TA CÓ:
\(Q=\frac{x\left(\sqrt{x+zy}-x\right)}{x+yz-x^2}+\frac{y\left(\sqrt{y+zx}-y\right)}{y+zx-y^2}+\frac{z\left(\sqrt{xy+z}-z\right)}{z+xy-z^2}\)

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

\(=\frac{x\left(\sqrt{\left(x+y\right)\left(z+x\right)}-x\right)}{xy+yz+zx}+\frac{y\left(\sqrt{\left(x+y\right)\left(y+z\right)}-y\right)}{xy+yz+zx}+\frac{z\left(\sqrt{\left(y+z\right)\left(z+x\right)}-z\right)}{xy+yz+za}\)

ÁP DỤNG BĐT CÔ-SI TA ĐƯỢC:

\(Q\le\frac{x\left(\frac{x+y+z+x}{2}-x\right)}{xy+zx+yz}+\frac{y\left(\frac{x+y+z+y}{2}-y\right)}{xy+yz+zx}+\frac{z\left(\frac{x+y+z+z}{2}-z\right)}{xy+yz+zx}\)

\(=\frac{xy+zx}{2\left(xy+yz+zx\right)}+\frac{xy+yz}{2\left(xy+yz+zx\right)}+\frac{yz+zx}{2\left(xy+yz+zx\right)}=1\)

DẤU BẰNG  XẢY RA \(\Leftrightarrow x=y=z=\frac{1}{3}\)

=

1/3

hok tốt

Từ \(\frac{1}{xy}+\frac{1}{yz}+\frac{1}{xz}=1\)

\(\Rightarrow\)\(x+y+z=xyz\)

Ta có : \(\sqrt{yz\left(1+x^2\right)}=\sqrt{yz+x^2yz}=\sqrt{yz+x\left(x+y+z\right)}=\sqrt{\left(x+y\right)\left(x+z\right)}\)

Tương tự : \(\sqrt{xy\left(1+z^2\right)}=\sqrt{\left(z+y\right)\left(z+x\right)}\); \(\sqrt{zx\left(1+y^2\right)}=\sqrt{\left(y+z\right)\left(y+x\right)}\)

Nên \(Q=\frac{x}{\sqrt{\left(x+y\right)\left(x+z\right)}}+\frac{y}{\sqrt{\left(y+z\right)\left(y+x\right)}}+\frac{z}{\sqrt{\left(z+x\right)\left(z+y\right)}}\)

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

Áp dụng BĐT \(\sqrt{A.B}\le\frac{A+B}{2}\left(A,B>0\right)\)

Dấu "=" xảy ra khi A = B :

Ta được :

\(Q\le\frac{1}{2}\left(\frac{x}{x+y}+\frac{x}{x+z}+\frac{y}{y+x}+\frac{y}{y+z}+\frac{z}{z+x}+\frac{z}{z+y}\right)=\frac{3}{2}\)

Vậy GTLN của \(Q=\frac{3}{2}\)khi \(x=y=z=\sqrt{3}\)

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

\(\ge x+\sqrt{2\sqrt{x^2yz}+x\left(y+z\right)}=x+\sqrt{x\cdot2\sqrt{yz}+x\left(y+z\right)}=x+\sqrt{x\left(y+z+2\sqrt{yz}\right)}\)

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

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

tương tự :

\(\frac{y}{y+\sqrt{y+xz}}\le\frac{\sqrt{y}}{\sqrt{y}+\sqrt{x}+\sqrt{z}}\)

\(\frac{z}{z+\sqrt{z+xy}}\le\frac{\sqrt{z}}{\sqrt{z}+\sqrt{x}+\sqrt{y}}\) 

cộng vế theo vế ta được 

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

dấu "=" xảy tra khi x=y=z=1/3

cái này thì chịu

khó muốn chết luôn làm sao làm đc

Câu hỏi của phan tuấn anh - Toán lớp 9 - Học toán với OnlineMath cái này y hệt, tham khảo đi nếu vẫn chưa làm dc thì nhắn cho mk

\(xy+yz+zx\le3xyz\Rightarrow\frac{1}{x}+\frac{1}{y}+\frac{1}{z}\le3\)

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

\(P\le\frac{1}{\sqrt{x^2+3xy}}+\frac{1}{\sqrt{y^2+3yz}}+\frac{1}{\sqrt{z^2+3zx}}=\frac{4}{2\sqrt{4x\left(x+3y\right)}}+\frac{4}{2\sqrt{4y\left(y+3z\right)}}+\frac{1}{2\sqrt{4z\left(z+3x\right)}}\)

\(P\le4\left(\frac{1}{4x+x+3y}+\frac{1}{4y+y+3z}+\frac{1}{4z+z+3x}\right)=4\left(\frac{1}{5x+3y}+\frac{1}{5y+3z}+\frac{1}{5z+3x}\right)\)

\(P\le\frac{4}{64}\left(\frac{5}{x}+\frac{3}{y}+\frac{5}{y}+\frac{3}{z}+\frac{5}{z}+\frac{3}{x}\right)=\frac{1}{2}\left(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}\right)\le\frac{3}{2}\)

