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chứng minh 1/ căn 1 + 1/ căn 2 + ..... + 1/ căn 64 > 14

I - Vy Nguyễn · Accepted Answer

Nhận thấy với mọi k $\in$ N* ta có : $\left(\sqrt{k+1}-\sqrt{k}\right).\left(\sqrt{k+1}+\sqrt{k}\right)=\left(\sqrt{k+1}\right)^2-\left(\sqrt{k}\right)^2=k+1-k=1$$	\implies$$\frac{\left(\sqrt{k+1}-\sqrt{k}\right).\left(\sqrt{k+1}+\sqrt{k}\right)}{\sqrt{k+1}+\sqrt{k}}=\frac{1}{\sqrt{k+1}+\sqrt{k}}$$	\implies$ $\frac{1}{\sqrt{k+1}+\sqrt{k}}=\sqrt{k+1}-\sqrt{k}$Thật vậy : $\frac{1}{\sqrt{k}}=\frac{2}{2.\sqrt{k}}>\frac{2}{\sqrt{k}+\sqrt{k+1}}=2.\left(\sqrt{k+1}-\sqrt{k}\right)$Thay k = 1 ; 2 ; 3 ; ....; 64 ta được :$\frac{1}{\sqrt{1}}>2.\left(\sqrt{1+1}-\sqrt{1}\right)=2.\left(\sqrt{2}-\sqrt{1}\right)=2.\sqrt{2}-2.\sqrt{1}$$\frac{1}{\sqrt{2}}>2.\left(\sqrt{2+1}-\sqrt{2}\right)=2.\left(\sqrt{3}-\sqrt{2}\right)=2.\sqrt{3}-2.\sqrt{2}$$\frac{1}{\sqrt{3}}>2.\left(\sqrt{3+1}-\sqrt{3}\right)=2.\left(\sqrt{4}-\sqrt{3}\right)=2.\sqrt{4}-2.\sqrt{3}$                                                  . . . . . . . . . . . . . . . . . . . . . .$\frac{1}{\sqrt{64}}>2.\left(\sqrt{64+1}-\sqrt{64}\right)=2.\left(\sqrt{65}-\sqrt{64}\right)=2.\sqrt{65}-2.\sqrt{64}$Cộng vế với vế ta được : $\frac{1}{\sqrt{1}}+\frac{1}{\sqrt{2}}+...+\frac{1}{\sqrt{64}}>2.\sqrt{2}-2.\sqrt{1}+2.\sqrt{3}-2.\sqrt{2}+....+2.\sqrt{65}-2.\sqrt{64}$ $	\implies$   $\frac{1}{\sqrt{1}}+\frac{1}{\sqrt{2}}+...+\frac{1}{\sqrt{64}}>2.\sqrt{65}-2.\sqrt{1}=2.\left(\sqrt{65}-\sqrt{1}\right)$ ( * ) Ta thấy : $\sqrt{65}>\sqrt{64}$$	\implies$ $\sqrt{65}-\sqrt{1}>\sqrt{64}-\sqrt{1}$$	\implies$ $\sqrt{65}-\sqrt{1}>7$$	\implies$ $2.\left(\sqrt{65}-\sqrt{1}\right)>2.7$$	\implies$ $2.\left(\sqrt{65}-\sqrt{1}\right)>14$ ( ** )Từ ( * ) ; ( ** ) $	\implies$ $\frac{1}{\sqrt{1}}+\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{3}}+...+\frac{1}{\sqrt{64}}>14\left(đpcm\right)$ Câu trả lời: Nhận thấy với mọi k $\in$ N* ta có : $\left(\sqrt{k+1}-\sqrt{k}\right).\left(\sqrt{k+1}+\sqrt{k}\right)=\left(\sqrt{k+1}\right)^2-\left(\sqrt{k}\right)^2=k+1-k=1$$	\implies$$\frac{\left(\sqrt{k+1}-\sqrt{k}\right).\left(\sqrt{k+1}+\sqrt{k}\right)}{\sqrt{k+1}+\sqrt{k}}=\frac{1}{\sqrt{k+1}+\sqrt{k}}$$	\implies$ $\frac{1}{\sqrt{k+1}+\sqrt{k}}=\sqrt{k+1}-\sqrt{k}$Thật vậy : $\frac{1}{\sqrt{k}}=\frac{2}{2.\sqrt{k}}>\frac{2}{\sqrt{k}+\sqrt{k+1}}=2.\left(\sqrt{k+1}-\sqrt{k}\right)$Thay k = 1 ; 2 ; 3 ; ....; 64 ta được :$\frac{1}{\sqrt{1}}>2.\left(\sqrt{1+1}-\sqrt{1}\right)=2.\left(\sqrt{2}-\sqrt{1}\right)=2.\sqrt{2}-2.\sqrt{1}$$\frac{1}{\sqrt{2}}>2.\left(\sqrt{2+1}-\sqrt{2}\right)=2.\left(\sqrt{3}-\sqrt{2}\right)=2.\sqrt{3}-2.\sqrt{2}$$\frac{1}{\sqrt{3}}>2.\left(\sqrt{3+1}-\sqrt{3}\right)=2.\left(\sqrt{4}-\sqrt{3}\right)=2.\sqrt{4}-2.\sqrt{3}$                                                  . . . . . . . . . . . . . . . . . . . . . .$\frac{1}{\sqrt{64}}>2.\left(\sqrt{64+1}-\sqrt{64}\right)=2.\left(\sqrt{65}-\sqrt{64}\right)=2.\sqrt{65}-2.\sqrt{64}$Cộng vế với vế ta được : $\frac{1}{\sqrt{1}}+\frac{1}{\sqrt{2}}+...+\frac{1}{\sqrt{64}}>2.\sqrt{2}-2.\sqrt{1}+2.\sqrt{3}-2.\sqrt{2}+....+2.\sqrt{65}-2.\sqrt{64}$ $	\implies$   $\frac{1}{\sqrt{1}}+\frac{1}{\sqrt{2}}+...+\frac{1}{\sqrt{64}}>2.\sqrt{65}-2.\sqrt{1}=2.\left(\sqrt{65}-\sqrt{1}\right)$ ( * ) Ta thấy : $\sqrt{65}>\sqrt{64}$$	\implies$ $\sqrt{65}-\sqrt{1}>\sqrt{64}-\sqrt{1}$$	\implies$ $\sqrt{65}-\sqrt{1}>7$$	\implies$ $2.\left(\sqrt{65}-\sqrt{1}\right)>2.7$$	\implies$ $2.\left(\sqrt{65}-\sqrt{1}\right)>14$ ( ** )Từ ( * ) ; ( ** ) $	\implies$ $\frac{1}{\sqrt{1}}+\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{3}}+...+\frac{1}{\sqrt{64}}>14\left(đpcm\right)$

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