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....

Lời giải:

Áp dụng BĐT Cô-si ngược dấu:

\(\sqrt{x-2010}=\frac{1}{2}\sqrt{4(x-2010)}\leq \frac{4+(x-2010)}{4}\)

\(\Rightarrow \sqrt{x-2010}-1\leq \frac{4+(x-2010)}{4}-1=\frac{x-2010}{4}\)

\(\Rightarrow \frac{\sqrt{x-2010}-1}{x-2010}\leq \frac{1}{4}\)

Hoàn toàn tương tự với những phân thức còn lại:

\(\Rightarrow \frac{\sqrt{x-2010}-1}{x-2010}+\frac{\sqrt{y-2011}-1}{y-2011}+\frac{\sqrt{z-2012}-1}{z-2012}\leq \frac{1}{4}+\frac{1}{4}+\frac{1}{4}=\frac{3}{4}\)

Dấu "=" xảy ra khi \(\left\{\begin{matrix} x-2010=4\\ y-2011=4\\ z-2012=4\end{matrix}\right.\Leftrightarrow \left\{\begin{matrix} x=2014\\ y=2015\\ z=2016\end{matrix}\right.\)

\(\dfrac{\sqrt{x-2009}-1}{x-2009}+\dfrac{\sqrt{y-2010}-1}{y-2010}+\dfrac{\sqrt{z-2011}-1}{z-2011}=\dfrac{3}{4}\)\(\left(\left\{{}\begin{matrix}x>2009\\y>2010\\z>2011\end{matrix}\right.\right)\)

\(\Leftrightarrow\dfrac{1}{4}-\dfrac{\sqrt{x-2009}-1}{x-2009}+\dfrac{1}{4}-\dfrac{\sqrt{y-2010}-1}{y-2010}+\dfrac{1}{4}-\dfrac{\sqrt{z-2011}-1}{z-2011}=0\)

\(\Leftrightarrow\dfrac{x-2009-4\sqrt{x-2009}+4}{x-2009}+\dfrac{y-2010-4\sqrt{y-2010}+4}{y-2010}+\dfrac{z-2011-4\sqrt{z-2011}+4}{z-2011}=0\)

Nhận xét: \(\left\{{}\begin{matrix}\dfrac{\left(\sqrt{x-2009}-2\right)^2}{x-2009}\ge0\\\dfrac{\left(\sqrt{y-2010}-2\right)^2}{y-2010}\ge0\\\dfrac{\left(\sqrt{z-2011}-2\right)^2}{z-2011}\ge0\end{matrix}\right.\)

\(\Rightarrow\left\{{}\begin{matrix}\sqrt{x-2009}-2=0\\\sqrt{y-2010}-2=0\\\sqrt{z-2011}-2=0\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}x=2013\\y=2014\\z=2015\end{matrix}\right.\)

Vậy \(\left(x;y;z\right)=\left(2013;2014;2015\right)\)

Đặt \(\hept{\begin{cases}a=x+2011\\b=y+2011\\c=z+2011\end{cases}}\) Ta có Hệ:

\(\hept{\begin{cases}\sqrt{a}+\sqrt{b+1}+\sqrt{c+2}\left(A\right)=\sqrt{b}+\sqrt{c+1}+\sqrt{a+2}\left(B\right)\\\sqrt{b}+\sqrt{c+1}+\sqrt{a+2}\left(B\right)=\sqrt{c}+\sqrt{a+1}+\sqrt{b+2}\left(C\right)\end{cases}}\)

Vai trò \(x,y,z\) bình đẳng

Giả sử \(c=Max\left(a;b;c\right)\) vì \(A=C\) ta có:

\(\sqrt{a}+\sqrt{b+1}+\sqrt{c+2}=\sqrt{c}+\sqrt{a+1}+\sqrt{b+2}\)

\(\Leftrightarrow\left(\sqrt{a+1}-\sqrt{a}\right)+\left(\sqrt{b+2}-\sqrt{b+1}\right)\)

\(=\sqrt{c+2}-\sqrt{c}=\left(\sqrt{c+2}-\sqrt{c+1}\right)+\left(\sqrt{c+1}-\sqrt{c}\right)\)

\(\Leftrightarrow\frac{1}{\sqrt{a+1}+\sqrt{a}}+\frac{1}{\sqrt{b+2}+\sqrt{b+1}}\)

\(=\frac{1}{\sqrt{c+2}+\sqrt{c+1}}+\frac{1}{\sqrt{c+1}+\sqrt{c}}\left(1\right)\)

Mặt khác \(\hept{\begin{cases}c\ge a\Rightarrow\frac{1}{\sqrt{a+1}+\sqrt{a}}\le\frac{1}{\sqrt{c+1}+\sqrt{c}}\\c\ge b\Rightarrow\frac{1}{\sqrt{b+2}+\sqrt{b+1}}\le\frac{1}{\sqrt{c+2}+\sqrt{c+1}}\end{cases}}\)

Suy ra \(\left(1\right)\) xảy ra khi \(a=b=c\Leftrightarrow x=y=z\) (Đpcm)

