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Áp dụng Bất đẳng thức: \(\left(a+b+c\right)\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)\ge9\) (Tự chứng minh)
\(\Rightarrow C=\frac{1}{x^2+2yz}+\frac{1}{y^2+2xz}+\frac{1}{z^2+2xy}\ge\frac{9}{x^2+y^2+z^2+2xy+2yz+2xz}=\frac{9}{\left(x+y+z\right)^2}\ge\frac{9}{3^2}=1\)Dấu "=" xảy ra \(\Leftrightarrow x=y=z=1\)
\(C=\frac{1}{x^2+2yz}+\frac{1}{y^2+2xz}+\frac{1}{z^2+2xy}\ge\frac{9}{x^2+y^2+z^2+2xy+2yz+2zx}=\frac{9}{\left(x+y+z\right)^2}\ge\frac{9}{3^2}=1\)
Dấu "=" xảy ra khi \(x=y=z=1\)
ÁP dụng bất đẳng thức AM-GM ta có:
\(P=\dfrac{x^2}{x^2+2yz}+\dfrac{y^2}{y^2+2xz}+\dfrac{z^2}{z^2+2xy}\ge\dfrac{\left(x+y+z\right)^2}{x^2+y^2+z^2+2\left(xy+yz+xz\right)}\)\(=\dfrac{\left(x+y+z\right)^2}{\left(x+y+z\right)^2}=1\)
Dấu "=" xảy ra\(\Leftrightarrow x=y=z>0\)
Vậy \(MinP=1\Leftrightarrow x=y=z>0\)
Ta có: \(\dfrac{1}{x}+\dfrac{1}{y}+\dfrac{1}{z}=0\)\(\Rightarrow xy+yz+xz=0\)
\(\Rightarrow\left\{{}\begin{matrix}xy=-yz-xz\\yz=-xy-xz\\xz=-xy-xz\end{matrix}\right.\)
\(\Rightarrow\dfrac{yz}{x^2+2yz}=\dfrac{yz}{x^2+yz-xy-xz}=\dfrac{yz}{\left(x-y\right)\left(x-z\right)}\)
Tương tự:
\(\Rightarrow\left\{{}\begin{matrix}\dfrac{xz}{y^2+2xz}=\dfrac{xz}{\left(x-y\right)\left(x-z\right)}\\\dfrac{xy}{z^2+2xy}=\dfrac{xy}{\left(x-y\right)\left(x-z\right)}\\\dfrac{yz}{x^2+2yz}=\dfrac{yz}{\left(x-y\right)\left(x-z\right)}\end{matrix}\right.\)
\(\Rightarrow A=\dfrac{xz}{\left(x-y\right)\left(x-z\right)}+\dfrac{xy}{\left(x-y\right)\left(x-z\right)}+\dfrac{yz}{\left(x-y\right)\left(x-z\right)}=\dfrac{xz+xy+yz}{\left(x-y\right)\left(x-z\right)}=\dfrac{0}{\left(x-y\right)\left(x-z\right)}=0\)
Vậy \(A=0.\)
Lời giải:
Từ \(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}=0\Rightarrow \frac{xy+yz+xz}{xyz}=0\Rightarrow xy+yz+xz=0\)
Suy ra \(yz=-xy-xz\)
\(\Rightarrow x^2+2yz=x^2+yz-xy-xz=x(x-y)-z(x-y)\)
\(\Leftrightarrow x^2+2yz=(x-z)(x-y)\)
\(\Rightarrow \frac{yz}{x^2+2yz}=\frac{yz}{(x-z)(x-y)}\)
Hoàn toàn tương tự với các phân thức còn lại và cộng theo vế:
\(A=\frac{yz}{(x-y)(x-z)}+\frac{xz}{(y-x)(y-z)}+\frac{xy}{(z-x)(z-y)}\)
