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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n ejmfjnhcy

a) Có \(\sqrt{25}=5;\sqrt{45}< \sqrt{49}=7\)

\(\Rightarrow\sqrt{25}+\sqrt{45}< 5+7=12\)

Vậy \(\sqrt{25}+\sqrt{45}< 12.\)

b) có \(\left(\sqrt{2013}+\sqrt{2015}\right)^2=2013+2015+2\sqrt{2013}.\sqrt{2015}\)\(=4028+2\sqrt{2013.2015}\)

\(\left(2\sqrt{2014}\right)^2=4.2014=4028+2.2014=4028+2\sqrt{2014^2}\)

Xét \(2014^2-2013.2015=2014.\left(2013+1\right)-2013\left(2014+1\right)\)

\(=2013.2014+2014-2013.2014-2013=1>0\)

\(\Rightarrow2\sqrt{2013.2015}< 2\sqrt{2014^2}\)

Hay \(\left(\sqrt{2013}+\sqrt{2015}\right)^2< \left(2\sqrt{2014}\right)^2\)

\(\Rightarrow\sqrt{2013}+\sqrt{2015}< 2\sqrt{2014}\)
Vậy \(\sqrt{2013}+\sqrt{2015}< 2\sqrt{2014}.\)

c) Có \(\left(\sqrt{2014}-\sqrt{2013}\right)\left(\sqrt{2014}+\sqrt{2013}\right)=2014-2013=1\)\(\rightarrow\sqrt{2014}-\sqrt{2013}=\dfrac{1}{\sqrt{2014}+\sqrt{2013}}\)

Mà \(\sqrt{2014}>\sqrt{2013};\sqrt{2013}>\sqrt{2012}\)

\(\rightarrow\sqrt{2014}+\sqrt{2013}>\sqrt{2013}+\sqrt{2012}\)

Hay \(\dfrac{1}{\sqrt{2014}+\sqrt{2013}}< \dfrac{1}{\sqrt{2013}+\sqrt{2012}}\)

Tương tự, ta có \(\dfrac{1}{\sqrt{2013}+\sqrt{2012}}=\sqrt{2013}-\sqrt{2012}\)

\(\Rightarrow\sqrt{2014}-\sqrt{2013}< \sqrt{2013}-\sqrt{2012}\)

Vậy \(\sqrt{2014}-\sqrt{2013}< \sqrt{2013}-\sqrt{2012}.\)

lop8. thi ap bdt nhu thanh song,

a)

VT=√25+√45<√2(25+45)=√140<√144=12=VP

b)

VT=√2013+√2015<√[2(2013+2015)]=√[4.2014]=2√(2014)=VP.

c) C=VT-VP

√2014+√2012-2√2012

kq(b)=> C<0

VT<VP

\(\frac{1}{\sqrt{2013}-\sqrt{2014}}-\frac{1}{\sqrt{2014}-\sqrt{2015}}\)

\(=\frac{\sqrt{2013}+\sqrt{2014}}{\left(\sqrt{2013}-\sqrt{2014}\right)\left(\sqrt{2014}-\sqrt{2015}\right)}-\frac{\sqrt{2014}+\sqrt{2015}}{\left(\sqrt{2014}-\sqrt{2015}\right)\left(\sqrt{2014}+\sqrt{2015}\right)}\)

\(=\frac{\sqrt{2013}+\sqrt{2014}}{2013-2014}-\frac{\sqrt{2014}+\sqrt{2015}}{2014-2015}\)

\(=-\sqrt{2013}-\sqrt{2014}+\sqrt{2014}+\sqrt{2015}\)

\(=\sqrt{2015}-\sqrt{2013}\)

\(=\frac{\sqrt{2013}+\sqrt{2014}}{2013-2014}-\frac{\sqrt{2014}+\sqrt{2015}}{2014-2015}\)

\(=-\sqrt{2013}-\sqrt{2014}+\sqrt{2014}+\sqrt{2015}\)

\(=\sqrt{2015}-\sqrt{2013}\)

Điều kiện: \(x\ge2012;y\ge2013;z\ge2014\)

Áp dụng bất đẳng thức Cauchy, ta có:

\(\left\{{}\begin{matrix}\dfrac{\sqrt{x-2012}-1}{x-2012}=\dfrac{\sqrt{4\left(x-2012\right)}-2}{2\left(x-2012\right)}\le\dfrac{\dfrac{4+x-2012}{2}-2}{2\left(x-2012\right)}=\dfrac{1}{4}\\\dfrac{\sqrt{y-2013}-1}{y-2013}=\dfrac{\sqrt{4\left(y-2013\right)}-2}{2\left(y-2013\right)}\le\dfrac{\dfrac{4+y-2013}{2}-2}{2\left(y-2013\right)}=\dfrac{1}{4}\\\dfrac{\sqrt{z-2014}-1}{z-2014}=\dfrac{\sqrt{4\left(z-2014\right)}-2}{2\left(z-2014\right)}\le\dfrac{\dfrac{4+z-2014}{2}-2}{2\left(z-2014\right)}=\dfrac{1}{4}\end{matrix}\right.\)

Cộng vế theo vế, ta được:

\(\dfrac{\sqrt{x-2012}-1}{x-2012}+\dfrac{\sqrt{y-2013}-1}{y-2013}+\dfrac{\sqrt{z-2014}-1}{z-2014}\le\dfrac{3}{4}\)

Đẳng thức xảy ra khi \(x=2016;y=2017;z=2018\)

Vậy....

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 \(\frac{\sqrt{2013}+\sqrt{2015}}{2}\le\sqrt{\frac{2013+2015}{2}}\Leftrightarrow\sqrt{2013}+\sqrt{2015}\le2\sqrt{2014}\)

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