Ta có \(x^2+y^2+z^2\ge xy+yz+zx\)

Đẳng thức xảy ra khi x = y = z 

Bạn áp dụng vào nhé.

Ngọc cứ làm tắt thì vài người hiểu chứ vài bạn không biết đâu :)

Ta có :

\(x^2+y^2+z^2=xy+xz+yz\)

\(\Rightarrow x^2+y^2+z^2-xy-xz-yz=0\)

\(\Rightarrow2\left(x^2+y^2+z^2-xy-xz-yz\right)=0\)

\(\Rightarrow x^2+y^2-2xy+y^2+z^2-2yz+x^2+z^2-2xz=0\)

\(\Rightarrow\left(x-y\right)^2+\left(y-z\right)^2+\left(x-z\right)^2=0\)

Mà \(\hept{\begin{cases}\left(x-y\right)^2\ge0\\\left(x-z\right)^2\ge0\\\left(y-z\right)^2\ge0\end{cases}}\)

\(\Rightarrow x-y=x-z=y-z=0\)

\(\Rightarrow x=y=z\)

\(\Rightarrow x^{2016}=y^{2016}=z^{2016}\)

Mà \(x^{2016}+y^{2016}+z^{2016}=3^{2016}\)

\(\Rightarrow x^{2016}=y^{2016}=z^{2016}=\frac{3^{2016}}{3}=3^{2015}\)

\(\Rightarrow x=y=z=\sqrt[2016]{3^{2015}}=\sqrt[2016]{\frac{3^{2016}}{3}}=\frac{3}{\sqrt[2016]{3}}\)

Mình chưa học cách làm như thế. Chẳng hiểu gì hết. các bạn có thể làm theo cách khác không?

ko bít làm à

k bik nên mới hỏi

Ta có: \(\left(xy+2016z\right)\left(yz+2016z\right)\left(zx+2016y\right)\\ =\left(xy+\left(x+y+z\right)z\right)\left(yz+\left(x+y+z\right)x\right)\left(zx+\left(x+y+z\right)y\right)\\ =\left(xy+zx+zy+z^2\right)\left(yz+x^2+xy+xz\right)\left(zx+xỹ+y^2+yz\right)\\ =\left(y+z\right)\left(x+z\right)\left(x+z\right)\left(y+x\right)\left(z+y\right)\left(x+y\right)\\ =\left(y+z\right)^2\left(x+y\right)^2\left(z+x\right)^2\\ \Rightarrow\frac{\left(xy+2016z\right)\left(yz+2016z\right)\left(zx+2016y\right)}{\left(x+y\right)^2\left(y+z\right)^2\left(z+x\right)^2}\\ =\frac{\left(y+z\right)^2\left(x+y\right)^2\left(z+x\right)^2}{\left(x+y\right)^2\left(y+z\right)^2\left(z+x\right)^2}\\ =1\)

\(M=\frac{x^3+y^3+z^3-3xyz}{x^2+y^2+z^2-xy-yz-zx}\)

Đặt \(N=x^3+y^3+z^3-3xyz\)

\(=\left(x+y\right)^3-3x^2y-3xy^2+z^3-3xyz\)

\(=\left(x+y\right)^3+z^3-3x^2y-3xy^2-3xyz\)

\(=\left(x+y+z\right)\left[\left(x+y\right)^2-\left(x+y\right).z+z^2\right]-3xy\left(x+y+z\right)\)

\(=\left(x+y+z\right)\left(x^2+2xy+y^2-zx-yz+z^2\right)-3xy\left(x+y+z\right)\)

\(=\left(x+y+z\right)\left(x^2+2xy+y^2-zx-yz+z^2-3xy\right)\)

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

Vậy \(M=\frac{N}{x^2+y^2+z^2-xy-yz-zx}=\frac{\left(x+y+z\right)\left(x^2+y^2+z^2-xy-yz-zx\right)}{x^2+y^2+z^2-xy-yz-zx}=x+y+z=2016\)

(*) bn ghi sai đề 1 chỗ nhé:ở mẫu thức của M phải là  \(x^2+y^2+z^2-xy-yz-zx\) nhé!