\(P_{max}=\frac{3}{2}\) khi \(x=y=z=1\)

Ta đặt \(x=tanA;y=tanB;z=tanC\) với \(ABC\) là các góc tam giá từ đây cần c/m

\(sinA+sinB+sinC\le\frac{3\sqrt{3}}{2}\)

tài liệu c/m BĐT này đầy trên mạng bn có thể tham tham khảo

VD:Cm : sinA+sinB+sinC bé hơn hoặc bằng (3* căn3)/2? | Yahoo Hỏi & Đáp

Dự đoán khi \(x=y=z=\frac{1}{\sqrt{3}}\) thì ta tìm được \(P=\frac{3\sqrt{3}}{2}\)

Ta sẽ chứng minh nó là GTNN

Thật vậy, ta cần chứng minh 

\(Σ\frac{1}{\sqrt{x^2+xy+xz+yz}}\le\frac{3\sqrt{3}}{2\sqrt{xy+xz+yz}}\left(xy+yz+xz=1\right)\)

\(\LeftrightarrowΣ\sqrt{x+y}\le\frac{3\sqrt{3\left(x+y\right)\left(x+z\right)\left(y+z\right)}}{2\sqrt{xy+xz+yz}}\)

Nhưng theo BĐT Cauchy-Schwarz ta có: 

\(\left(Σ\sqrt{x+y}\right)^2\le\left(1+1+1\right)Σ\left(x+y\right)=6\left(x+y+z\right)\)

Như vậy, ta còn phải chứng minh :

\(\sqrt{6\left(x+y+z\right)}\le\frac{3\sqrt{3\left(x+y\right)\left(x+z\right)\left(y+z\right)}}{2\sqrt{xy+xz+yz}}\)

\(\Leftrightarrow9\left(x+y\right)\left(x+z\right)\left(y+z\right)\ge8\left(x+y+z\right)\left(xy+xz+yz\right)\)

\(\LeftrightarrowΣz\left(x-y\right)^2\ge0\) luôn đúng. Nên \(P_{Min}=\frac{3\sqrt{3}}{2}\Leftrightarrow x=y=z=\frac{1}{\sqrt{3}}\)

Áp dụng BĐT Cô-si ta có:

\(1+x^3+y^3\ge3\sqrt[3]{1.x^3.y^3}=3xy\Rightarrow\sqrt{1+x^3+y^3}\ge\sqrt{3xy}\Rightarrow\frac{\sqrt{1+x^3+y^3}}{xy}\ge\frac{\sqrt{3xy}}{xy}\)

Tương tự:\(\frac{\sqrt{1+y^3+z^3}}{yz}\ge\frac{\sqrt{3yz}}{yz};\frac{\sqrt{1+z^3+x^3}}{zx}\ge\frac{\sqrt{3zx}}{zx}\)

Công vế với vế của 3 BĐT trên ta đươc:

\(P\ge\frac{\sqrt{3xy}}{xy}+\frac{\sqrt{3yz}}{yz}+\frac{\sqrt{3zx}}{zx}=\sqrt{3}\left(\frac{1}{\sqrt{xy}}+\frac{1}{\sqrt{yz}}+\frac{1}{\sqrt{zx}}\right)\) \(=\sqrt{3}.\left(\sqrt{x}+\sqrt{y}+\sqrt{z}\right)\ge3\sqrt{3}\)

Dấu '='xảy ra khi \(\hept{\begin{cases}x=y=z\\xyz=1\end{cases}\Leftrightarrow x=y=z=1}\)

Vậy \(P_{min}=3\sqrt{3}\)khi \(x=y=z=1\)

:))

Thay  \(1=\sqrt{xy}+\sqrt{yz}+\sqrt{zx}\)  ta có

\(1+x=x+\sqrt{xy}+\sqrt{yz}+\sqrt{zx}=\left(\sqrt{x}+\sqrt{y}\right)\left(\sqrt{y}+\sqrt{z}\right)\)

Tương tự  \(1+y=\left(\sqrt{y}+\sqrt{x}\right)\left(\sqrt{y}+\sqrt{z}\right)\)  và  \(1+z=\left(\sqrt{z}+\sqrt{x}\right)\left(\sqrt{z}+\sqrt{y}\right)\)

\(\Rightarrow\sqrt{\left(1+x\right)\left(1+y\right)\left(1+z\right)}=\left(\sqrt{x}+\sqrt{y}\right)\left(\sqrt{y}+\sqrt{z}\right)\left(\sqrt{z}+\sqrt{x}\right)\)

và  \(\frac{\sqrt{x}}{1+x}+\frac{\sqrt{y}}{1+y}+\frac{\sqrt{z}}{1+z}\)

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

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

\(=\frac{2\left(\sqrt{xy}+\sqrt{yz}+\sqrt{zx}\right)}{\left(\sqrt{x}+\sqrt{y}\right)\left(\sqrt{y}+\sqrt{z}\right)\left(\sqrt{z}+\sqrt{x}\right)}\)

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

Do đó P = 2