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\(5x^2+2xy+2y^2-\left(4x^2+4xy+y^2\right)=\left(x-y\right)^2\ge0\\ \Leftrightarrow5x^2+2xy+2y^2\ge4x^2+4xy+y^2=\left(2x+y\right)^2\)
\(\Leftrightarrow P\le\dfrac{1}{2x+y}+\dfrac{1}{2y+z}+\dfrac{1}{2z+x}=\dfrac{1}{9}\left(\dfrac{9}{x+x+y}+\dfrac{9}{y+y+z}+\dfrac{9}{z+z+x}\right)\\ \Leftrightarrow P\le\dfrac{1}{9}\left(\dfrac{1}{x}+\dfrac{1}{x}+\dfrac{1}{y}+\dfrac{1}{y}+\dfrac{1}{y}+\dfrac{1}{z}+\dfrac{1}{z}+\dfrac{1}{z}+\dfrac{1}{x}\right)\\ \Leftrightarrow P\le\dfrac{1}{9}\left(\dfrac{3}{x}+\dfrac{3}{y}+\dfrac{3}{z}\right)=\dfrac{1}{3}\left(\dfrac{1}{x}+\dfrac{1}{y}+\dfrac{1}{z}\right)=1\)
Dấu \("="\Leftrightarrow x=y=z=1\)
Ta có: \(5x^2+6xy+5y^2=4\left(x+y\right)^2+\left(x-y\right)^2\ge4\left(x+y\right)^2\)
tương tự: \(5y^2+6yz+5z^2\ge4\left(y+z\right)^2\) ;\(5z^2+6xz+5z^2\ge4\left(x+z\right)^2\)
\(\Rightarrow P\ge\dfrac{2\left(x+y\right)}{x+y+2z}+\dfrac{2\left(y+z\right)}{y+z+2x}+\dfrac{2\left(x+z\right)}{x+z+2y}\)
\(\Leftrightarrow\dfrac{P}{2}\ge\dfrac{x+y}{x+y+2z}+\dfrac{y+z}{y+z+2x}+\dfrac{x+z}{x+z+2y}\)
\(\Leftrightarrow\dfrac{P}{2}\ge\dfrac{x+y}{\left(x+z\right)+\left(y+z\right)}+\dfrac{y+z}{\left(x+y\right)+\left(x+z\right)}+\dfrac{x+z}{\left(x+y\right)+\left(y+z\right)}\)Theo BDT Nesbit
\(\dfrac{x+y}{\left(x+z\right)+\left(y+z\right)}+\dfrac{y+z}{\left(x+y\right)+\left(x+z\right)}+\dfrac{x+z}{\left(x+y\right)+\left(y+z\right)}\ge\dfrac{3}{2}\)
Vậy \(\dfrac{P}{2}\ge\dfrac{3}{2}\Leftrightarrow P\ge3\)
Min P = 3 khi x = y = z
Lời giải:
Ta có: \(5x^2+6xy+5y^2=3(x^2+y^2+2xy)+2(x^2+y^2)\)
\(=3(x+y)^2+2(x^2+y^2)\geq 3(x+y)^2+(x+y)^2\) (theo BĐT AM-GM)
\(\Leftrightarrow 5x^2+6xy+5y^2\geq 4(x+y)^2\Rightarrow \sqrt{5x^2+6xy+5y^2}\geq 2(x+y)\)
Thực hiện tương tự với những biểu thức còn lại suy ra:
\(P\geq \frac{2(x+y)}{x+y+2z}+\frac{2(y+z)}{y+z+2x}+\frac{2(z+x)}{z+x+2y}\)
\(P\geq 2\left(\frac{x+y}{x+y+2z}+\frac{y+z}{y+z+2x}+\frac{z+x}{z+x+2y}\right)=2\left(\frac{(x+y)^2}{(x+y+2z)(x+y)}+\frac{(y+z)^2}{(y+z+2x)(y+z)}+\frac{(z+x)^2}{(z+x+2y)(z+x)}\right)\)
Áp dụng BĐT Cauchy-Schwarz:
\(P\geq 2.\frac{(x+y+y+z+z+x)^2}{(x+y+2z)(x+y)+(y+z+2x)(y+z)+(z+x+2y)(z+x)}\)
\(\Leftrightarrow P\geq 2. \frac{4(x+y+z)^2}{2(x+y+z)^2+2(xy+yz+xz)}=\frac{4(x+y+z)^2}{(x+y+z)^2+xy+yz+xz}\)
\(\geq \frac{4(x+y+z)^2}{(x+y+z)^2+\frac{(x+y+z)^2}{3}}=3\) (theo AM-GM \(xy+yz+xz\leq \frac{(x+y+z)^2}{3}\))
Vậy \(P\geq 3\Leftrightarrow P_{\min}=3\)
Dấu bằng xảy ra khi \(x=y=z\)
đặt\(A=\dfrac{x^3}{2x+3y+5z}+\dfrac{y^3}{2y+3z+5x}+\dfrac{z^3}{2z+3x+5y}\)
