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Lời giải:
Áp dụng BĐT AM-GM ta có:
\(\text{VT}=x-\frac{x}{x^2+z}+y-\frac{y}{y^2+x}+z-\frac{z}{z^2+y}=(x+y+z)-\left(\frac{x}{x^2+z}+\frac{y}{y^2+x}+\frac{z}{z^2+y}\right)\)
\(\geq (x+y+z)-\left(\frac{x}{2\sqrt{x^2z}}+\frac{y}{2\sqrt{y^2x}}+\frac{z}{2\sqrt{z^2y}}\right)=(x+y+z)-\frac{1}{2}\left(\frac{1}{\sqrt{x}}+\frac{1}{\sqrt{y}}+\frac{1}{\sqrt{z}}\right)(1)\)
Từ giả thiết \(xy+yz+xz=3xyz\Rightarrow \frac{1}{x}+\frac{1}{y}+\frac{1}{z}=3\)
Cauchy-Schwarz:
\(3=\frac{1}{x}+\frac{1}{y}+\frac{1}{z}\geq \frac{9}{x+y+z}\Rightarrow x+y+z\geq 3(2)\)
\(\left(\frac{1}{\sqrt{x}}+\frac{1}{\sqrt{y}}+\frac{1}{\sqrt{z}}\right)^2\leq (\frac{1}{x}+\frac{1}{y}+\frac{1}{z})(1+1+1)=9\)
\(\Rightarrow \left(\frac{1}{\sqrt{x}}+\frac{1}{\sqrt{y}}+\frac{1}{\sqrt{z}}\right)\leq 3(3)\)
Từ \((1);(2);(3)\Rightarrow \text{VT}\geq 3-\frac{1}{2}.3=\frac{3}{2}\)
Mặt khác: \(\text{VP}=\frac{1}{2}(\frac{1}{x}+\frac{1}{y}+\frac{1}{z})=\frac{3}{2}\)
Do đó \(\text{VT}\geq \text{VP}\) (đpcm)
Dấu "=" xảy ra khi $x=y=z=1$
\(x+y+z=0\)=>\(\frac{1}{yz}+\frac{1}{xz}+\frac{1}{xy}=0\)(*)
ta co :
\(\sqrt{\frac{1}{x^2}+\frac{1}{y^2}+\frac{1}{z^2}}^2=\left|\frac{1}{x}+\frac{1}{y}+\frac{1}{z}\right|^2\)
\(\frac{1}{x^2}+\frac{1}{y^2}+\frac{1}{z^2}=\frac{1}{x^2}+\frac{1}{y^2}+\frac{1}{z^2}+\frac{2}{xy}+\frac{2}{yz}+\frac{2}{xz}\)
\(\frac{2}{xy}+\frac{2}{xz}+\frac{2}{yz}=0\) luon dung vi (*)
=> dpcm
ban sua lai de di dau "-"=>"+"
1,theo giả thiết => \(x^2+y^2+z^2=x+y+z\)
mà \(3\left(x^2+y^2+z^2\right)>=\left(x+y+z\right)^2\)(bunhiacopxki)
=>\(x+y+z=< 3\)
ta có:\(\frac{1}{x+2}+\frac{1}{y+2}+\frac{1}{z+2}>=\frac{9}{x+y+z+6}=1\)(cauchy schwarz)
Áp dụng bất đẳng thức Cauchy-Schwarz :
\(VT=\frac{x^2}{y+z}+\frac{y^2}{z+x}+\frac{z^2}{x+y}\ge\frac{\left(x+y+z\right)^2}{2\left(x+y+z\right)}=\frac{6^2}{2\cdot6}=3\)
Dấu "=" xảy ra \(\Leftrightarrow x=y=z=2\)
p/s: Đề sai nha bạn. Dạng tổng quát của bài toán :
Cho \(a,b,c>0;a+b+c=p\). Chứng minh rằng :
\(\frac{a^2}{b+c}+\frac{b^2}{c+a}+\frac{c^2}{a+b}\ge\frac{p}{2}\)
Theo AM-GM: \(x^3+y^2\ge2\sqrt{x^3y^2}=2xy\sqrt{x}\)
\(\Rightarrow\frac{2\sqrt{x}}{x^3+y^2}\le\frac{2\sqrt{x}}{2xy\sqrt{x}}=\frac{1}{xy}\)
Tương tự: \(\frac{2\sqrt{y}}{y^3+z^2}\le\frac{1}{yz}\)
\(\frac{2\sqrt{z}}{z^3+x^2}\le\frac{1}{zx}\)
Cộng vế với vế => \(VT\le\frac{1}{xy}+\frac{1}{yz}+\frac{1}{zx}\)
