right)\right]=\sqrt{5}\left(xy+yz+zx\right)\).

\(\Leftrightarrow\frac{1}{xyz}\left(x\sqrt{2y^2+yz+z^2}+y\sqrt{2z^2+zx+2x^2}+z\sqrt{2x^2+xy+2y^2}\right)\)\(\ge\)\(\frac{\sqrt{5}\left(xy+yz+zx\right)}{xyz}=\sqrt{5}\left(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}\right)\).

\(\Leftrightarrow P\ge\frac{\sqrt{5}}{3}.3\left(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}\right)=\frac{\sqrt{5}}{3}\left(1^2+1^2+1^2\right)\left[\left(\frac{1}{\sqrt{x}}\right)^2+\left(\frac{1}{\sqrt{y}}\right)^2+\left(\frac{1}{\sqrt{z}}\right)^2\right]\)

\(\left(4\right)\).

Vì \(x,y,z>0\)nên áp dụng bất đẳng thức Bu-nhi-a-cốp-xki, ta được:
\(\left(1^2+1^2+1^2\right)\left[\left(\frac{1}{\sqrt{x}}\right)^2+\left(\frac{1}{\sqrt{y}}\right)^2+\left(\frac{1}{\sqrt{z}}\right)^2\right]\ge\)\(\left(1.\frac{1}{\sqrt{x}}+1.\frac{1}{\sqrt{y}}+1.\frac{1}{\sqrt{z}}\right)^2\).

\(\Leftrightarrow\left(1^2+1^2+1^2\right)\left[\left(\frac{1}{\sqrt{x}}\right)^2+\left(\frac{1}{\sqrt{y}}\right)^2+\left(\frac{1}{\sqrt{z}}\right)^2\right]\ge\left(\frac{1}{\sqrt{x}}+\frac{1}{\sqrt{y}}+\frac{1}{\sqrt{z}}\right)^2=1^2=1\)

(vì\(\frac{1}{\sqrt{x}}+\frac{1}{\sqrt{y}}+\frac{1}{\sqrt{z}}=1\)).

\(\Leftrightarrow\frac{\sqrt{5}}{3}\left(1^2+1^2+1^2\right)\left[\left(\frac{1}{\sqrt{x}}\right)^2+\left(\frac{1}{\sqrt{y}}\right)^2+\left(\frac{1}{\sqrt{z}}\right)^2\right]\ge\frac{\sqrt{5}}{3}\)\(\left(5\right)\).

Từ \(\left(4\right)\)và \(\left(5\right)\), ta được:

\(P\ge\frac{\sqrt{5}}{3}\).

Dấu bằng xảy ra.

\(\Leftrightarrow\hept{\begin{cases}x=y=z>0\\\sqrt{xy}+\sqrt{yz}+\sqrt{zx}=\sqrt{xyz}\end{cases}}\Leftrightarrow x=y=z=9\).

Vậy \(minP=\frac{\sqrt{5}}{3}\Leftrightarrow x=y=z=9\).

a)\(A=\sqrt{2}-\sqrt{12-8\sqrt{2}}\)

\(A=\sqrt{2}-\sqrt{\left(2\sqrt{2}-2\right)^2}\)

\(A=\sqrt{2}-2\sqrt{2}+2\)

\(A=2-\sqrt{2}\)

c)\(C=\dfrac{2\sqrt{3-\sqrt{5}}}{\sqrt{10}-\sqrt{2}}=\dfrac{2\sqrt{3-\sqrt{5}}}{\sqrt{2}\left(\sqrt{5}-1\right)}=\dfrac{\sqrt{2}\sqrt{3-\sqrt{5}}}{\sqrt{5}-1}=\dfrac{\sqrt{6-2\sqrt{5}}}{\sqrt{5}-1}=\dfrac{\sqrt{\left(\sqrt{5}-1\right)^2}}{\sqrt{5}-1}=\dfrac{\sqrt{5}-1}{\sqrt{5}-1}=1\)

d)với x,y,x>0 xyz=100 =>\(\sqrt{xyz}=\sqrt{100}=10\)

\(D=\dfrac{\sqrt{x}}{\sqrt{xy}+\sqrt{x}+10}+\dfrac{\sqrt{y}}{\sqrt{yz}+\sqrt{y}+1}+\dfrac{10\sqrt{z}}{\sqrt{xz}+10\sqrt{z}+10}\)

\(D=\dfrac{\sqrt{x}}{\sqrt{xy}+\sqrt{x}+\sqrt{xyz}}+\dfrac{\sqrt{y}}{\sqrt{yz}+\sqrt{y}+1}+\dfrac{\sqrt{xyz^2}}{\sqrt{xz}+\sqrt{xyz^2}+\sqrt{xyz}}\)

\(D=\dfrac{1}{\sqrt{y}+1+\sqrt{yz}}+\dfrac{\sqrt{y}}{\sqrt{yz}+\sqrt{y}+1}+\dfrac{\sqrt{yz}}{1+\sqrt{yz}+\sqrt{y}}\)

\(D=\dfrac{1+\sqrt{y}+\sqrt{yz}}{\sqrt{yz}+\sqrt{y}+1}=1\)

mình chỉ giải được câu a,c,d còn câu b mình nghĩ sai đề

Nhân thêm và, dùng Cauchy

\(1\sqrt{x-1}=\sqrt{1\left(x-1\right)}\le\frac{x}{2}\). Tương tự với y thì nhân 2; với z thì nhân 3

\(\frac{yz\sqrt{x-1}+xz\sqrt{y-2}+xy\sqrt{z-3}}{xyz}\)

\(=\frac{\sqrt{x-1}}{x}+\frac{\sqrt{y-2}}{y}+\frac{\sqrt{z-3}}{z}\)

Ta có: \(\sqrt{x-1}\le\frac{1+x-1}{2}=\frac{x}{2}\)

\(\Rightarrow\frac{\sqrt{x-1}}{x}\le\frac{1}{2}\)

Chứng minh tương tự ta được: \(\frac{\sqrt{y-2}}{y}\le\frac{1}{2\sqrt{2}}\)

                                                 \(\frac{\sqrt{z-3}}{z}\le\frac{1}{2\sqrt{3}}\)

Suy ra: \(\frac{\sqrt{x-1}}{x}+\frac{\sqrt{y-2}}{y}+\frac{\sqrt{z-3}}{z}\le\frac{1}{2}+\frac{1}{2\sqrt{2}}+\frac{1}{2\sqrt{3}}=\frac{1}{2}\left(1+\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{3}}\right)\)

Vậy GTLN của biểu thức = \(\frac{1}{2}.\left(1+\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{3}}\right)\Leftrightarrow\hept{\begin{cases}x=2\\y=4\\z=6\end{cases}}\)

thay xyz=100 vào tính rồi rút gọn thôi

cụ thể ra là gì