\(=>A=\dfrac{x^4}{2x^2+3xy+5xz}+\dfrac{y^4}{2y^2+3yz+5xy}+\dfrac{z^4}{2z^2+3xz+5yz}\)
BBDT AM-GM
\(=>A\ge\dfrac{\left(x^2+y^2+z^2\right)^2}{2\left(x^2+y^2+z^2\right)+8\left(xy+yz+xz\right)}\)
theo BDT AM -GM ta chứng minh được \(xy+yz+xz\le x^2+y^2+z^2\)
vì \(x^2+y^2\ge2xy\)
\(y^2+z^2\ge2yz\)
\(x^2+z^2\ge2xz\)
\(=>2\left(x^2+y^2+z^2\right)\ge2\left(xy+yz+xz\right)< =>xy+yz+xz\le x^2+y^2+z^2\)
\(=>2\left(x^2+y^2+z^2\right)+8\left(xy+yz+xz\right)\le10\left(x^2+y^2+z^2\right)\)
\(=>A\ge\dfrac{\left(x^2+y^2+z^2\right)^2}{10\left(x^2+y^2+z^2\right)}=\dfrac{x^2+y^2+z^2}{10}=\dfrac{\dfrac{1}{3}}{10}=\dfrac{1}{30}\left(đpcm\right)\)
dấu"=" xảy ra<=>x=y=z=1/3
cm bđt phụ \(5x^2+6xy+5y^2\ge4\left(x+y\right)^2\)nhé
Ta có: \(\sqrt{5x^2+6xy+5y^2}=\sqrt{4\left(x+y\right)^2+\left(x-y\right)^2}\ge\sqrt{4\left(x+y\right)^2}=2\left(x+y\right)\)
\(\Rightarrow\frac{\sqrt{5x^2+6xy+5y^2}}{x+y+2z}\ge\frac{2\left(x+y\right)}{x+y+2z}\)(1)
Tương tự, ta có: \(\frac{\sqrt{5y^2+6yz+5z^2}}{y+z+2x}\ge\frac{2\left(y+z\right)}{y+z+2x}\)(2); \(\frac{\sqrt{5z^2+6zx+5x^2}}{z+x+2y}\ge\frac{2\left(z+x\right)}{z+x+2y}\)(3)
Cộng theo vế của 3 BĐT (1), (2), (3), ta được: \(\frac{\sqrt{5x^2+6xy+5y^2}}{x+y+2z}+\frac{\sqrt{5y^2+6yz+5z^2}}{y+z+2x}+\frac{\sqrt{5z^2+6zx+5x^2}}{z+x+2y}\)\(\ge2\left[\frac{x+y}{\left(y+z\right)+\left(z+x\right)}+\frac{y+z}{\left(z+x\right)+\left(x+y\right)}+\frac{z+x}{\left(x+y\right)+\left(y+z\right)}\right]\)
Đặt \(x+y=a;y+z=b;z+x=c\)thì \(\frac{x+y}{\left(y+z\right)+\left(z+x\right)}+\frac{y+z}{\left(z+x\right)+\left(x+y\right)}+\frac{z+x}{\left(x+y\right)+\left(y+z\right)}\)\(=\frac{a}{b+c}+\frac{b}{c+a}+\frac{c}{a+b}\)
Nhưng ta có BĐT Nesbitt quen thuộc sau: \(\frac{a}{b+c}+\frac{b}{c+a}+\frac{c}{a+b}\ge\frac{3}{2}\)
Thật vậy: 
(Bài này mình đã làm nhiều rồi nha nên ngại đánh lại, đây là bất đẳng thức có rất nhiều cách chứng minh nhưng mình nghĩ dồn biến là cách hay và đẹp nhất nha! Có thể tham khảo nhiều cách khác trên mạng, vô thống kê hỏi đáp của mình xem ảnh)
Như vậy: \(\frac{\sqrt{5x^2+6xy+5y^2}}{x+y+2z}+\frac{\sqrt{5y^2+6yz+5z^2}}{y+z+2x}+\frac{\sqrt{5z^2+6zx+5x^2}}{z+x+2y}\)\(\ge2\left[\frac{x+y}{\left(y+z\right)+\left(z+x\right)}+\frac{y+z}{\left(z+x\right)+\left(x+y\right)}+\frac{z+x}{\left(x+y\right)+\left(y+z\right)}\right]\)\(\ge2.\frac{3}{2}=3\)
Đẳng thức xảy ra khi x = y = z
Ta có x2-xy+y2=\(\left(\dfrac{x+y}{2}\right)^2+3\left(\dfrac{x-y}{2}\right)^2\)\(\ge\)\(\left(\dfrac{x+y}{2}\right)^2\)
=>\(\dfrac{\sqrt{x^2-xy+y^2}}{x+y+2z}\ge\dfrac{x+y}{2\left(x+y+2z\right)}\)(1) . Tương tự ...
