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Áp dụng BĐT Cô-si,ta có :
x4 + yz \(\ge\)\(2\sqrt{x^4yz}=2x^2\sqrt{yz}\); \(y^4+xz\ge2y^2\sqrt{xz}\); \(z^4+xy\ge2z^2\sqrt{xy}\)
\(\Rightarrow\frac{x^2}{x^4+yz}+\frac{y^2}{y^4+xz}+\frac{z^2}{z^4+xy}\le\frac{x^2}{2x^2\sqrt{yz}}+\frac{y^2}{2y^2\sqrt{xz}}+\frac{z^2}{2z^2\sqrt{xy}}=\frac{1}{2\sqrt{yz}}+\frac{1}{2\sqrt{xz}}+\frac{1}{2\sqrt{xy}}\)
CM : x + y + z \(\ge\sqrt{xy}+\sqrt{yz}+\sqrt{xz}\)
\(\frac{x^2}{x^4+yz}+\frac{y^2}{y^4+xz}+\frac{z^2}{z^4+xy}\le\frac{1}{2}\left(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}\right)=\frac{1}{2}.\frac{yz+xz+xy}{xyz}=\frac{1}{2}.\frac{3xyz}{xyz}=\frac{3}{2}\)
Áp dụng BĐT Cauchy cho các cặp số dương, ta có: \(\Sigma\frac{x^2}{x^4+yz}\le\Sigma\frac{x^2}{2x^2\sqrt{yz}}=\Sigma\frac{1}{2\sqrt{yz}}\)
\(\le\frac{1}{4}\Sigma\left(\frac{1}{y}+\frac{1}{z}\right)=\frac{1}{2}\left(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}\right)\)
\(=\frac{1}{2}.\frac{xy+yz+zx}{xyz}\le\frac{1}{2}.\frac{x^2+y^2+z^2}{xyz}=\frac{1}{2}.\frac{3xyz}{xyz}=\frac{3}{2}\)
Đẳng thức xảy ra khi x = y = z = 1
![](https://rs.olm.vn/images/avt/0.png?1311)
\(VT=\sum\frac{x^2}{x^4+yz}\le\sum\frac{x^2}{2x^2\sqrt{yz}}=\frac{1}{2}\sum\frac{1}{\sqrt{yz}}\le\frac{1}{4}\sum\left(\frac{1}{y}+\frac{1}{z}\right)=\frac{1}{2}\left(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}\right)\)
\(\Rightarrow VT\le\frac{1}{2}\left(\frac{xy+yz+zx}{xyz}\right)\le\frac{1}{2}\left(\frac{x^2+y^2+z^2}{xyz}\right)=\frac{3}{2}\)
Dấu "=" xảy ra khi \(x=y=z=1\)
![](https://rs.olm.vn/images/avt/0.png?1311)
Ta có: \(\frac{x^2}{x^4+yz}\le\frac{x^2}{2\sqrt{x^4.yz}}=\frac{x^2}{2x^2\sqrt{yz}}=\frac{1}{2\sqrt{yz}}\)(BĐt cosi) (1)
CMTT: \(\frac{y^2}{y^4+xz}\le\frac{1}{2\sqrt{xz}}\) (2)
\(\frac{z^2}{z^4+xy}\le\frac{1}{2\sqrt{xy}}\)(3)
Từ (1); (2) và (3) =>A = \(\frac{x^2}{x^4+yz}+\frac{y^2}{y^4+xz}+\frac{z^2}{z^4+xy}\le\frac{1}{2}\left(\frac{1}{\sqrt{xz}}+\frac{1}{\sqrt{yz}}+\frac{1}{\sqrt{xy}}\right)\)
Áp dụng bđt \(ab+bc+ac\le a^2+b^2+c^2\)
cmt đúng: <=> \(\left(a-b\right)^2+\left(b-c\right)^2+\left(a-c\right)^2\ge0\)(luôn đúng)
Khi đó: A \(\le\frac{1}{2}\cdot\left(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}\right)=\frac{1}{2}\cdot\frac{xy+yz+xz}{xyz}\le\frac{1}{2}\cdot\frac{x^2+y^2+z^2}{xyz}=\frac{3xyz}{2xyz}=\frac{3}{2}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(P\le\frac{1}{2}\left(\Sigma\frac{1}{\sqrt{xy}}\right)\le\frac{\left(xy+yz+zx\right)^2}{6x^2y^2z^2}\le\frac{\left(x^2+y^2+z^2\right)^2}{6x^2y^2z^2}=\frac{3}{2}\)
