C.hóa \(x+y=1\) và dùng C-S:

\(VT^2\le\frac{2x}{\left(y+1\right)^2}+\frac{2y}{\left(x+1\right)^2}\le\frac{8}{9}=VP^2\)

\(BDT\Leftrightarrow\frac{x}{\left(2-x\right)^2}+\frac{y}{\left(2-y\right)^2}\le\frac{4}{9}\left(1\right)\)

Ta có BĐT phụ \(\frac{x}{\left(2-x\right)^2}\le\frac{20}{27}x-\frac{4}{27}\)

\(\Leftrightarrow-\frac{\left(2x-1\right)^2\left(5x-16\right)}{27\left(x-2\right)^2}\le0\) *Đúng*

Tương tự cho 2 BĐT còn lại rồi cộng theo vế:

\(VT_{\left(1\right)}\le\frac{20}{27}\left(x+y\right)-\frac{4}{27}\cdot2=\frac{4}{9}=VP_{\left(1\right)}\)

"=" khi \(x=y=\frac{1}{2}\)

khó mình mới lớp 6

\(a_n=\frac{1+\left(\frac{n}{n+2}\right)^n}{1-\left(\frac{n}{n+2}\right)^n}\)

\(a_n=\frac{\left(\frac{n}{n+2}\right)^2-\left(-1\right)}{\left(1-\frac{n}{n+2}\right)\left(1+\frac{n}{n+2}\right)}\)

\(a_n=\frac{\left(\frac{n}{n+2}-1\right)\left(\frac{n}{n+2}+1\right)}{\left(1-\frac{n}{n+2}\right)\left(1+\frac{n}{n+2}\right)}\)

\(a_n=\frac{\left(\frac{n}{n+2}-1\right)}{\left(1-\frac{n}{n+2}\right)}\)

\(a_n=\frac{-\left(1-\frac{n}{n+2}\right)}{\left(1-\frac{n}{n+2}\right)}\)

\(a_n=1\)

\(\Rightarrow\hept{\begin{cases}a=1\\n=1\end{cases}}\)

vậy \(\hept{\begin{cases}a=1\\n=1\end{cases}}\)