Vì xyz=1\(\Rightarrow x^2\left(y+z\right)\ge2x^2\sqrt{yz}=2x\sqrt{x}\)

Tương tự \(y^2\left(z+x\right)\ge2y\sqrt{y};z^2=\left(x+y\right)\ge2z\sqrt{z}\)

\(\Rightarrow P\ge\frac{2x\sqrt{x}}{y\sqrt{y}+2z\sqrt{z}}+\frac{2y\sqrt{y}}{z\sqrt{z}+2x\sqrt{x}}+\frac{2z\sqrt{z}}{x\sqrt{x}+2y\sqrt{y}}\)

Đặt \(x\sqrt{x}+2y\sqrt{y}=a;y\sqrt{y}+2z\sqrt{z}=b;z\sqrt{z}+2x\sqrt{x}=c\)

\(\Rightarrow x\sqrt{x}=\frac{4c+a-2b}{9};y\sqrt{y}=\frac{4a+b-2c}{9};z\sqrt{z}=\frac{4b+c-2a}{9}\)

\(\Rightarrow P\ge\frac{2}{9}\left(\frac{4c+a-2b}{b}+\frac{4a+b-2c}{a}+\frac{4b+c-2a}{b}\right)\)

\(=\frac{2}{9}\text{ }\left[4\left(\frac{c}{b}+\frac{a}{c}+\frac{b}{a}\right)+\left(\frac{a}{b}+\frac{b}{c}+\frac{c}{a}\right)-6\right]\ge\frac{2}{9}\left(4.3+2-6\right)=2\)

Min P =2 khi và chỉ khi a=b=c khi va chỉ khi x=y=z=1

Ta có: \(\left(\sqrt{x+y}\right)^2=\left(\sqrt{x-z}+\sqrt{y-z}\right)^2\)

\(\Leftrightarrow\)\(x+y=x+y-2z+2\sqrt{\left(x-z\right)\left(y-z\right)}\)

\(\Leftrightarrow2z=2\sqrt{\left(x-z\right)\left(y-z\right)}\)

Theo giả thiết, ta có: 

theo giả thiết, ta có: \(\frac{1}{x}+\frac{1}{y}-\frac{1}{z}=0\Rightarrow\frac{1}{z}-\frac{1}{x}=\frac{1}{y}\)\(\Rightarrow\frac{x-z}{zx}=\frac{1}{y}\Rightarrow x-z=\frac{zx}{y}\)

Tương tự, ta có: \(y-z=\frac{zy}{x}\)

Do đó: \(2\sqrt{\left(x-z\right)\left(y-z\right)}=2\sqrt{\frac{zx}{y}.\frac{zy}{x}}=2z\) (1)

ta có: \(\left(\sqrt{x+y}\right)^2=\left(\sqrt{x-z}+\sqrt{y-z}\right)^2\)

\(\Leftrightarrow2z=2\sqrt{\left(x-z\right)\left(y-z\right)}\)(2)

Thay (2) vào (1) ta thấy (2) luôn đúng

Suy ra ĐPCM

Vì \(x>0,y>0\Rightarrow\frac{1}{x}>0;\frac{1}{y}>0\)

mà \(\frac{1}{x}+\frac{1}{y}-\frac{1}{z}=0\Rightarrow\frac{1}{z}=\frac{1}{x}+\frac{1}{y}\Rightarrow\frac{1}{z}>0\Rightarrow z>0\)

Ta có: \(\frac{1}{x}+\frac{1}{y}-\frac{1}{z}=0\Leftrightarrow yz+zx-xy=0\)

\(\Leftrightarrow-z^2=-z^2+yz+zx-xy=-\left(x-z\right)\left(y-z\right)\)

\(\Leftrightarrow z^2=\left(x-z\right)\left(y-z\right)>0\)

\(\Rightarrow z=\sqrt{\left(x-z\right)\left(y-z\right)}\left(z>0\right)\)

Lại có: \(x+y=x-z+y-z+2z\)

\(=\left(x-z\right)+\left(y-z\right)+2\sqrt{\left(x-z\right)\left(y-z\right)}=\left(\sqrt{x-z}+\sqrt{y-z}\right)^2\)

Suy ra \(\sqrt{x+y}=\sqrt{x-z}+\sqrt{y-z}\) (ĐPCM)

\(1=\frac{1}{2}\left(\frac{x}{y}+\frac{y}{z}\right)+\frac{1}{2}\left(\frac{y}{z}+\frac{z}{x}\right)+\frac{1}{2}\left(\frac{z}{x}+\frac{x}{y}\right)\)

\(\ge\sqrt{\frac{x}{y}.\frac{y}{z}}+\sqrt{\frac{y}{z}.\frac{z}{x}}+\sqrt{\frac{z}{x}.\frac{x}{y}}=VP\) (rút gọn lại thôi:v)

\(P=\frac{x^2}{x\sqrt{y}}+\frac{y^2}{y\sqrt{z}}+\frac{z^2}{z\sqrt{x}}\ge\frac{\left(x+y+z\right)^2}{x\sqrt{y}+y\sqrt{z}+z\sqrt{x}}\)

\(P\ge\frac{\left(x+y+z\right)^2}{\frac{x\left(y+4\right)}{4}+\frac{y\left(z+4\right)}{4}+\frac{z\left(x+4\right)}{4}}=\frac{4\left(x+y+z\right)^2}{xy+yz+zx+4\left(x+y+z\right)}\)

\(P\ge\frac{4\left(x+y+z\right)^2}{\frac{\left(x+y+z\right)^2}{3}+4\left(x+y+z\right)}=\frac{12\left(x+y+z\right)}{x+y+z+12}=12-\frac{144}{x+y+z+12}\ge12-\frac{144}{12+12}=6\)

\(\Rightarrow P_{min}=6\) khi \(x=y=z=4\)

\(\hept{\begin{cases}x,y,z>0\\x+y+z=xyz\end{cases}}\)

\(\Rightarrow\frac{1}{xy} +\frac{1}{yz}+\frac{1}{zx}=1\)

Có : \(\frac{1}{\sqrt{1+x^2}}=\frac{1}{\sqrt{\frac{1}{xy}+\frac{1}{yz}+\frac{1}{zx}+x^2}}\le\frac{1}{2.\sqrt{\frac{x^2y}{xyz}}}\le\frac{1}{2}\)

\(\frac{1}{\sqrt{1+y^2}}=\frac{1}{\sqrt{\frac{1}{xy}+\frac{1}{yz}+\frac{1}{zx}+y^2}}\le\frac{1}{2\sqrt{\frac{y^2z}{xyz}}}\le\frac{1}{2}\)

\(\frac{1}{\sqrt{1+z^2}}=\frac{1}{\sqrt{\frac{1}{xy}+\frac{1}{yz}+\frac{1}{zx}+z^2}}\le\frac{1}{2\sqrt{\frac{z^2x}{xyz}}}\le\frac{1}{2}\)

\(\Rightarrow\frac{1}{\sqrt{1+x^2}}+\frac{1}{\sqrt{1+y^2}}+\frac{1}{\sqrt{1+z^2}}\le\frac{3}{2}\)

Vậy P max = 3/2

bn tìm đề thi hsg tỉnh thanh hóa lớp 9 năm nào đó là thấy

bài này dài,ngại làm

đặt là được

Câu hỏi của Hoàng Gia Anh Vũ - Toán lớp 9 - Học toán với OnlineMath