dấu "=" xảy ra \(\Leftrightarrow\)\(x=y=z=1\)
mình nhầm :) làm lại nhé
\(P\le\frac{1}{2}\left(\Sigma\frac{1}{\sqrt{xy}}\right)\le\frac{\left(\sqrt{xy}+\sqrt{yz}+\sqrt{zx}\right)^2}{6xyz}\le\frac{xy+yz+zx}{2xyz}\le\frac{x^2+y^2+z^2}{2xyz}=\frac{3}{2}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
Bunhiacopxki: \(\left(x^2+yz+zx\right)\left(y^2+yz+zx\right)\ge\left(xy+yz+zx\right)^2\)
\(\Rightarrow\frac{xy}{x^2+yz+zx}\le\frac{xy\left(y^2+yz+zx\right)}{\left(xy+yz+zx\right)^2}\)
Thiết lập tương tự và cộng lại:
\(\Rightarrow VT\le\frac{xy\left(y^2+yz+zx\right)+yz\left(z^2+xy+zx\right)+zx\left(x^2+yz+xy\right)}{\left(xy+yz+zx\right)^2}\)
\(VT\le\frac{xy^3+xy^2z+x^2yz+yz^3+xy^2z+xyz^2+x^3z+xyz^2+x^2yz}{\left(xy+yz+zx\right)^2}\)
Ta chỉ cần chứng minh: \(\frac{xy^3+xy^2z+x^2yz+yz^3+xy^2z+xyz^2+x^3z+xyz^2+x^2yz}{\left(xy+yz+zx\right)^2}\le\frac{x^2+y^2+z^2}{xy+yz+zx}\)
\(\Leftrightarrow xy^3+xy^2z+x^2yz+yz^3+xy^2z+xyz^2+x^3z+xyz^2+x^2yz\le\left(x^2+y^2+z^2\right)\left(xy+yz+zx\right)\)
\(\Leftrightarrow x^2yz+xy^2z+xyz^2\le x^3y+y^3z+z^3x\)
\(\Leftrightarrow\frac{x^2}{z}+\frac{y^2}{x}+\frac{z^2}{y}\ge x+y+z\) (đúng theo Cauchy-Schwarz)
Dấu "=" xảy ra khi \(x=y=z\)
![](https://rs.olm.vn/images/avt/0.png?1311)
http://diendantoanhoc.net/topic/160455-%C4%91%E1%BB%81-to%C3%A1n-v%C3%B2ng-2-tuy%E1%BB%83n-sinh-10-chuy%C3%AAn-b%C3%ACnh-thu%E1%BA%ADn-2016-2017/
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v~~ ko thằng admin :(( t làm cái bài này mất gần 30 phút mà bây giờ nó éo hiện câu trả lời của tao ???? hận quá đi
bài này easy lắm bạn ơi :((
áp dụng BDT (Am-ag) mẫu ta có
\(\left(x^2+y^2\right)\ge2\sqrt{x^2y^2}=2xy\) rồi thay vào
suy ra \(\frac{1}{x^2+y^2+2}\le\frac{1}{2xy+2}\)
\(\left(y^2+z^2\right)\ge2yz\)
suy ra \(\frac{1}{y^2+z^2+2}\le\frac{1}{2yz+2}\)
tượng tự vs BDT con lại rồi + vế vs vế ta được
\(VT\le\frac{1}{2xy+2}+\frac{1}{2yz+2}+\frac{1}{2xz+2}=\frac{1}{xy+xy+1+1}+\frac{1}{yz+yz+1+1}+\frac{1}{xz+xz+1+1}\)
gọi cái \(\frac{1}{yz+yz+1+1}+.........=Pain\)
áp dụng cosi sáp cho 4 số ta được
\(\frac{1}{xy+xy+1+1}\le\frac{1}{16}\left(\frac{1}{xy}+\frac{1}{xy}+\frac{1}{1}+\frac{1}{1}\right)\)
