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Ta có:\(\sqrt{\dfrac{yz}{x^2+2017}}=\sqrt{\dfrac{yz}{x^2+xy+yz+zx}}=\sqrt{\dfrac{yz}{\left(x+y\right)\left(x+z\right)}}\)
\(=\sqrt{\dfrac{y}{x+y}\cdot\dfrac{z}{x+z}}\le\dfrac{\dfrac{y}{x+y}+\dfrac{z}{x+z}}{2}\)
Tương tự ta có:\(\sqrt{\dfrac{zx}{y^2+2017}}\le\dfrac{\dfrac{x}{x+y}+\dfrac{z}{y+z}}{2}\)
\(\sqrt{\dfrac{xy}{z^2+2017}}\le\dfrac{\dfrac{y}{z+y}+\dfrac{x}{x+z}}{2}\)
Cộng vế với vế ta có:
\(\sqrt{\dfrac{yz}{x^2+2017}}+\sqrt{\dfrac{zx}{y^2+2017}}+\sqrt{\dfrac{xy}{z^2+2017}}\)
\(\le\dfrac{\dfrac{y}{x+y}+\dfrac{z}{x+z}+\dfrac{z}{z+y}+\dfrac{x}{x+y}+\dfrac{y}{z+y}+\dfrac{x}{x+z}}{2}\)
\(=\dfrac{\dfrac{x+y}{x+y}+\dfrac{y+z}{y+z}+\dfrac{z+x}{z+x}}{2}=\dfrac{1+1+1}{2}=\dfrac{3}{2}\)
Dấu "=" xảy ra \(\Leftrightarrow x=y=z=\dfrac{\sqrt{2017}}{\sqrt{3}}\)
Lời giải:
Áp dụng BĐT AM-GM:
$1=xy+yz+xz+2xyz\leq \frac{(x+y+z)^2}{3}+2.\frac{(x+y+z)^3}{27}$
$\Leftrightarrow 1\leq \frac{t^2}{3}+\frac{2t^3}{27}$ (đặt $x+y+z=t$)
$\Leftrightarrow 2t^3+9t^2-27\geq 0$
$\Leftrightarrow (t+3)^2(2t-3)\geq 0$
$\Leftrightarrow 2t-3\geq 0$
$\Leftrightarrow t\geq \frac{3}{2}$ hay $x+y+z\geq \frac{3}{2}$ (đpcm)
Dấu "=" xảy ra khi $x=y=z=\frac{1}{2}$
Có \(VT=\dfrac{x^2}{x^3-xyz+2013x}+\dfrac{y^2}{y^3-xyz+2013y}+\dfrac{z^2}{z^3-xyz+2013z}\)
\(\ge\dfrac{\left(x+y+z\right)^2}{x^3+y^3+z^3-3xyz+2013\left(x+y+z\right)}\)
\(=\dfrac{\left(x+y+z\right)^2}{\left(x+y+z\right)\left[x^2+y^2+z^2-\left(xy+yz+zx\right)\right]+2013\left(x+y+z\right)}\)
\(=\dfrac{x+y+z}{x^2+y^2+z^2-\left(xy+yz+zx\right)+3\left(xy+yz+zx\right)}\)
(vì \(2013=3.671=3\left(xy+yz+zx\right)\))
\(=\dfrac{x+y+z}{x^2+y^2+z^2+2\left(xy+yz+zx\right)}\)
\(=\dfrac{x+y+z}{\left(x+y+z\right)^2}\)
\(=\dfrac{1}{x+y+z}\)
ĐTXR \(\Leftrightarrow\dfrac{1}{x^2-yz+2013}=\dfrac{1}{y^2-zx+2013}=\dfrac{1}{z^2-xy+2013}\)
\(\Leftrightarrow x^2-yz=y^2-zx=z^2-xy\)
\(\Leftrightarrow x=y=z\) (với \(x,y,z>0\))
Vậy ta có đpcm.
