Câu a:

Có dạng tổng quát:\(\frac{1}{\left(k+1\right)\sqrt{k}+k\sqrt{x+1}}=\frac{1}{\sqrt{\left(k+1\right)k}\left(\sqrt{k+1}+\sqrt{k}\right)}=\frac{\sqrt{k+1}-\sqrt{k}}{\sqrt{\left(k+1\right)k}}=\frac{1}{\sqrt{k}}-\frac{1}{\sqrt{k-1}}\)

Áp dụng kết quả trên suy ra câu a

Ta có

\(\hept{\begin{cases}\sqrt{2008}+\sqrt{2005}< \sqrt{2015}+\sqrt{2009}\left(1\right)\\\sqrt{2010}+\sqrt{2007}< \sqrt{2015}+\sqrt{2009}\left(2\right)\end{cases}}\)

\(\Rightarrow\frac{1}{\sqrt{2008}+\sqrt{2005}}+\frac{1}{\sqrt{2010}+\sqrt{2007}}>\frac{2}{\sqrt{2015}+\sqrt{2009}}\)

\(\Leftrightarrow\frac{\sqrt{2008}-\sqrt{2005}}{3}+\frac{\sqrt{2010}-\sqrt{2007}}{3}>\frac{\sqrt{2015}-\sqrt{2009}}{3}\)

\(\Leftrightarrow\sqrt{2008}+\sqrt{2009}+\sqrt{2010}>\sqrt{2005}+\sqrt{2007}+\sqrt{2015}\)

A=√2008+√2009+√2010A=2008+2009+2010 và B=√2005+√2007+√2015

k và kb với mình nha !!!

tính cụ thể từng cái ra

\(A-B=\sqrt{2009}-\sqrt{2007}+\sqrt{2010}-\sqrt{2008}+\sqrt{2011}-\sqrt{2015}\)

\(=\frac{2}{\sqrt{2009}+\sqrt{2007}}+\frac{2}{\sqrt{2010}+\sqrt{2008}}-\frac{4}{\sqrt{2011}+\sqrt{2015}}\)

Ta có \(\left\{{}\begin{matrix}\sqrt{2009}+\sqrt{2007}< \sqrt{2011}+\sqrt{2015}\\\sqrt{2010}+\sqrt{2008}< \sqrt{2011}+\sqrt{2015}\end{matrix}\right.\)

\(\Rightarrow\frac{2}{\sqrt{2009}+\sqrt{2007}}+\frac{2}{\sqrt{2010}+\sqrt{2008}}>\frac{2}{\sqrt{2011}+\sqrt{2015}}+\frac{2}{\sqrt{2011}+\sqrt{2015}}=\frac{4}{\sqrt{2011}+\sqrt{2015}}\)

\(\Rightarrow\frac{2}{\sqrt{2009}+\sqrt{2007}}+\frac{2}{\sqrt{2010}+\sqrt{2008}}-\frac{4}{\sqrt{2011}+\sqrt{2015}}>0\)

\(\Rightarrow A-B>0\Rightarrow A>B\)

Bài 1: 

Ta có: \(a+b\ge2\sqrt{ab}\)

\(b+c\ge2\sqrt{bc}\)

\(a+c\ge2\sqrt{ac}\)

Do đó: \(2\left(a+b+c\right)\ge2\left(\sqrt{ab}+\sqrt{bc}+\sqrt{ac}\right)\)

hay \(a+b+c\ge\sqrt{ab}+\sqrt{cb}+\sqrt{ac}\)