\(\frac{1}{yz+yz+1+1}\le\frac{1}{16}\left(\frac{1}{yz}+\frac{1}{yz}+\frac{1}{1}+\frac{1}{1}\right)\)
\(\frac{1}{xz+xz+1+1}\le\frac{1}{16}\left(\frac{1}{xz}+\frac{1}{xz}+\frac{1}{1}+\frac{1}{1}\right)\)
+ vế với vế ta được
\(VT\le Pain\le\frac{1}{16}\left(\frac{2}{xz}+\frac{2}{yz}+\frac{2}{xy}+\frac{2}{2}+\frac{2}{2}+\frac{2}{2}\right)\)
\(VT\le PAIN\le\frac{1}{8}\left(\frac{1}{xz}+\frac{1}{yz}+\frac{1}{xy}+1+1+1\right)\)
bây giờ m đi chứng minh cái \(\frac{1}{zy}+\frac{1}{yz}+\frac{1}{xy}\ge3\) chắc là m làm được
áp dụng BDT cô si ta có
\(\frac{1}{xz}+xz\ge2\)
\(\frac{1}{yz}+yz\ge2\)
\(\frac{1}{xz}+zx\ge2\)
+ vế với vế ta được
\(\frac{1}{xy}+\frac{1}{yz}+\frac{1}{zx}+xy+yz+zx\ge6\)
mà đề bài cho xy+yz+xz=3 suy ra
\(\frac{1}{xy}+\frac{1}{yz}+\frac{1}{zx}\ge3\)
nhưng mà nó trái dấu oy :(( kệ nhé cứ thay vào nhé không sao hết bạn oy :)
thay vào ta được
\(VT\le PAIN\le\frac{1}{8}\left(3+3\right)=\frac{3}{4}\)
ĐIỀU CẦN PHẢI CHỨNG MINH :((
![](https://rs.olm.vn/images/avt/0.png?1311)
Lời giải:
Ta có:
\(\text{VT}=\frac{1}{x^2+y^2+2}+\frac{1}{y^2+z^2+2}+\frac{1}{z^2+x^2+2}\)
\(\Rightarrow 2\text{VT}=\frac{2}{x^2+y^2+2}+\frac{2}{y^2+z^2+2}+\frac{2}{z^2+x^2+2}\)
\(2\text{VT}=1-\frac{x^2+y^2}{x^2+y^2+2}+1-\frac{y^2+z^2}{y^2+z^2+2}+1-\frac{z^2+x^2}{z^2+x^2+2}\)
\(2\text{VT}=3-\left(\frac{x^2+y^2}{x^2+y^2+2}+\frac{y^2+z^2}{y^2+z^2+2}+\frac{z^2+x^2}{z^2+x^2+2}\right)=3-A\)
Áp dụng BĐT Cauchy-Schwarz:
\(A\geq \frac{(\sqrt{x^2+y^2}+\sqrt{y^2+z^2}+\sqrt{z^2+x^2})^2}{2(x^2+y^2+z^2)+6}=\frac{(\sqrt{x^2+y^2}+\sqrt{y^2+z^2}+\sqrt{z^2+x^2})^2}{2(x^2+y^2+z^2+xy+yz+xz)}(*)\)
Xét tử số:
\((\sqrt{x^2+y^2}+\sqrt{y^2+z^2}+\sqrt{z^2+x^2})^2\)
\(=2(x^2+y^2+z^2)+2(\sqrt{(x^2+y^2)(x^2+z^2)}+\sqrt{(x^2+y^2)(y^2+z^2)}+\sqrt{(y^2+z^2)(z^2+x^2)})\)
Áp dụng BĐT Bunhiacopxky:
\(\sqrt{(x^2+y^2)(x^2+z^2)}\geq \sqrt{(x^2+yz)^2}=x^2+yz\)
\(\sqrt{(x^2+y^2)(y^2+z^2)}\geq \sqrt{(xz+y^2)^2}=xz+y^2\)
\(\sqrt{(y^2+z^2)(z^2+x^2)}\geq \sqrt{(z^2+xy)^2}=z^2+xy\)
\(\Rightarrow \sum \sqrt{(x^2+y^2)(x^2+z^2)}\geq x^2+y^2+z^2+xy+yz+xz\)
\(\Rightarrow (\sqrt{x^2+y^2}+\sqrt{y^2+z^2}+\sqrt{z^2+x^2})^2\geq 4(x^2+y^2+z^2)+2(xy+yz+xz)\)
\(\geq 3(x^2+y^2+z^2)+3(xy+yz+xz)=3(x^2+y^2+z^2+xy+yz+xz)\)
(theo BĐT AM-GM)
Do đó: Từ \((*)\Rightarrow A\geq \frac{3(x^2+y^2+z^2+xy+yz+xz)}{2(x^2+y^2+z^2+xy+yz+xz)}=\frac{3}{2}\)