\(VT=\sqrt{\dfrac{yz}{x^2+xy+yz+xz}}+\sqrt{\dfrac{xy}{y^2+xy+yz+xz}}+\sqrt{\dfrac{xz}{z^2+xy+yz+xz}}\)

\(VT=\sqrt{\dfrac{yz}{\left(x+y\right)\left(x+z\right)}}+\sqrt{\dfrac{xy}{\left(y+z\right)\left(x+y\right)}}+\sqrt{\dfrac{xz}{\left(x+z\right)\left(y+z\right)}}\)

Áp dụng bất đẳng thức Cauchy - Schwarz

\(\Rightarrow\left\{{}\begin{matrix}\sqrt{\dfrac{yz}{\left(x+y\right)\left(x+z\right)}}\le\dfrac{\dfrac{y}{x+y}+\dfrac{z}{x+z}}{2}\\\sqrt{\dfrac{xy}{\left(y+z\right)\left(x+y\right)}}\le\dfrac{\dfrac{x}{x+y}+\dfrac{y}{y+z}}{2}\\\sqrt{\dfrac{xz}{\left(x+z\right)\left(y+z\right)}}\le\dfrac{\dfrac{x}{x+z}+\dfrac{z}{y+z}}{2}\end{matrix}\right.\)

\(\Rightarrow VT\le\dfrac{\left(\dfrac{x}{x+y}+\dfrac{y}{x+y}\right)+\left(\dfrac{y}{y+z}+\dfrac{z}{y+z}\right)+\left(\dfrac{z}{x+z}+\dfrac{x}{x+z}\right)}{2}\)

\(\Rightarrow VT\le\dfrac{\dfrac{x+y}{x+y}+\dfrac{y+z}{y+z}+\dfrac{x+z}{x+z}}{2}=\dfrac{3}{2}\)

\(\Leftrightarrow\sqrt{\dfrac{yz}{x^2+2016}}+\sqrt{\dfrac{xy}{y^2+2016}}+\sqrt{\dfrac{xz}{z^2+2016}}\le\dfrac{3}{2}\) ( đpcm )

Dấu " = " xảy ra khi \(x=y=z=4\sqrt{42}\)

Sửa đề:\(\sqrt{\dfrac{yz}{x^2+2016}}+\sqrt{\dfrac{xy}{z^2+2016}}+\sqrt{\dfrac{xz}{y^2+2016}}\le\dfrac{3}{2}\)

Giải

Ta có:

\(\sqrt{\dfrac{xy}{z^2+2016}}=\sqrt{\dfrac{xy}{z^2+xy+xz+yz}}=\sqrt{\dfrac{xy}{\left(x+z\right)\left(y+z\right)}}\)

Áp dụng BĐT AM-GM ta có:

\(\sqrt{\dfrac{xy}{z^2+2016}}=\sqrt{\dfrac{xy}{\left(x+z\right)\left(y+z\right)}}\le\dfrac{1}{2}\left(\dfrac{x}{x+z}+\dfrac{y}{y+z}\right)\)

Tương tự cho 2 BĐT còn lại ta có:

\(\sqrt{\dfrac{yz}{x^2+2016}}\le\dfrac{1}{2}\left(\dfrac{y}{x+y}+\dfrac{z}{x+z}\right);\sqrt{\dfrac{xz}{y^2+2016}}\le\dfrac{1}{2}\left(\dfrac{x}{x+y}+\dfrac{z}{y+z}\right)\)

Cộng theo vế 3 BĐT trên ta có:

\(\Sigma\sqrt{\dfrac{xy}{z^2+2016}}\le\dfrac{1}{2}\Sigma\left(\dfrac{x}{x+z}+\dfrac{y}{y+z}\right)=\dfrac{1}{2}\Sigma\left(\dfrac{x}{x+z}+\dfrac{z}{x+z}\right)=\dfrac{3}{2}\)

Đẳng thức xảy ra khi \(x=y=z=4\sqrt{42}\)

xí bài này nhé, lát nữa hoặc mai giải