Đặt \(a=\sqrt{\frac{yz}{x}},b=\sqrt{\frac{zx}{y}},c=\sqrt{\frac{xy}{z}}\) \(\Rightarrow ab+bc+ac=1\)
Suy ra bài toán trở về dạng chứng minh \(\frac{1}{a^2+1}+\frac{1}{b^2+1}+\frac{1}{c^2+1}\le\frac{9}{4}\)
\(\Leftrightarrow1-\frac{a^2}{a^2+1}+1-\frac{b^2}{b^2+1}+1-\frac{c^2}{c^2+1}\le\frac{9}{4}\)
\(\Leftrightarrow\frac{a^2}{a^2+1}+\frac{b^2}{b^2+1}+\frac{c^2}{c^2+1}\ge\frac{3}{4}\)(*)
Áp dụng bất đẳng thức AM-GM ta có :
\(\frac{a^2}{a^2+1}+\frac{b^2}{b^2+1}+\frac{c^2}{c^2+1}\ge\frac{\left(a+b+c\right)^2}{a^2+b^2+c^2+3}\)
Đặt t = a+b+c \(\Rightarrow a^2+b^2+c^2=t^2-2\)
Ta cần chứng minh \(\frac{t^2}{t^2+1}\ge\frac{3}{4}\Leftrightarrow4t^2\ge3t^2+3\Rightarrow t^2\ge3\)(Luôn đúng vì \(\left(a+b+c\right)^2\ge3\left(ab+bc+ac\right)=3\))
Vậy ta có đpcm
Ta cần chứng minh:\(\dfrac{1}{\sqrt{x+y+xy}}+\dfrac{1}{\sqrt{y+z+yz}}+\dfrac{1}{\sqrt{z+x+zx}}\ge\sqrt{3}\)
Áp dụng bất đẳng thức Bunhiacopxki, ta được:
\(\dfrac{1}{\sqrt{x+y+xy}}+\dfrac{1}{\sqrt{y+z+yz}}+\dfrac{1}{\sqrt{z+x+zx}}\ge\dfrac{9}{\sqrt{x+y+xy}+\sqrt{y+z+yz}+\sqrt{z+x+zx}}\)
Mặt khác, ta có:
\(\left(\sqrt{x+y+xy}+\sqrt{y+z+yz}+\sqrt{z+x+zx}\right)^2\le3\left(\left(x+y+xy\right)+\left(y+z+yz\right)+\left(z+x+zx\right)\right)\)
\(\Leftrightarrow\left(\sqrt{x+y+xy}+\sqrt{y+z+yz}+\sqrt{z+x+zx}\right)^2\le3\left(6+xy+yz+zx\right)\)Lại có:
\(xy+yz+zx\le\dfrac{\left(x+y+z\right)^2}{3}=\dfrac{9}{3}=3\)
\(\Rightarrow\left(\sqrt{x+y+xy}+\sqrt{y+z+yz}+\sqrt{z+x+zx}\right)^2\le3\left(6+3\right)=27\)
\(\Rightarrow\sqrt{x+y+xy}+\sqrt{y+z+yz}+\sqrt{z+x+zx}\le3\sqrt{3}\)
\(\Rightarrow\dfrac{9}{\sqrt{x+y+xy}+\sqrt{y+z+yz}+\sqrt{z+x+zx}}\ge\dfrac{9}{3\sqrt{3}}=\sqrt{3}\)
Do đó \(\dfrac{1}{\sqrt{x+y+xy}}+\dfrac{1}{\sqrt{y+z+yz}}+\dfrac{1}{\sqrt{z+x+zx}}\ge\sqrt{3}\)
Dấu bằng xảy ra \(\Leftrightarrow x=y=z=1\).
a) Giả sử \(x^2-xy+y^2\ge\frac{1}{3}\left(x^2+xy+y^2\right)\)
\(\Leftrightarrow3\left(x^2-xy+y^2\right)\ge\frac{1}{3}.3\left(x^2+xy+y^2\right)\)
\(\Leftrightarrow3\left(x^2-xy+y^2\right)\ge x^2+xy+y^2\)
\(\Leftrightarrow3x^2-3xy+3y^2-x^2-xy-y^2\ge0\)
\(\Leftrightarrow2x^2-4xy+2y^2\ge0\)
\(\Leftrightarrow2\left(x^2-2xy+y^2\right)\ge0\)
\(\Leftrightarrow2\left(x-y\right)^2\ge0\)(luôn đúng với mọi \(x,y\in R\)).
Dấu bằng xảy ra\(\Leftrightarrow x=y\).
Vậy \(x^2-xy+y^2\ge\frac{1}{3}\left(x^2+xy+y^2\right)\)với \(x,y\in R\).
