1/

Đề \(\Rightarrow z^{15}+x^{15}-\left(y^{15}+z^{15}\right)=2\left(y^{2016}-x^{2016}\right)\)

\(\Rightarrow x^{15}-y^{15}=2\left(y^{2016}-x^{2016}\right)\)

+Nếu \(x=y\text{ thì }VT=0=VP\)

+Nếu \(x>y\text{ thì }VT>0>VP\)

+Nếu \(x<\)\(y\) thì \(VT<0\)\(<\)\(VP\)

Vậy \(x=y\)

Làm tương tự, ta có: \(y=z\)

\(\Rightarrow x=y=z\)

\(\Rightarrow x^{15}+x^{15}=2x^{2016}\Leftrightarrow x^{2016}=x^{15}\Leftrightarrow x^{15}\left(x^{2001}-1\right)=0\)

\(\Leftrightarrow x^{2001}=1\text{ (do }x>0\text{)}\)

\(\Leftrightarrow x=1\)

Vậy \(x=y=z=1\)

\(1=x+y+xy\le x+y+\frac{\left(x+y\right)^2}{4}=\left(\frac{x+y}{2}+1\right)^2-1\)

\(\Rightarrow\left(\frac{x+y}{2}+1\right)^2\ge2\Rightarrow\frac{x+y}{2}+1\ge\sqrt{2}\Rightarrow x+y\ge2\sqrt{2}-2\)

\(1=x+y+xy\ge2\sqrt{xy}+xy=\left(\sqrt{xy}+1\right)^2-1\)

\(\Rightarrow\left(\sqrt{xy}+1\right)^2\le2\Rightarrow\sqrt{xy}+1\le\sqrt{2}\Rightarrow\sqrt{xy}\le\sqrt{2}-1\)

\(\Rightarrow xy\le3-2\sqrt{2}\)

\(P=\frac{1}{x+y}+\frac{1}{x}+\frac{1}{y}=\frac{x+y+xy}{x+y}+\frac{x+y}{xy}\)

\(=1+\left(\frac{xy}{x+y}+\frac{\left(\sqrt{2}-1\right)^2}{4}.\frac{x+y}{xy}\right)+\frac{1+2\sqrt{2}}{4}.\frac{x+y}{xy}\)

\(\ge1+2\sqrt{\frac{xy}{x+y}.\frac{\left(\sqrt{2}-1\right)^2}{4}\frac{x+y}{xy}}+\frac{1+2\sqrt{2}}{4}.\frac{2\sqrt{2}-2}{3-2\sqrt{2}}=\frac{5+5\sqrt{2}}{2}\)

Dấu bằng xảy ra khi và chỉ khi \(x=y=\sqrt{2}-1\)

\(\frac{x}{1+y^2}=\frac{x\left(1+y^2\right)-xy^2}{1+y^2}=x-\frac{xy^2}{1+y^2}\)

Áp dụng Côsi: \(1+y^2\ge2y\Rightarrow\frac{xy^2}{1+y^2}\le\frac{xy^2}{2y}=\frac{xy}{2}\Rightarrow-\frac{xy^2}{1+y^2}\ge-\frac{xy}{2}\)

Do đó: \(\frac{x}{1+y^2}\ge x-\frac{xy}{2}\)

Tương tự ta có: \(\frac{y}{1+z^2}\ge y-\frac{yz}{2};\frac{z}{1+x^2}\ge z-\frac{zx}{2}\)

Mà \(\left(x+y+z\right)^2=x^2+y^2+z^2+2\left(xy+yz+zy\right)\ge xy+yz+zx+2\left(xy+yz+zy\right)\)

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

 \(\Rightarrow\frac{x}{1+y^2}+\frac{y}{1+z^2}+\frac{z}{1+x^2}\ge x+y+z-\frac{1}{2}\left(xy+yz+zx\right)\ge3-\frac{1}{2}.3=\frac{3}{2}\)

Dấu "=" xảy ra khi và chỉ khi x = y = z = 1

Vậy GTNN của P là 1

tôi ko bt người ae ạ

Gỉa sử (n+1).(n+2)....(n+n) chia hết cho \(2^{n+1}\) => (n+1).(n+2)...(n+n) - \(2^n=2^{n+1}-2^n=2^n\) mà \(2^n\) chia hết cho \(2^n\) => (n+1).(n+2)....(n+n)

chia hết cho \(2^n\)( mâu thuẫn) => đpcm