\(\Rightarrow 2\text{VT}\leq 3-\frac{3}{2}=\frac{3}{2}\)
\(\Rightarrow \text{VT}\leq \frac{3}{4}\) (đpcm)
Dấu bằng xảy ra khi \(x=y=z=1\)
We have: \(\dfrac{1}{x^2+y^2+2}=\dfrac{1}{x^2+y^2+z^2+2-z^2}\le\dfrac{1}{5-z^2}\)
Similarly and by adding them:
\(\dfrac{1}{5-x^2}+\dfrac{1}{5-y^2}+\dfrac{1}{5-z^2}\le\dfrac{3}{4}\left(\circledast\right)\)
We know that \(\dfrac{1}{5-x^2}\le\dfrac{3\left(x^2+x\right)}{8\left(x^2+x+1\right)}\)
\(\Leftrightarrow-\dfrac{\left(x-1\right)^2\left(3x^2+9x+8\right)}{8\left(x^2-5\right)\left(x^2+x+1\right)}\le0\) It's obviously
\(\Rightarrow L.H.S_{\left(\circledast\right)}\le\dfrac{3}{8}\left(\dfrac{x^2+x}{x^2+x+1}+\dfrac{y^2+y}{y^2+y+1}+\dfrac{z^2+z}{z^2+z+1}\right)\le\dfrac{3}{4}\)
The equality occur when \(x=y=z=1\)
Done!
Áp dụng bất đẳng thức Bunyakovsky
\(\Rightarrow\left(x^4+yz\right)\left(1+1\right)\ge\left(x^2+\sqrt{yz}\right)^2\)
\(\Rightarrow\frac{x^2}{x^4+yz}\le\frac{2x^2}{\left(x^2+\sqrt{yz}\right)^2}\)
Tương tự ta có : \(\hept{\begin{cases}\frac{y^2}{y^4+xz}\le\frac{2y^2}{\left(y^2+\sqrt{xz}\right)^2}\\\frac{z^2}{z^4+xy}\le\frac{2z^2}{\left(z^2+\sqrt{xy}\right)^2}\end{cases}}\)
\(\Rightarrow VT\le2\left[\frac{x^2}{\left(x^2+\sqrt{yz}\right)^2}+\frac{y^2}{\left(y^2+\sqrt{xz}\right)^2}+\frac{z^2}{\left(z^2+\sqrt{xy}\right)}\right]\)
Chứng minh rằng :
\(2\left[\frac{x^2}{\left(x^2+\sqrt{yz}\right)^2}+\frac{y^2}{\left(y^2+\sqrt{xz}\right)^2}+\frac{z^2}{\left(z^2+\sqrt{xy}\right)}\right]\le\frac{3}{2}\)
\(\Leftrightarrow\frac{x^2}{\left(x^2+\sqrt{yz}\right)^2}+\frac{y^2}{\left(y^2+\sqrt{xz}\right)^2}+\frac{z^2}{\left(z^2+\sqrt{xy}\right)^2}\le\frac{3}{4}\)
Áp dụng bất đẳng thức Cauchy
\(\Rightarrow x^2+\sqrt{yz}\ge2\sqrt{x^2\sqrt{yz}}=2x\sqrt{\sqrt{yz}}\)
\(\Rightarrow\left(x^2+\sqrt{yz}\right)^2\ge4x^2\sqrt{yz}\)
\(\Rightarrow\frac{x^2}{\left(x^2+\sqrt{yz}\right)^2}\le\frac{x^2}{4x^2\sqrt{yz}}=\frac{1}{4\sqrt{yz}}\)
Tương tự ta có : \(\hept{\begin{cases}\frac{y^2}{\left(y^2+\sqrt{xz}\right)^2}\le\frac{1}{4\sqrt{xz}}\\\frac{z^2}{\left(z^2+\sqrt{zy}\right)^2}\le\frac{1}{4\sqrt{xy}}\end{cases}}\)
\(\Leftrightarrow\frac{x^2}{\left(x^2+\sqrt{yz}\right)^2}+\frac{y^2}{\left(y^2+\sqrt{xz}\right)^2}+\frac{z^2}{\left(z^2+\sqrt{xy}\right)^2}\)
\(\le\frac{1}{4}\left(\frac{1}{\sqrt{xy}}+\frac{1}{\sqrt{yz}}+\frac{1}{xz}\right)\)