Đặt \(A=\frac{x\sqrt{x}}{x+\sqrt{xy}+y}+\frac{y\sqrt{y}}{y+\sqrt{yz}+z}+\frac{z\sqrt{z}}{z+\sqrt{zx}+x}\left(x,y,z>0\right)\)
Và đặt \(B=\frac{y\sqrt{y}}{x+\sqrt{xy}+y}+\frac{z\sqrt{z}}{y+\sqrt{yz}+z}+\frac{x\sqrt{x}}{z+\sqrt{zx}+x}\left(x,y,z>0\right)\)
Đặt \(\sqrt{x}=m,\sqrt{y}=n,\sqrt{z}=p\left(m,n,p>0\right)\)thì theo đề bài : \(m+n+p=2\)
Lúc đó:
\(A=\frac{m^2.m}{m^2+mn+n^2}+\frac{n^2.n}{n^2+np+p^2}+\frac{p^2.p}{p^2+pm+m^2}\)
\(A=\frac{m^3}{m^2+mn+n^2}+\frac{n^3}{n^2+np+p^2}+\frac{p^3}{p^2+pm+m^2}\)
Và \(B=\frac{n^3}{m^2+mn+n^2}+\frac{p^3}{n^2+np+p^2}+\frac{m^3}{p^2+pm+m^2}\)
Xét hiệu \(A-B=\frac{m^3-n^3}{m^2+mn+n^2}+\frac{n^3-p^3}{n^2+np+p^2}+\frac{p^3-m^3}{p^2+pm+m^2}\)
\(\Leftrightarrow A-B=\frac{\left(m-n\right)\left(m^2+mn+n^2\right)}{m^2+mn+n^2}+\frac{\left(n-p\right)\left(n^2+np+p^2\right)}{n^2+np+p^2}\)\(+\frac{\left(p-m\right)\left(p^2+pm+m^2\right)}{p^2+pm+m^2}\)
\(\Leftrightarrow A-B=\left(m-n\right)+\left(n-p\right)+\left(p-m\right)\)
\(\Leftrightarrow A-B=m-n+n-p+p-m=0\)
\(\Leftrightarrow A=B\)
Xét \(A+B=\frac{m^3+n^3}{m^2+mn+n^2}+\frac{n^3+p^3}{n^2+np+p^2}+\frac{p^3+m^3}{p^2+pm+m^2}\)
\(\Leftrightarrow A+A=2A=\frac{\left(m+n\right)\left(m^2-mn+n^2\right)}{m^2+m+n^2}+\frac{\left(n+p\right)\left(n^2-np+p^2\right)}{n^2+np+p^2}\)\(\frac{\left(p+m\right)\left(p^2-pm+m^2\right)}{p^2+pm+m^2}\)
Theo câu a), ta có \(x^2-xy+y^2\ge\frac{1}{3}\left(x^2+xy+y^2\right)\)với \(x,y\in R\)
\(\Leftrightarrow\frac{x^2-xy+y^2}{x^2+xy+y^2}\ge\frac{1}{3}\left(1\right)\)
Dấu bằng xảy ra \(\Leftrightarrow x=y\)
Áp dụng bất đẳng thức (1) (với \(m,n>0\)), ta được:
\(\frac{m^2-mn+n^2}{m^2+mn+n^2}\ge\frac{1}{3}\)
\(\Leftrightarrow\frac{\left(m+n\right)\left(m^2-mn+n^2\right)}{m^2+mn+n^2}\ge\frac{m+n}{3}\left(2\right)\)
Dấu bằng xảy ra \(\Leftrightarrow m=n>0\)
Chứng minh tương tự, ta được:
\(\frac{\left(n+p\right)\left(n^2-np+p^2\right)}{n^2+np+p^2}\ge\frac{n+p}{3}\left(3\right)\)
Dấu bằng xảy ra\(\Leftrightarrow n=p>0\)
\(\frac{\left(p+m\right)\left(p^2-pm+m^2\right)}{p^2+pm+m^2}\ge\frac{p+m}{2}\left(4\right)\)
Dấu bằng xảy ra\(\Leftrightarrow p=m>0\)
Từ \(\left(2\right),\left(3\right),\left(4\right)\), ta được:
\(\frac{\left(m+n\right)\left(m^2-mn+n^2\right)}{m^2+mn+n^2}+\frac{\left(n+p\right)\left(n^2-np+p^2\right)}{n^2+np+p^2}\)\(+\frac{\left(p+m\right)\left(p^2-pm+m^2\right)}{p^2-pm+m^2}\ge\frac{m+n}{3}+\frac{n+p}{3}+\frac{p+m}{3}\)
\(\Leftrightarrow2A\ge\frac{m+n+n+p+p+m}{3}\)
\(\Leftrightarrow2A\ge\frac{2\left(m+n+p\right)}{3}\)
\(\Leftrightarrow A\ge\frac{m+n+p}{3}\)
\(\Leftrightarrow A\ge\frac{2}{3}\)(vì \(m+n+p=2\)) (điều phải chứng minh).
Dấu bằng xảy ra.
\(\Leftrightarrow\hept{\begin{cases}m=n=p>0\\m+n+p=2\end{cases}}\Leftrightarrow m=n=p=\frac{2}{3}\)\(\Leftrightarrow\sqrt{x}=\sqrt{y}=\sqrt{z}=\frac{2}{3}\Leftrightarrow x=y=z=\frac{4}{9}\)
Vậy nếu \(x,y,z>0\) và \(\sqrt{x}+\sqrt{y}+\sqrt{z}=2\)thì: \(\frac{x\sqrt{x}}{x+\sqrt{xy}+y}+\frac{y\sqrt{y}}{y+\sqrt{yz}+z}+\frac{z\sqrt{z}}{z+\sqrt{zx}+x}\ge\frac{2}{3}\).