\(A=\frac{-yz(y-z)}{(x-y)(y-z)(z-x)}+\frac{-xz(z-x)}{(x-y)(y-z)(z-x)}+\frac{-xy(x-y)}{x-y)(y-z)(z-x)}\)
\(A=\frac{xy^2+yz^2+zx^2-(x^2y+y^2z+z^2x)}{(x-y)(y-z)(z-x)}\)
\(A=\frac{xy^2+yz^2+zx^2-(x^2y+y^2z+z^2x)}{xy^2+yz^2+zx^2-(x^2y+y^2z+z^2x)}=1\)
Ta có :
\(x+y+z=1\)
\(\Rightarrow\left(x+y+z\right)^2=1\)
Áp dụng BĐT Cauchy-schwar dưới dạng engel ta có :
\(\dfrac{1}{x^2+2yz}+\dfrac{1}{y^2+2zx}+\dfrac{1}{z^2+2xy}\ge\dfrac{\left(1+1+1\right)^2}{x^2+y^2+z^2+2xy+2yz+2zx}=\dfrac{9}{1}=9\)
\(\text{Ta có : }x+y+z=1\\ \Rightarrow\left(x+y+z\right)^2=1\\ \Rightarrow x^2+y^2+z^2+2xy+2xz+2yz=1\\ \Rightarrow\dfrac{1}{x^2+2yz}+\dfrac{1}{y^2+2xz}+\dfrac{1}{z^2+2xy}\\ =\dfrac{x^2+y^2+z^2+2xy+2xz+2yz}{x^2+2yz}+\dfrac{x^2+y^2+z^2+2xy+2xz+2yz}{y^2+2xz}+\dfrac{x^2+y^2+z^2+2xy+2xz+2yz}{z^2+2xy}\\ =\dfrac{x^2+2yz}{x^2+2yz}+\dfrac{y^2+2xz}{x^2+2yz}+\dfrac{z^2+2xy}{x^2+2yz}+\dfrac{x^2+2yz}{y^2+2xz}+\dfrac{y^2+2xz}{y^2+2xz}+\dfrac{z^2+2xy}{y^2+2xz}+\dfrac{x^2+2yz}{z^2+2xy}+\dfrac{y^2+2xz}{z^2+2xy}+\dfrac{z^2+2xy}{z^2+2xy}\\ =1+\left(\dfrac{y^2+2xz}{x^2+2yz}+\dfrac{x^2+2yz}{y^2+2xz}\right)+\left(\dfrac{z^2+2xy}{x^2+2yz}+\dfrac{x^2+2yz}{z^2+2xy}\right)+1+\left(\dfrac{y^2+2xz}{z^2+2xy}+\dfrac{z^2+2xy}{y^2+2xz}\right)+1\)Áp dụng \(BDT:\dfrac{a}{b}+\dfrac{b}{a}\ge2\)
\(\Rightarrow1+\left(\dfrac{y^2+2xz}{x^2+2yz}+\dfrac{x^2+2yz}{y^2+2xz}\right)+\left(\dfrac{z^2+2xy}{x^2+2yz}+\dfrac{x^2+2yz}{z^2+2xy}\right)+1+\left(\dfrac{y^2+2xz}{z^2+2xy}+\dfrac{z^2+2xy}{y^2+2xz}\right)+1\\ \ge1+2+2+1+2+1\ge9\left(đpcm\right)\)
Dấu \("="\) xảy ra khi: \(\left\{{}\begin{matrix}y^2+2xz=x^2+2yz\\z^2+2xy=x^2+2yz\\y^2+2xz=z^2+2xy\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}y^2-2yz=x^2-2xz\\z^2-2yz=x^2-2xy\\y^2-2xy=z^2-2xz\end{matrix}\right.\\ \Leftrightarrow\left\{{}\begin{matrix}y^2-2yx+z^2=x^2-2xz+z^2\\z^2-2yz+y^2=x^2-2xy+y^2\\y^2-2xy+x^2=z^2-2xz+x^2\end{matrix}\right.\\ \Leftrightarrow\left\{{}\begin{matrix}\left(y-z\right)^2=\left(x-z\right)^2\\\left(z-y\right)^2=\left(x-y\right)^2\\\left(y-x\right)^2=\left(z-x\right)^2\end{matrix}\right.\\ \Leftrightarrow\left\{{}\begin{matrix}y-z=x-z\\z-y=x-y\\y-x=z-x\end{matrix}\right.\\ \Leftrightarrow\left\{{}\begin{matrix}x=y\\z=x\\y=z\end{matrix}\right.\Leftrightarrow x=y=z\\\text{Mà } x+y+z=1\\ \Leftrightarrow3x=1\\ \Leftrightarrow x=\dfrac{1}{3}\\ \Leftrightarrow x=y=z=\dfrac{1}{3}\)
