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\(ĐK:x,y,z>\frac{1}{2}\)
Ta có: \(\left(x+2y\right)^2=\left(\frac{3y}{2}+\frac{y+2x}{2}\right)^2\ge4.\frac{3y}{2}.\frac{y+2x}{2}=3y\left(2x+y\right)\)\(\Rightarrow\frac{2x+y}{x+2y}\le\frac{x+2y}{3y}\Rightarrow\frac{2x+y}{x\left(x+2y\right)}\le\frac{x+2y}{3xy}=\frac{1}{3}\left(\frac{2}{x}+\frac{1}{y}\right)\)
Tương tự: \(\frac{2y+z}{y\left(y+2z\right)}\le\frac{1}{3}\left(\frac{2}{y}+\frac{1}{z}\right)\); \(\frac{2z+x}{z\left(z+2x\right)}\le\frac{1}{3}\left(\frac{2}{z}+\frac{1}{x}\right)\)
Cộng theo vế ba bất đẳng thức trên, ta được: \(VT\le\frac{1}{x}+\frac{1}{y}+\frac{1}{z}\le\frac{1}{\sqrt{2x-1}}+\frac{1}{\sqrt{2y-1}}+\frac{1}{\sqrt{2z-1}}=3\)
Đẳng thức xảy ra khi x = y = z = 1
Ta có: \(x^2\left(y+z\right)\ge x^2.2\sqrt{yz}=2\sqrt{x^4}.\sqrt{\frac{1}{x}}=2x\sqrt{x}\)(Áp dụng BĐT Cô - si cho 2 số dương y,z và sử dụng giả thiết xyz = 1)
Hoàn toàn tương tự: \(y^2\left(z+x\right)\ge2y\sqrt{y};z^2\left(x+y\right)\ge2z\sqrt{z}\)
Do đó \(P=\frac{x^2\left(y+z\right)}{y\sqrt{y}+2z\sqrt{z}}+\frac{y^2\left(z+x\right)}{z\sqrt{z}+2x\sqrt{x}}+\frac{z^2\left(x+y\right)}{x\sqrt{x}+2y\sqrt{y}}\)
\(\ge\frac{2x\sqrt{x}}{y\sqrt{y}+2z\sqrt{z}}+\frac{2y\sqrt{y}}{z\sqrt{z}+2x\sqrt{x}}+\frac{2z\sqrt{z}}{x\sqrt{x}+2y\sqrt{y}}\)
Đặt \(a=x\sqrt{x}+2y\sqrt{y}\), \(b=y\sqrt{y}+2z\sqrt{z}\), \(c=z\sqrt{z}+2x\sqrt{x}\)
Suy ra: \(x\sqrt{x}=\frac{4c+a-2b}{9}\), \(y\sqrt{y}=\frac{4a+b-2c}{9}\), \(z\sqrt{z}=\frac{4b+c-2a}{9}\)
Do đó \(P\ge\frac{2}{9}\left(\frac{4c+a-2b}{b}+\frac{4a+b-2c}{c}+\frac{4b+c-2a}{a}\right)\)
\(=\frac{2}{9}\left[4\left(\frac{c}{b}+\frac{a}{c}+\frac{b}{a}\right)+\left(\frac{a}{b}+\frac{b}{c}+\frac{c}{a}\right)-6\right]\)
\(\ge\frac{2}{9}\left[4.3\sqrt[3]{\frac{c}{b}.\frac{a}{c}.\frac{b}{a}}+3\sqrt[3]{\frac{a}{b}.\frac{b}{c}.\frac{c}{a}}-6\right]\)(Áp dụng BĐT Cô - si cho 3 số dương)
\(=\frac{2}{9}\left[4.3+3-6\right]=2\)
Vậy \(P\ge2\)
Đẳng thức xảy ra khi x = y = z = 1
gt\(\Leftrightarrow\sqrt{xy}+\sqrt{x}+\sqrt{y}+1=9\Leftrightarrow\sqrt{xy}+\sqrt{x}+\sqrt{y}=8\)
Ta có:\(\sqrt{xy}\le\frac{x+y}{2}\Leftrightarrow\left(\sqrt{x}-\sqrt{y}\right)^2\ge0\)(đúng);
\(\sqrt{x}\le\frac{x+4}{4}\Leftrightarrow x-4\sqrt{x}+4\ge0\Leftrightarrow\left(\sqrt{x}-2\right)^2\ge0\)(đúng)
\(\sqrt{y}\le\frac{y+4}{4}\Leftrightarrow\left(\sqrt{y}-2\right)^2\ge0\)(đúng)
Cộng theo vế ba BĐT ta có:\(8\le\sqrt{xy}+\sqrt{x}+\sqrt{y}\le\frac{x+y}{2}+\frac{x+4}{4}+\frac{y+4}{4}=\frac{3\left(x+y\right)}{4}+2\)
\(\Leftrightarrow\frac{3}{4}\left(x+y\right)\ge6\Leftrightarrow x+y\ge8\)
Lại có:\(\frac{x^2}{y}+\frac{y^2}{x}\ge\frac{\left(x+y\right)^2}{y+x}=x+y\ge8\)
Nên GTNN của P là 8 đạt được khi \(x=y=4\)
Ta có: \(\left(\sqrt{x}+1\right)\left(\sqrt{y}+1\right)\ge9\)
\(\Leftrightarrow\sqrt{xy}+\sqrt{x}+\sqrt{y}\ge8\)
