Toán 9 Căn bậc hai

ĐK: a, b, c[tex]\geq[/tex] 0, b[tex]\neq[/tex]
Ta có: [tex]2(\sqrt{a}-\sqrt{b})=\frac{2(\sqrt{a}-\sqrt{b})(\sqrt{a}+\sqrt{b})}{\sqrt{a}+\sqrt{b}}=\frac{2(a-b)}{\sqrt{a}+\sqrt{b}}[/tex] [tex]=\frac{2(b+1-b)}{\sqrt{b+1}+\sqrt{b}}=\frac{2}{\sqrt{b+1}+\sqrt{b}}[/tex]
Lại có: [tex]\sqrt{b+1}> \sqrt{b}< = > \sqrt{b+1}+\sqrt{b}> 2\sqrt{b}[/tex]

[tex]< = > \frac{1}{\sqrt{b+1}+\sqrt{b}}< \frac{1}{2\sqrt{b}}[/tex]
[tex]< = > \frac{2}{\sqrt{b+1}+\sqrt{b}}< \frac{1}{\sqrt{b}}[/tex]
[tex]< = > 2(\sqrt{a}-\sqrt{b})< \frac{1}{\sqrt{b}}[/tex] (1)
Tương tự với [tex]2(\sqrt{a}-\sqrt{c})[/tex] ta có:
[tex]\sqrt{2(c+2)+2c}> \sqrt{c+2}[/tex]
<=>[tex]2\sqrt{c+1}> \sqrt{c+2}[/tex]
<=>[tex]4\sqrt{c+1}>2\sqrt{c+1}> \sqrt{c+1}+\sqrt{c}[/tex]
Do đó [tex]\frac{1}{\sqrt{c+1}}< \frac{4}{\sqrt{c+2}+\sqrt{c}}[/tex]
<=>[tex]\frac{1}{\sqrt{b}}< \frac{4}{\sqrt{a}+\sqrt{c}}[/tex] =[tex]2(\sqrt{a}-\sqrt{c})[/tex] (1)
Từ (1) và (2) suy ra ĐPCM