Vậy \(\dfrac{1}{x^2+2yz}+\dfrac{1}{y^2+2xz}+\dfrac{1}{z^2+2xy}\ge9\) với \(x;y;z>0\) và \(x+y+z=1\)
đẳng thức xảy ra khi : \(x=y=z=\dfrac{1}{3}\)
\(\dfrac{1}{x}+\dfrac{1}{y}+\dfrac{1}{z}=0\Leftrightarrow xy+yz+zx=0\)
\(\Rightarrow yz=-xy-zx\Rightarrow\dfrac{yz}{x^2+2yz}=\dfrac{yz}{x^2+yz-xy-zx}=\dfrac{yz}{\left(x-y\right)\left(x-z\right)}\)
Tương tự: \(\dfrac{xz}{y^2+2xz}=\dfrac{xz}{\left(y-x\right)\left(y-z\right)}\) ; \(\dfrac{xy}{z^2+2xy}=\dfrac{xy}{\left(x-z\right)\left(y-z\right)}\)
\(\Rightarrow A=\dfrac{-yz\left(y-z\right)-zx\left(z-x\right)-xy\left(x-y\right)}{\left(x-y\right)\left(y-z\right)\left(z-x\right)}=1\)
Từ \(\dfrac{1}{x}\) + \(\dfrac{1}{y}\) + \(\dfrac{1}{z}\) = 0
\(=>yz+xz+xy=0\)
\(=>yz=-xz-xy\)
Ta có : \(x^2+2yz=x^2+yz+yz=x^2+yz-yx-xz=\left(y-x\right)\left(z-x\right)=-\left(x-y\right)\left(z-x\right)\)
Tương tư :
\(y^2+2xz=y^2+xz+xz=y^2+xz-xy-yz=\left(y-x\right)\left(y-z\right)=-\left(x-y\right)\left(y-z\right)\)
\(z^2+2xy=z^2+xy+xy=z^2+xy-yz-xz=\left(z-y\right)\left(z-x\right)=-\left(y-z\right)\left(z-x\right)\)Nên A = \(\dfrac{yz}{x^2+2yz}+\dfrac{xz}{y^2+2xz}+\dfrac{xy}{z^2+2xy}\)
=\(\dfrac{-yz}{\left(x-y\right)\left(z-x\right)}+\dfrac{-xz}{\left(x-y\right)\left(y-z\right)}+\dfrac{-xy}{\left(y-z\right)\left(z-x\right)}\)
=\(\dfrac{-yz\left(y-z\right)-xz\left(z-x\right)-xy\left(x-y\right)}{\left(x-y\right)\left(y-z\right)\left(z-x\right)}\)
=\(\dfrac{(-y^2z+yz^2-z^2x+x^2z-x^2y+xy^2)+(xyz-xyz)}{\left(x-y\right)\left(y-z\right)\left(z-x\right)}\)
=\(\dfrac{\left(xyz-y^2z-z^2x+yz^2\right)+\left(-x^2y+xy^2+x^2z-xyz\right)}{\left(x-y\right)\left(y-z\right)\left(z-x\right)}\)
=\(\dfrac{z\left(xy-y^2-xz+zy\right)-x\left(xy-y^2-xz+zy\right)}{\left(x-y\right)\left(y-z\right)\left(z-x\right)}\)
=\(\dfrac{\left(z-x\right)\left(xy-y^2-xz+zy\right)}{\left(x-y\right)\left(y-z\right)\left(z-x\right)}\)
=\(\dfrac{\left(z-x\right)\left(x-y\right)\left(y-z\right)}{\left(x-y\right)\left(y-z\right)\left(z-x\right)}\)
=1
= =" ... bài này làm dài ..bấm máy mỏi tay lắm...
