Xét $n=1$, ta có BĐT đúngXét $n>1$Ta có $\frac1{n^2+(n+1)^2}=\frac1{2n^2+2n+1}<\frac1{2n(n+1)}=\frac12(\frac1n-\frac1{n+1})$$\Rightarrow VT<\frac15+\frac12(\frac12-\frac1{n+1})=\frac9{20}$
Xét $n=1$, ta có BĐT đúngXét $n>1$Ta có $\frac1{n^2+(n+1)^2}=\frac1{2n^2+2n+1}<\frac1{2n(n+1)}=\frac12(\frac1n-\frac1{n+1})$$\Rightarrow VT<\frac15+\frac12(\frac12-\frac1{n+1})=\frac9{20}$
Xét $n=1$, ta có BĐT đúngXét $n>1$Ta có $\frac1{n^2+(n+1)^2}=\frac1{2n^2+2n+1}<\frac1{2n(n+1)}=\frac12(\frac1n-\frac1{n+1})$$\Rightarrow VT<\frac15+\frac12(\frac12-\frac1{n+1})=\frac9{20}$