Theo AM-GM; \(VT\le\frac{1}{xy}+\frac{1}{yz}+\frac{1}{zx}\le\frac{\frac{1}{x^2}+\frac{1}{y^2}+\frac{1}{y^2}+\frac{1}{z^2}+\frac{1}{z^2}+\frac{1}{x^2}}{2}=\frac{1}{x^2}+\frac{1}{y^2}+\frac{1}{z^2}\)
Dấu " = " xảy ra <=> x=y=z=1
Áp dụng bất đẳng thức Cacuhy - Schwarz
\(\Rightarrow\hept{\begin{cases}x^3+y^2\ge2\sqrt{x^3y^2}=2xy\sqrt{x}\\y^3+z^2\ge2\sqrt{y^3z^2}=2yz\sqrt{y}\\z^3+x^2\ge2\sqrt{z^3x^2}=2xz\sqrt{z}\end{cases}}\)
\(\Rightarrow\hept{\begin{cases}\frac{2\sqrt{x}}{x^3+y^2}\le\frac{2\sqrt{x}}{2xy\sqrt{x}}=\frac{1}{xy}\\\frac{2\sqrt{y}}{y^3+z^2}\le\frac{2\sqrt{y}}{2yz\sqrt{y}}=\frac{1}{yz}\\\frac{2\sqrt{z}}{z^3+x^2}\le\frac{2\sqrt{z}}{2xz\sqrt{z}}=\frac{1}{xz}\end{cases}}\)
\(\Rightarrow VT\le\frac{1}{xy}+\frac{1}{yz}+\frac{1}{xz}\left(1\right)\)
Áp dụng bất đẳng thức Cacuchy Schwarz
\(\Rightarrow\hept{\begin{cases}\frac{1}{x^2}+\frac{1}{y^2}\ge2\sqrt{\frac{1}{x^2y^2}}=\frac{2}{xy}\\\frac{1}{y^2}+\frac{1}{z^2}\ge2\sqrt{\frac{1}{y^2z^2}}=\frac{2}{yz}\\\frac{1}{z^2}+\frac{1}{x^2}\ge2\sqrt{\frac{1}{z^2x^2}}=\frac{2}{xz}\end{cases}}\)
\(\Rightarrow2\left(\frac{1}{x^2}+\frac{1}{y^2}+\frac{1}{z^2}\right)\ge2\left(\frac{1}{xy}+\frac{1}{yz}+\frac{1}{xz}\right)\)
\(\Rightarrow\frac{1}{x^2}+\frac{1}{y^2}+\frac{1}{z^2}\ge\frac{1}{xy}+\frac{1}{yz}+\frac{1}{xz}\left(2\right)\)
Từ (1) và (2)
\(\Rightarrow VT\le\frac{1}{x^2}+\frac{1}{y^2}+\frac{1}{z^2}\)
\(\Leftrightarrow\frac{2\sqrt{x}}{x^3+y^2}+\frac{2\sqrt{y}}{y^3+z^2}+\frac{2\sqrt{z}}{z^3+x^2}\le\frac{1}{x^2}+\frac{1}{y^2}+\frac{1}{z^2}\left(đpcm\right)\)
Xét: \(\frac{1-x^2}{x+yz}+\frac{1-y^2}{y+xz}+\frac{1-z^2}{z+xy}\)
Thay thế \(x+y+z=1\)
\(\Leftrightarrow\frac{\left(x+y+z\right)^2-x^2}{x\left(x+y+z\right)+yz}+\frac{\left(x+y+z\right)^2-y^2}{y\left(x+y+z\right)+xz}+\frac{\left(x+y+z\right)^2-z^2}{z\left(x+y+z\right)+xy}\)
Áp dụng hằng đẳng thức hiệu 2 bình phương: \(a^2-b^2=\left(a-b\right)\left(a+b\right)\)
\(\Leftrightarrow\frac{\left(y+z\right)\left(2x+y+z\right)}{x^2+xy+xz+yz}+\frac{\left(x+z\right)\left(x+2y+z\right)}{xy+y^2+yz+xz}+\frac{\left(x+y\right)\left(x+y+2z\right)}{xz+zy+z^2+xy}\)
\(\Leftrightarrow\frac{\left(y+z\right)\left(2x+y+z\right)}{\left(x+y\right)\left(x+z\right)}+\frac{\left(x+z\right)\left(x+2y+z\right)}{\left(x+y\right)\left(y+z\right)}+\frac{\left(x+y\right)\left(x+y+2z\right)}{\left(x+z\right)\left(y+z\right)}\)
