Bạn chú ý nhận xét: nếu $0<a<b$ thì $\dfrac{1}{a}>\dfrac{1}{b}$
1.
$\dfrac{1}{2\sqrt{n}}+\dfrac{1}{2\sqrt{n+1}}>\dfrac{1}{\sqrt{n}+\sqrt{n+1}}+\dfrac{1}{\sqrt{n+1}+\sqrt{n+2}}$
$\dfrac{1}{\sqrt{n}+\sqrt{n+1}}+\dfrac{1}{\sqrt{n+1}+\sqrt{n+2}}\\=\dfrac{(n+1)-n}{\sqrt{n}+\sqrt{n+1}}+\dfrac{(n+2)-(n+1)}{\sqrt{n+1}+\sqrt{n+2}}\\=\dfrac{(n+1)-(n)}{\sqrt{n}+\sqrt{n+1}}+\dfrac{(n+2)-(n+1)}{\sqrt{n+1}+\sqrt{n+2}}\\=\dfrac{(\sqrt{n}+\sqrt{n+1})(\sqrt{n+1}-\sqrt{n}}{\sqrt{n}+\sqrt{n+1}}+\dfrac{(\sqrt{n+1}+\sqrt{n+2})(\sqrt{n+2}-\sqrt{n+1})}{\sqrt{n+1}+\sqrt{n+2}}\\=\sqrt{n+1}-\sqrt{n}+\sqrt{n+2}-\sqrt{n+1}=\sqrt{n+2}-\sqrt{n}$
Suy ra $\dfrac{1}{2\sqrt{n}}+\dfrac{1}{2\sqrt{n+1}}>\sqrt{n+2}-\sqrt{n}\Leftrightarrow \dfrac{1}{\sqrt{n}}+\dfrac{1}{\sqrt{n+1}}>2(\sqrt{n+2}-\sqrt{n})$
2.
$\dfrac12\left (\dfrac{1}{\sqrt{n}}-\dfrac{1}{\sqrt{n+1}}\right)=\dfrac{\sqrt{n+1}-\sqrt{n}}{2\sqrt{n(n+1)}}$
Cần so sánh $2n+1$ và $2\sqrt{n(n+1)}$
Ta có $n(n+1)=n^2+n<n^2+n+\dfrac14=\left(n+\dfrac12\right)^2 \Rightarrow 2\sqrt{n(n+1)}<2\sqrt{\left(n+\dfrac12\right)^2}=2n+1$
Suy ra $\dfrac{\sqrt{n+1}-\sqrt{n}}{2n+1}<\dfrac{\sqrt{n+1}-\sqrt{n}}{2\sqrt{n(n+1)}}=\dfrac12\left (\dfrac{1}{\sqrt{n}}-\dfrac{1}{\sqrt{n+1}}\right)$
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