Áp dụng bđt \(\frac{\sqrt{a}+\sqrt{b}}{2}\le\sqrt{\frac{a+b}{2}}\) :

Xét : \(N-M=2\sqrt{2014}-\left(\sqrt{2015}+\sqrt{2013}\right)\)

Theo bđt trên thì \(\frac{\sqrt{2013}+\sqrt{2015}}{2}\le\sqrt{\frac{2013+2015}{2}}\Leftrightarrow\sqrt{2013}+\sqrt{2015}\le2\sqrt{2014}\)

\(\Rightarrow N-M>0\Rightarrow N>M\)

1) Ta có bđt sau : \(\frac{\sqrt{a}+\sqrt{b}}{2}< \sqrt{\frac{a+b}{2}}\) (bạn tự c/m)

Áp dụng : \(\frac{\sqrt{2005}+\sqrt{2007}}{2}< \sqrt{\frac{2005+2007}{2}}\)

\(\Rightarrow\sqrt{2005}+\sqrt{2007}< 2\sqrt{2006}\)

2) Xét : \(A-B=2\sqrt{2014}-\left(\sqrt{2013}+\sqrt{2015}\right)\)

Theo câu 1) , ta dễ dàng c/m được \(2\sqrt{2014}>\sqrt{2013}+\sqrt{2015}\)

Do đó A - B > 0 => A > B

2) Bình phương 2 vế ta có:

 \(A^2=2014-2013=1\)

\(B^2=2015-2014=1\)

=>A=B

Ta có: \(\sqrt{2015}-\sqrt{2014}=\dfrac{2015-2014}{\sqrt{2015}+\sqrt{2014}}>\dfrac{2016-2015}{\sqrt{2016}+\sqrt{2015}}=\sqrt{2016}-\sqrt{2015}\)

Ta có:  √2015−√2014=2015−2014√2015+√2014>2016−2015√2016+√2015=√2016−√2015

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

Chứng minh \(\frac{1}{\left(n+1\right)\sqrt{n}+n\sqrt{n+1}}=\frac{1}{\sqrt{n}}-\frac{1}{\sqrt{n+1}}\) rồi áp dụng với n = 1,2,....,2014

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

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

bạn bingf phương lên nha rồi so sánh \(2014\)và \(\sqrt{2015.2013}\)

dễ thấy\(\sqrt{2015.2013}=\sqrt{\left(2014+1\right)\left(2014-1\right)}=\sqrt{2014^2-1}\le\sqrt{2014^2}=2014\)

cụ thể rồi còn j