Giả sử z là số lớn nhất trong 3 số 

Từ đề bài ta có:

\(\sqrt{x+2011}+\sqrt{y+2012}+\sqrt{z+2013}=\sqrt{z+2011}+\sqrt{x+2012}+\sqrt{y+2013}\)

\(\Leftrightarrow\sqrt{x+2012}-\sqrt{x+2011}+\sqrt{y+2013}-\sqrt{y+2012}=\sqrt{z+2012}-\sqrt{z+2011}+\sqrt{z+2013}-\sqrt{z+2012}\)

\(\Leftrightarrow\frac{1}{\sqrt{x+2012}+\sqrt{x+2011}}+\frac{1}{\sqrt{y+2013}+\sqrt{y+2012}}=\frac{1}{\sqrt{z+2012}+\sqrt{z+2011}}+\frac{1}{\sqrt{z+2013}+\sqrt{z+2012}}\)

Ta lại có:

\(\hept{\begin{cases}\frac{1}{\sqrt{x+2012}+\sqrt{x+2011}}\ge\frac{1}{\sqrt{z+2012}+\sqrt{z+2011}}\\\frac{1}{\sqrt{y+2013}+\sqrt{y+2012}}\ge\frac{1}{\sqrt{z+2013}+\sqrt{z+2012}}\end{cases}}\)

Dấu = xảy ra khi x = y = z

Tương tự cho trường hợp x lớn nhất với y lớn nhất.

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\(\Leftrightarrow\dfrac{4\sqrt{x-2009}-4}{x-2009}-1+\dfrac{4\sqrt{x-2009}-4}{x-2009}-1+\dfrac{4\sqrt{x-2009}-4}{x-2009}-1=0\)\(\Leftrightarrow-\dfrac{\left(\sqrt{x-2009}-2\right)^2}{x-2009}-\dfrac{\left(\sqrt{y-2010}-2\right)^2}{y-2010}-\dfrac{\left(\sqrt{z-2011}-2\right)^2}{z-2011}=0\)

VT <=0 đẳng thức khi và chỉ khi \(\left\{{}\begin{matrix}x-2009=4=>x=2013\\y=2014\\z=2015\end{matrix}\right.\)

Đặt a = \(\sqrt{x-2009}\)

b = \(\sqrt{y-2010}\)

c = \(\sqrt{z-2011}\)

\(\Leftrightarrow\dfrac{a-1}{a^2}+\dfrac{b-1}{b^2}+\dfrac{c-1}{c^2}=\dfrac{3}{4}\)

\(\Leftrightarrow\dfrac{1}{a}-\dfrac{1}{a^2}+\dfrac{1}{b}-\dfrac{1}{b^2}+\dfrac{1}{c}-\dfrac{1}{c^2}=\dfrac{3}{4}\)

\(\Leftrightarrow\dfrac{1}{a}-\dfrac{1}{a^2}-\dfrac{1}{4}+\dfrac{1}{b}-\dfrac{1}{b^2}-\dfrac{1}{4}+\dfrac{1}{c}-\dfrac{1}{c^2}-\dfrac{1}{4}=0\)

\(\Leftrightarrow-(\dfrac{1}{a}-\dfrac{1}{2})^2-\left(\dfrac{1}{b}-\dfrac{1}{2}\right)^2-\left(\dfrac{1}{c}-\dfrac{1}{2}\right)^2=0\)

Dấu = xảy ra khi
a = 2

b = 2

c = 2

\(\Leftrightarrow\sqrt{x-2009}=2\)

\(\sqrt{y-2010}=2\)

\(\sqrt{z-2011}=2\)

\(\Leftrightarrow x-2009=4\)

\(y-2010=4\)

\(z-2011=4\)

=> x = 2013

y = 2014

z = 2015

\(\sqrt{x-2}+\sqrt{y+2012}+\sqrt{z-2013}=\dfrac{1}{2}\left(x+y+z\right)\Leftrightarrow2\sqrt{x-2}+2\sqrt{y+2012}+2\sqrt{z-2013}=x+y+z\Leftrightarrow x-2\sqrt{x-2}+y-2\sqrt{y+2012}+z-2\sqrt{z-2013}=0\Leftrightarrow x-2-2\sqrt{x-2}+1+y+2012-2\sqrt{y+2012}+1+z-2013-2\sqrt{z-2013}+1=0\Leftrightarrow\left(\sqrt{x-2}-1\right)^2+\left(\sqrt{y+2012}-1\right)^2+\left(\sqrt{z-2013}-1\right)^2=0\Leftrightarrow\)\(\left\{{}\begin{matrix}\sqrt{x-2}-1=0\\\sqrt{y+2012}-1=0\\\sqrt{z-2013}-1=0\end{matrix}\right.\)\(\Leftrightarrow\)\(\left\{{}\begin{matrix}x-2=1\\y+2012=1\\z-2013=1\end{matrix}\right.\)\(\Leftrightarrow\)\(\left\{{}\begin{matrix}x=3\\y=-2011\\z=2014\end{matrix}\right.\)

Vậy x=3;y=-2011;z=2014