Đoạn cuối của cô Nguyễn Linh Chi em có 1 cách biến đổi tương đương cũng khá ngắn gọn ạ
\(RHS\ge2\cdot\frac{\left(x+y+z\right)^2}{x^2+y^2+z^2-\left(x+y+z\right)+18}\)
Theo đánh giá của cô Nguyễn Linh Chi thì \(xy+yz+zx\ge x+y+z\ge3\)
Ta cần chứng minh:\(\frac{\left(x+y+z\right)^2}{x^2+y^2+z^2-\left(x+y+z\right)+18}\ge\frac{1}{2}\)
Thật vậy,BĐT tương đương với:
\(2\left(x+y+z\right)^2\ge x^2+y^2+z^2-x-y-z+18\)
\(\Leftrightarrow\left(x+y+z\right)^2+x+y+z-12\ge0\)
\(\Leftrightarrow\left(x+y+z+4\right)\left(x+y+z-3\right)\ge0\) ( luôn đúng với \(x+y+z\ge3\) )
=> đpcm
Áp dụng: \(AB\le\frac{\left(A+B\right)^2}{4}\)với mọi A, B
Ta có:
\(x^3+8=\left(x+2\right)\left(x^2-2x+4\right)\le\frac{\left(x+2+x^2-2x+4\right)^2}{4}\)
=> \(\sqrt{x^3+8}\le\frac{x^2-x+6}{2}\)
=> \(\frac{x^2}{\sqrt{x^3+8}}\ge\frac{2x^2}{x^2-x+6}\)
Tương tự
=> \(\frac{x^2}{\sqrt{x^3+8}}+\frac{y^2}{\sqrt{y^3+8}}+\frac{z^2}{\sqrt{z^3+8}}\)
\(\ge\frac{2x^2}{x^2-x+6}+\frac{2y^2}{y^2-y+6}+\frac{2z^2}{z^2-z+6}\)
\(=2\left(\frac{x^2}{x^2-x+6}+\frac{y^2}{y^2-y+6}+\frac{z^2}{z^2-z+6}\right)\)
\(\ge2\frac{\left(x+y+z\right)^2}{x^2-x+6+y^2-y+6+z^2-z+6}\)
\(=2\frac{\left(x+y+z\right)^2}{x^2+y^2+z^2-\left(x+y+z\right)+18}\)(1)
Ta có: \(x+y+z\le xy+yz+zx\le\frac{\left(x+y+z\right)^2}{3}\) với mọi x, y, z
=> \(\left(x+y+z\right)^2-3\left(x+y+z\right)\ge0\)
=> \(\left(x+y+z\right)\left(x+y+z-3\right)\ge0\)
=> \(x+y+z\ge3\)với mọi x, y, z dương
Và \(x^2+y^2+z^2=\left(x+y+z\right)^2-2\left(xy+yz+zx\right)\le\left(x+y+z\right)^2-2\left(x+y+z\right)\)
Do đó: \(\frac{\left(x+y+z\right)^2}{x^2+y^2+z^2-\left(x+y+z\right)+18}\)
\(\ge\frac{\left(x+y+z\right)^2}{\left(x+y+z\right)^2-3\left(x+y+z\right)+18}\)
Đặt: x + y + z = t ( t\(\ge3\))
Xét hiệu: \(\frac{t^2}{t^2-3t+18}-\frac{1}{2}=\frac{t^2+3t-18}{t^2-3t+18}=\frac{\left(t-3\right)\left(t+6\right)}{\left(t-\frac{3}{2}\right)^2+\frac{63}{4}}\ge0\)với mọi t \(\ge3\)
Do đó: \(\frac{\left(x+y+z\right)^2}{\left(x+y+z\right)^2-3\left(x+y+z\right)+18}\ge\frac{1}{2}\)(2)
Từ (1); (2)
=> \(\frac{x^2}{\sqrt{x^3+8}}+\frac{y^2}{\sqrt{y^3+8}}+\frac{z^2}{\sqrt{z^3+8}}\ge2.\frac{1}{2}=1\)
Dấu "=" xảy ra <=> x= y = z = 1
ĐẶt \(A=x^2+y^2+z^2\Rightarrow4A-12=4\left(x^2+y^2+z^2\right)-2\left(x+y+z+xy+yz+zx\right)\)
\(\Rightarrow3A-12=\left(x-1\right)^2+\left(y-1\right)^2+\left(z-1\right)^2+\left(x-y\right)^2+\left(y-z\right)^2+\left(z-x\right)^2-3\)
\(\Rightarrow3A\ge9\Rightarrow A\ge3\)
dấu= xảy ra khi x=y=z=1
Sử dụng các bđt cơ bản
\(x^2+y^2+z^2\ge\frac{1}{3}\left(x+y+z\right)^2\ge xy+yz+zx\)