Chứng minh rằng : \(\frac{1}{4}\left(\frac{1}{\sqrt{xy}}+\frac{1}{\sqrt{yz}}+\frac{1}{\sqrt{xz}}\right)\le\frac{3}{4}\)
\(\Leftrightarrow\frac{1}{\sqrt{xy}}+\frac{1}{\sqrt{yz}}+\frac{1}{\sqrt{xz}}\le3\)
Theo đề bài ta có : \(x^2+y^2+z^2=3xyz\)
\(\frac{x}{yz}+\frac{y}{xz}+\frac{z}{xy}=3\)
\(\Rightarrow\frac{1}{\sqrt{xy}}+\frac{1}{\sqrt{yz}}+\frac{1}{\sqrt{xz}}\le3\)
\(\Leftrightarrow\frac{1}{\sqrt{xy}}+\frac{1}{\sqrt{yz}}+\frac{1}{\sqrt{xz}}\le\frac{x}{yz}+\frac{y}{xz}+\frac{z}{xy}\)
Áp dụng bất đẳng thức Cauchy
\(\Rightarrow\frac{1}{\sqrt{xy}}\le\frac{\frac{1}{x}+\frac{1}{y}}{2}\)
Tương tự ta có : \(\hept{\begin{cases}\frac{1}{\sqrt{xz}}\le\frac{\frac{1}{x}+\frac{1}{z}}{2}\\\frac{1}{\sqrt{xy}}\le\frac{\frac{1}{z}+\frac{1}{y}}{2}\end{cases}}\)
\(\Rightarrow\frac{1}{\sqrt{xy}}+\frac{1}{\sqrt{yz}}+\frac{1}{\sqrt{xz}}\le\frac{1}{x}+\frac{1}{y}+\frac{1}{z}\left(1\right)\)
Áp dụng bất đẳng thức Cauchy
\(\Rightarrow\frac{x}{yz}+\frac{y}{xz}\ge2\sqrt{\frac{1}{z^2}}=\frac{2}{z}\)
Tương tự ta có :
\(\hept{\begin{cases}\frac{y}{xz}+\frac{z}{xy}\ge\frac{2}{x}\\\frac{x}{zy}+\frac{z}{xy}\ge\frac{2}{y}\end{cases}}\)
\(\Rightarrow2\left(\frac{x}{yz}+\frac{y}{zx}+\frac{z}{xy}\right)\ge2\left(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}\right)\)
\(\Leftrightarrow\frac{x}{yz}+\frac{y}{xz}+\frac{z}{xy}\ge\frac{1}{x}+\frac{1}{y}+\frac{1}{z}\left(2\right)\)
Từ (1) và (2)
\(\Rightarrow\frac{1}{\sqrt{xy}}+\frac{1}{\sqrt{yz}}+\frac{1}{\sqrt{xz}}\le3\left(đpcm\right)\)
Vậy \(\frac{1}{4}\left(\frac{1}{\sqrt{xy}}+\frac{1}{\sqrt{yz}}+\frac{1}{\sqrt{xz}}\right)\le\frac{3}{4}\)
\(\Rightarrow2\left[\frac{x^2}{\left(x^2+\sqrt{yz}\right)^2}+\frac{y^2}{\left(y^2+\sqrt{xz}\right)^2}+\frac{z^2}{\left(z^2+\sqrt{xy}\right)}\right]\le\frac{3}{2}\)
Mà \(VT\le2\left[\frac{x^2}{\left(x^2+\sqrt{yz}\right)^2}+\frac{y^2}{\left(y^2+\sqrt{xz}\right)^2}+\frac{z^2}{\left(z^2+\sqrt{xy}\right)}\right]\)
\(\Rightarrow VT\le\frac{3}{2}\) ( đpcm)
Dấu "=" xảy ra khi \(x=y=z=1\)
Chúc bạn học tốt !!!
\(\text{Σ}\frac{x^2}{x^4+yz}\le\text{Σ}\frac{x^2}{2x^2\sqrt{yz}}=\text{Σ}\frac{1}{2\sqrt{yz}}\le\text{Σ}\frac{\frac{1}{y}+\frac{1}{z}}{4}=\frac{\frac{1}{x}+\frac{1}{y}+\frac{1}{z}}{2}=\frac{\frac{xy+yz+xz}{xyz}}{2}=\frac{\frac{3\left(xy+yz+xz\right)}{x^2+y^2+z^2}}{2}\)(1)
Dễ dàng CM được: \(x^2+y^2+z^2\ge xy+yz+xz\)
Thay vào (1) -> dpcm