Đặt \(\left\{{}\begin{matrix}y+z=a\\x+z=b\\x+y=c\end{matrix}\right.\)(a,b,c>0). Khi đó ta có :
S=\(\dfrac{1}{2}\left(\dfrac{c}{a+b}+\dfrac{b}{a+c}+\dfrac{a}{b+c}\right)\ge\dfrac{3}{4}\) (Netbit)
Ta có 5x2+2xy+2y2=(2x+y)2+(x-y)2>=(2x+y)2
Khi đó P<=\(\frac{1}{2x+y}+\frac{1}{2y+z}+\frac{1}{2z+x}\)
Lại có \(\frac{1}{2x+y}=\frac{1}{x+x+y}\le\frac{1}{9}\left(\frac{1}{x}+\frac{1}{y}+\frac{1}{x}\right)\)
Tương tự \(\frac{1}{2y+z}\le\frac{1}{9}\left(\frac{1}{y}+\frac{1}{z}+\frac{1}{y}\right)\)
\(\frac{1}{2z+x}\le\frac{1}{9}\left(\frac{1}{z}+\frac{1}{x}+\frac{1}{z}\right)\)
Khi đó P<=\(\frac{1}{3}\left(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}\right)\le\frac{1}{3}\sqrt{3\left(\frac{1}{x^2}+\frac{1}{y^2}+\frac{1}{z^2}\right)}\le\frac{\sqrt{3}}{3}\)
Dấu bằng xảy ra khi x=y=z=\(\frac{\sqrt{3}}{3}\)
HAY
\(P=\dfrac{x^3}{2x+3y+5z}+\dfrac{y^3}{2y+3z+5x}+\dfrac{z^3}{2z+3x+5y}\)
\(P=\dfrac{x^4}{2x^2+3xy+5xz}+\dfrac{y^4}{2y^2+3yz+5xy}+\dfrac{z^4}{2z^2+3xz+5yz}\)
\(P\ge\dfrac{\left(x^2+y^2+z^2\right)^2}{2\left(x^2+y^2+z^2\right)+8\left(xy+yz+zx\right)}\ge\dfrac{\left(x^2+y^2+z^2\right)^2}{2\left(x^2+y^2+z^2\right)+8\left(x^2+y^2+z^2\right)}\)
\(P\ge\dfrac{x^2+y^2+z^2}{10}\ge\dfrac{1}{30}\)
\(P_{min}=\dfrac{1}{30}\) khi \(x=y=z=\dfrac{1}{3}\)
\(2x^2+2xy+5y^2=\left(x+2y\right)^2+\left(x-y\right)^2\ge\left(x+2y\right)^2\)
\(\Rightarrow P\ge\dfrac{x+2y}{3x+y+5z}+\dfrac{y+2z}{3y+z+5x}+\dfrac{z+2x}{3x+x+5y}\)
\(\Rightarrow P\ge\dfrac{\left(x+2y\right)^2}{\left(x+2y\right)\left(3x+y+5z\right)}+\dfrac{\left(y+2z\right)^2}{\left(y+2z\right)\left(3y+z+5x\right)}+\dfrac{\left(z+2x\right)^2}{\left(z+2x\right)\left(3x+x+5y\right)}\)
\(\Rightarrow P\ge\dfrac{\left(x+2y\right)^2}{3x^2+2y^2+7xy+5xz+10yz}+\dfrac{\left(y+2z\right)^2}{3y^2+2z^2+7yz+5xy+10xz}+\dfrac{\left(z+2x\right)^2}{3z^2+2x^2+7xz+5yz+10xy}\)
\(\Rightarrow P\ge\dfrac{\left(x+2y+y+2z+z+2x\right)^2}{5\left(x^2+y^2+z^2\right)+22\left(xy+xz+yz\right)}\)
\(\Rightarrow P\ge\dfrac{9\left(x+y+z\right)^2}{5\left(x+y+z\right)^2+12\left(xy+xz+yz\right)}\ge\dfrac{9\left(x+y+z\right)^2}{5\left(x+y+z\right)^2+\dfrac{12\left(x+y+z\right)^2}{3}}\)
\(\Rightarrow P\ge1\)
\(\Rightarrow P_{min}=1\) khi \(x=y=z\)