nhanh gọn lẹ.... A = 0
@phynit em hiểu nguyên tắc đó. cái em càng không hiểu là các bạn bấm chọn. trong khi cái bước rất quan trọng thì đang bỏ lửng
Em suy nghĩ rất nhiều nhiều về cái đề này. không làm nổi-->theo dõi -->
A sẽ giải thích tại sao đặt được nhân tử vậy cho nhé
Ta có:
\(xy\left(x-y\right)+yz\left(y-z\right)+zx\left(z-x\right)\)
\(=xy\left(x-y\right)+y^2z-z^2y+z^2x-zx^2\)
\(=xy\left(x-y\right)+\left(y^2z-zx^2\right)+\left(z^2x-z^2y\right)\)
\(=\left(x-y\right)\left(xy-zx-zy+z^2\right)\)
\(=\left(x-y\right)\left(\left(xy-zx\right)+\left(z^2-zy\right)\right)\)
\(=\left(x-y\right)\left(y-z\right)\left(x-z\right)\)
Cậu ta làm sai thì làm sao ngonhuminh với thầy phynit hiểu được
Xét A= \(\dfrac{x}{\sqrt{x+2yz}}\).\(\dfrac{1}{\sqrt{2}}\)=\(\dfrac{x}{\sqrt{2x+4yz}}\)=\(\sqrt{\dfrac{x.x}{2x+4yz}}\)
ta có x+y+z=\(\dfrac{1}{2}\)=> 2x+2y+2z= 1=> 2x+4yz= 4yz+1-2y-2z=(2y-1)(2z-1)
từ đó A= \(\sqrt{\dfrac{x}{2y-1}.\dfrac{x}{2z-1}}\)=\(\sqrt{\dfrac{x}{2y-2x-2y-2z}.\dfrac{x}{2z-2x-2y-2z}}\)
=\(\sqrt{\dfrac{x}{-2\left(x+y\right)}\dfrac{x}{-2\left(x+z\right)}}\)=\(\sqrt{\dfrac{1}{4}.\dfrac{x}{x+z}.\dfrac{x}{x+y}}\)=\(\dfrac{1}{2}\sqrt{\dfrac{x}{x+y}.\dfrac{x}{x+z}}\)
Áp dụng cô si \(\sqrt{ab}\)≤\(\dfrac{a+b}{2}\) =>\(\dfrac{1}{2}\sqrt{ab}\)≤\(\dfrac{a+b}{4}\)ta được
A≤\(\dfrac{1}{4}\).(\(\dfrac{x}{x+y}\)+\(\dfrac{x}{x+z}\))
cmmt thì \(\dfrac{P}{\sqrt{2}}\)≤ \(\dfrac{1}{4}\).\(\left(\dfrac{x}{x+y}+\dfrac{x}{x+z}+\dfrac{y}{y+x}+\dfrac{y}{y+z}+\dfrac{z}{z+x}+\dfrac{z}{z+y}\right)\)
\(\dfrac{P}{\sqrt{2}}\)≤\(\dfrac{3}{4}\)=>P≤\(\dfrac{3.\sqrt{2}}{4}\)=\(\dfrac{3}{2\sqrt{2}}\)
Dấu"=" xảy ra <=> x=y=z=\(\dfrac{1}{6}\)
Lời giải:
Áp dụng BĐT Bunhiacopxky:
\(\left(\frac{x^2}{x^2+2yz}+\frac{y^2}{y^2+2xz}+\frac{z^2}{z^2+2xy}\right)(x^2+2yz+y^2+2xz+z^2+2xy)\geq (x+y+z)^2\)
\(\Leftrightarrow P(x+y+z)^2\geq (x+y+z)^2\)
\(\Rightarrow P\geq 1\)
Vậy \(P_{\min}=1\)
Dấu bằng xảy ra khi \(x=y=z\)
\(P=\dfrac{x^2}{x^2+2yz}+\dfrac{y^2}{y^2+2xz}+\dfrac{z^2}{z^2+2xy}\)
Áp dụng BDT Cô-si : \(a^2+b^2\ge2ab\)
\(\Rightarrow\left\{{}\begin{matrix}y^2+z^2\ge2yz\\x^2+z^2\ge2xz\\x^2+y^2\ge2xy\end{matrix}\right.\\ \Rightarrow\left\{{}\begin{matrix}x^2+y^2+z^2\ge x^2+2yz>0\\x^2+y^2+z^2\ge y^2+2xz>0\\x^2+y^2+z^2\ge z^2+2xy>0\end{matrix}\right.\\ \Rightarrow\left\{{}\begin{matrix}\dfrac{x^2}{x^2+y^2+z^2}\le\dfrac{x^2}{x^2+2yz}\\\dfrac{y^2}{x^2+y^2+z^2}\le\dfrac{y^2}{y^2+2xz}\\\dfrac{z^2}{x^2+y^2+z^2}\le\dfrac{z^2}{z^2+2xy}\end{matrix}\right.\\ \Rightarrow P=\dfrac{x^2}{x^2+2yz}+\dfrac{y^2}{y^2+2xz}+\dfrac{z^2}{z^2+2xy}\\ \ge\dfrac{x^2}{x^2+y^2+z^2}+\dfrac{y^2}{x^2+y^2+z^2}+\dfrac{z^2}{x^2+y^2+z^2}\\ \ge\dfrac{x^2+y^2+z^2}{x^2+y^2+z^2}\ge1\forall x;y;z\)
Dấu "=" xảy ra khi \(:\left\{{}\begin{matrix}y=z\\x=z\\x=y\end{matrix}\right.\Leftrightarrow x=y=z\)
Vậy \(P_{Min}=1\) khi \(x=y=z\)