Áp dụng bđt \(\frac{\sqrt{a}+\sqrt{b}}{2}< \sqrt{\frac{a+b}{2}}\) với a > 0; b > 0; a \(\ne\) b ta có:

\(\frac{\sqrt{2016}+\sqrt{2014}}{2}< \sqrt{\frac{2016+2014}{2}}\)

\(\Rightarrow\frac{\sqrt{2016}+\sqrt{2014}}{2}< \sqrt{\frac{4030}{2}}\)

\(\Rightarrow\frac{\sqrt{2016}+\sqrt{2014}}{2}< \sqrt{2015}\)

\(\Rightarrow\sqrt{2016}+\sqrt{2014}< 2.\sqrt{2015}\)

\(\Rightarrow\sqrt{2016}-\sqrt{2015}< \sqrt{2015}-\sqrt{2014}\)

\(\sqrt{2016}-\sqrt{2015}=\dfrac{1}{\sqrt{2016}+\sqrt{2015}}\)

\(\sqrt{2015}-\sqrt{2014}=\dfrac{1}{\sqrt{2015}+\sqrt{2014}}\)

mà \(\sqrt{2016}+\sqrt{2015}>\sqrt{2014}+\sqrt{2015}\)

nên \(\sqrt{2016}-\sqrt{2015}< \sqrt{2015}-\sqrt{2014}\)

\(\sqrt{2016}+\sqrt{2015}>\sqrt{2015}+\sqrt{2014}\)

=> \(\frac{1}{\sqrt{2016}+\sqrt{2015}}

Áp dụng bđt \(\frac{\sqrt{a}+\sqrt{b}}{2}< \sqrt{\frac{a+b}{2}}\) với a > 0; b > 0; a khác b ta có:

\(\frac{\sqrt{2016}+\sqrt{2014}}{2}< \sqrt{\frac{2016+2014}{2}}\)

\(\Rightarrow\frac{\sqrt{2016}+\sqrt{2014}}{2}< \sqrt{\frac{4030}{2}}\)

\(\Rightarrow\sqrt{2016}+\sqrt{2014}< \sqrt{2015}.2\)

\(\Rightarrow\sqrt{2016}-\sqrt{2015}< \sqrt{2015}-\sqrt{2014}\)

A = \(\frac{2016-2015}{\sqrt{2016}+\sqrt{2015}}=\frac{1}{\sqrt{2016}+\sqrt{2015}}\); B = \(\frac{2015-2014}{\sqrt{2015}+\sqrt{2014}}=\frac{1}{\sqrt{2015}+\sqrt{2014}}\)

Mà \(\sqrt{2016}+\sqrt{2015}>\sqrt{2015}+\sqrt{2014}\) ( Vì \(\sqrt{2016}>\sqrt{2014}\))

Nên \(\frac{1}{\sqrt{2016}+\sqrt{2015}}

Ta đặt \(x=2015\), khi đó \(2014=x-1,2016=x+1.\)  Ta có như sau

\(2014^2\cdot2016=\left(x-1\right)^2\left(x+1\right)=\left(x^2-1\right)\left(x-1\right)\)\(

Có Ta có\(VT=\frac{2014}{\sqrt{2015}}+\frac{2015}{\sqrt{2014}}=\frac{2015-1}{\sqrt{2015}}+\frac{2014+1}{\sqrt{2014}}=\sqrt{2015}-\frac{1}{\sqrt{2015}}+\sqrt{2014}+\frac{1}{\sqrt{2014}}.\)​​​\(20140\Leftrightarrow VT>VP\)

Có: \(\sqrt{2015}< \sqrt{2016}\)

=>\(\frac{1}{\sqrt{2015}}>\frac{1}{\sqrt{2016}}\)

=>\(\frac{1}{\sqrt{2015}}-\frac{1}{\sqrt{2016}}>0\)

=>\(\sqrt{2015}+\sqrt{2016}+\frac{1}{\sqrt{2015}}-\frac{1}{\sqrt{2016}}>\sqrt{2015}+\sqrt{2016}\)

=>\(\left(\sqrt{2015}+\frac{1}{\sqrt{2015}}\right)+\left(\sqrt{2016}-\frac{1}{\sqrt{2016}}\right)>\sqrt{2015}+\sqrt{2016}\)

=>\(\frac{2016}{\sqrt{2015}}+\frac{2015}{\sqrt{2016}}>\sqrt{2015}+\sqrt{2016}\)