Áp dụng bất đẳng thức Cauchy cho 2 bộ số thực không âm
\(\Rightarrow\left\{\begin{matrix}\left(x+y\right)\left(x+z\right)\le\left(\frac{2x+y+z}{2}\right)^2=\frac{\left(2x+y+z\right)^2}{4}\\\left(x+y\right)\left(y+z\right)\le\left(\frac{x+2y+z}{2}\right)^2=\frac{\left(x+2y+z\right)^2}{4}\\\left(x+z\right)\left(y+z\right)\le\left(\frac{x+y+2z}{2}\right)^2=\frac{\left(x+y+2z\right)^2}{4}\end{matrix}\right.\)
\(\Rightarrow\left\{\begin{matrix}\frac{\left(y+z\right)\left(2x+y+z\right)}{\left(x+y\right)\left(x+z\right)}\ge\frac{4\left(y+z\right)\left(2x+y+z\right)}{\left(2x+y+z\right)^2}=\frac{4\left(y+z\right)}{2x+y+z}\\\frac{\left(x+z\right)\left(x+2y+z\right)}{\left(x+y\right)\left(y+z\right)}\ge\frac{4\left(x+z\right)\left(x+2y+z\right)}{\left(x+2y+z\right)^2}=\frac{4\left(x+z\right)}{x+2y+z}\\\frac{\left(x+y\right)\left(x+y+2z\right)}{\left(x+z\right)\left(y+z\right)}\ge\frac{4\left(x+y\right)\left(x+y+2z\right)}{\left(x+y+2z\right)^2}=\frac{4\left(x+y\right)}{x+y+2z}\end{matrix}\right.\)
\(\Rightarrow VT\ge\frac{4\left(y+z\right)}{2x+y+z}+\frac{4\left(x+z\right)}{x+2y+z}+\frac{4\left(x+y\right)}{x+y+2z}\)
\(\Rightarrow VT\ge4\left(\frac{y+z}{2x+y+z}+\frac{x+z}{x+2y+z}+\frac{x+y}{x+y+2z}\right)\)
Ta có: \(x+y+z=1\)
\(\Rightarrow\left\{\begin{matrix}y+z=1-x\\x+z=1-y\\x+y=1-z\end{matrix}\right.\) ( 1 )
\(\Rightarrow\left\{\begin{matrix}2x+y+z=1+x\\x+2y+z=1+y\\x+y+2z=1+z\end{matrix}\right.\) ( 2 )
Từ ( 1 ) và ( 2 )
\(\Rightarrow VT\ge4\left(\frac{1-x}{1+x}+\frac{1-y}{1+y}+\frac{1-z}{1+z}\right)\)
\(\Rightarrow VT\ge4\left(\frac{1+x-2x}{1+x}+\frac{1+y-2y}{1+y}+\frac{1+z-2z}{1+z}\right)\)
\(\Rightarrow VT\ge4\left[3-\left(\frac{2x}{1+x}+\frac{2y}{1+y}+\frac{2z}{1+z}\right)\right]\)
\(\Rightarrow VT\ge12-4\left(\frac{2x}{1+x}+\frac{2y}{1+y}+\frac{2z}{1+z}\right)\)
Chứng minh rằng \(12-4\left(\frac{2x}{1+x}+\frac{2y}{1+y}+\frac{2z}{1+z}\right)\ge6\)
\(\Leftrightarrow4\left(\frac{2x}{1+x}+\frac{2y}{1+y}+\frac{2z}{1+z}\right)\le6\)
\(\Leftrightarrow\frac{2x}{1+x}+\frac{2y}{1+y}+\frac{2z}{1+z}\le\frac{3}{2}\)
\(\Leftrightarrow\frac{x}{1+x}+\frac{y}{1+y}+\frac{z}{1+z}\le\frac{3}{4}\)
\(\Leftrightarrow\frac{1+x-1}{1+x}+\frac{1+y-1}{1+y}+\frac{1+z-1}{1+z}\le\frac{3}{4}\)
\(\Leftrightarrow1-\frac{1}{1+x}+1-\frac{1}{1+y}+1-\frac{1}{1+z}\le\frac{3}{4}\)
\(\Leftrightarrow3-\left(\frac{1}{1+x}+\frac{1}{1+y}+\frac{1}{1+z}\right)\le\frac{3}{4}\)
Áp dụng bất đẳng thức cộng mẫu số
\(\Rightarrow\frac{1}{1+x}+\frac{1}{1+y}+\frac{1}{1+z}\ge\frac{\left(1+1+1\right)^2}{3+x+y+z}=\frac{9}{4}\)
\(\Rightarrow3-\left(\frac{1}{1+x}+\frac{1}{1+y}+\frac{1}{1+z}\right)\le3-\frac{9}{4}\)
\(\Rightarrow3-\left(\frac{1}{1+x}+\frac{1}{1+y}+\frac{1}{1+z}\right)\le\frac{3}{4}\) ( đpcm )
Vì \(12-4\left(\frac{2x}{1+x}+\frac{2y}{1+y}+\frac{2z}{1+z}\right)\ge6\)
\(\Rightarrow VT\ge6\)
\(\Leftrightarrow\)\(\frac{1-x^2}{x+yz}+\frac{1-y^2}{y+xz}+\frac{1-z^2}{z+xy}\ge6\) ( đpcm )
Cách khác:
\(A=\frac{1-x^2}{x+yz}+\frac{1-y^2}{y+xz}+\frac{1-z^2}{z+xy}=\frac{1-x^2}{x(x+y+z)+yz}+\frac{1-y^2}{y(x+y+z)+xz}+\frac{1-z^2}{z(x+y+z)+xy}\)
\(\Leftrightarrow A=\frac{1-x^2}{(x+y)(x+z)}+\frac{1-y^2}{(y+z)(y+x)}+\frac{1-z^2}{(z+x)(z+y)}=\frac{2(x+y+z)-[xy(x+y)+yz(y+z)+xz(x+z)]}{(x+y)(y+z)(x+z)}\)
Có \(A\geq 6\Leftrightarrow 2-[xy(x+y)+yz(y+z)+xz(x+z)]\ge 6(x+y)(y+z)(x+z)\)
\(\Leftrightarrow 2+9xyz\geq 7(x+y+z)(xy+yz+xz)\)
\(\Leftrightarrow 2+9xyz\geq 7(xy+yz+xz)\) \((\star)\)
Theo BĐT Schur bậc 3 kết hợp AM-GM:
\(xyz\geq (x+y-z)(y+z-x)(x+z-y)=(1-2x)(1-2y)(1-2z)\)
\(\Leftrightarrow 9xyz\geq 4(xy+yz+xz)-1\)
\(\Rightarrow 2+9(xy+yz+xz)\geq 1+4(xy+yz+xz)=(x+y+z)^2+4(xy+yz+xz)\)\(\geq 7(xy+yz+xz)\)
Do đó \((\star)\) được CM. Bài toán hoàn tất. Dấu bằng xảy ra khi \(x=y=z=\frac{1}{3}\)
Áp dụng bất đẳng thức AM - GM, ta được: \(2yz+2=x^2+\left(y^2+2yz+z^2\right)=x^2+\left(y+z\right)^2\ge2\sqrt{x^2.\left(y+z\right)^2}=2x\left(y+z\right)\Rightarrow yz+1\ge x\left(y+z\right)\)\(\Rightarrow VT\le\frac{x^2}{x^2+x+x\left(y+z\right)}+\frac{y+z}{x+y+z+1}+\frac{1}{xyz+3}=\frac{x+y+z}{x+y+z+1}+\frac{1}{xyz+3}\)
- Nếu \(x+y+z\le2\)thì \(VT\le1-\frac{1}{x+y+z+1}+\frac{1}{xyz+3}\le1-\frac{1}{3}+\frac{1}{3}=1\)
- Nếu \(x+y+z\ge2\), ta đặt x + y + z = p; xy + yz + zx = q; xyz = r thì áp dụng bất đẳng thức Schur, ta được \(VT\le\frac{p}{p+1}+\frac{1}{\frac{p\left(4q-p^2\right)}{9}+3}=\frac{p}{p+1}+\frac{9}{p^3-4p+27}\)
Khảo sát hàm trên với \(p\in\left[\sqrt{2};2\right]\)ta cũng có \(VT\le1\)
Vậy ta có: \(\frac{x^2}{x^2+yz+x+1}+\frac{y+z}{x+y+z+1}+\frac{1}{xyz+3}\le1\)
Đẳng thức xảy ra khi x = y = 1; z = 0
Ta thấy x và y ko đều
1 x 13
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