Theo bất đẳng thức CÔ-si:
\(8\le\sqrt{xy}+\sqrt{x}+\sqrt{y}\le\frac{x+y}{2}+\frac{x+4}{4}+\frac{y+4}{4}\)
\(\Rightarrow\frac{2x+2y+x+4+y+4}{4}=\frac{3x+3y+8}{4}=\frac{3\left(x+y\right)}{4}+\frac{8}{4}=\frac{3\left(x+y\right)}{4}+2\)
\(\Rightarrow\frac{3\left(x+y\right)}{4}+2\ge8\)
\(\Rightarrow\frac{3\left(x+y\right)}{4}\ge6\)
\(\Rightarrow x+y\ge8\)
Theo BĐT Cô si: \(\hept{\begin{cases}\frac{x^2}{y}+y\ge2x\\\frac{y^2}{x}+x\ge2y\end{cases}\Rightarrow\frac{x^2}{y}+y+\frac{y^2}{x}+x\ge2x+2y}\)
\(\Rightarrow P=\frac{x^2}{y}+\frac{y^2}{x}\ge x+y\ge8\)
Vậy Gía trị nhỏ nhất của P là 8 khi x = y = 4
a,\(A\ge\frac{9}{\sqrt{x}+\sqrt{y}+\sqrt{z}}\ge\frac{9}{\sqrt{3\left(x+y+z\right)}}=3\)=3
MInA=3<=>x=y=z=1
b)dùng cô si đi(đề thi chuyên bình phước năm 2016-2017)
\(ĐKXĐ:x,y,z\ge1\left(x,y,z\inℤ\right)\)
Ta có: \(\left(x+2y\right)^2=\left(\frac{2x+y}{2}+\frac{3y}{2}\right)^2\ge4.\frac{2x+y}{2}.\frac{3y}{2}=3y\left(2x+y\right)\)
\(\Rightarrow\frac{2x+y}{x+2y}\le\frac{x+2y}{3y}\Rightarrow\frac{2x+y}{x\left(x+2y\right)}\le\frac{1}{3}\left(\frac{2}{x}+\frac{1}{y}\right)\)
Tương tự: \(\frac{2y+z}{y\left(y+2x\right)}\le\frac{1}{3}\left(\frac{2}{y}+\frac{1}{z}\right)\);\(\frac{2z+x}{z\left(z+2x\right)}\le\frac{1}{3}\left(\frac{2}{z}+\frac{1}{x}\right)\)
\(\Rightarrow A\le\frac{1}{3}.3\left(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}\right)=\frac{1}{x}+\frac{1}{y}+\frac{1}{z}\)(*)
Ta có: \(\sqrt{2x-1}=\sqrt{\left(2x-1\right).1}\le\frac{2x-1+1}{2}=x\)(BĐT Cô - si)
\(\Rightarrow\frac{1}{x}\le\frac{1}{\sqrt{2x-1}}\)
Tương tự: \(\frac{1}{y}\le\frac{1}{\sqrt{2y-1}}\);\(\frac{1}{z}\le\frac{1}{\sqrt{2z-1}}\)
\(\Rightarrow\frac{1}{x}+\frac{1}{y}+\frac{1}{z}\le\frac{1}{\sqrt{2x-1}}+\frac{1}{\sqrt{2y-1}}+\frac{1}{\sqrt{2z-1}}=3\)(**)
Từ (*) và (**) suy ra \(A=\frac{2x+y}{x\left(x+2y\right)}+\frac{2y+z}{y\left(y+2z\right)}+\frac{2z+x}{z\left(z+2x\right)}\le3\)
Đẳng thức xảy ra khi x = y = z = 1
Từ đẳng thức đã cho suy ra \(x>\frac{1}{2};y>\frac{1}{2};z>\frac{1}{2}\)
Áp dụng\(\left(a+b\right)^2\ge4ab\)ta có \(\left(x+2y\right)^2=\left(\frac{2x+y}{2}+\frac{3y}{2}\right)^2\ge4\cdot\frac{2x+y}{2}\cdot\frac{3y}{2}\)
\(\Rightarrow\left(x+2y\right)^2\ge3y\left(2x+y\right)\)(Dấu "=" xảy ra <=> x=y)
=> \(\frac{2x+y}{x+2y}\le\frac{x+2y}{3y}\Rightarrow\frac{2x+y}{x\left(x+2y\right)}\le\frac{1}{3}\left(\frac{2}{x}+\frac{1}{y}\right)\)
Tương tự \(\hept{\begin{cases}\frac{2y+z}{y\left(y+2z\right)}\le\frac{1}{3}\left(\frac{2}{y}+\frac{1}{z}\right)\\\frac{2z+x}{z\left(z+2x\right)}\le\frac{1}{3}\left(\frac{2}{z}+\frac{1}{x}\right)\end{cases}}\)
=> \(A\le\frac{1}{x}+\frac{1}{y}+\frac{1}{z}\)(Dấu "=" xảy ra <=> x=y=z)
Ta có \(\sqrt{\left(2x-1\right)\cdot1}\le\frac{\left(2x-1\right)+1}{2}\Rightarrow\sqrt{2x-1}\le x\Rightarrow\frac{1}{x}\le\frac{1}{\sqrt{2x-1}}\)
Tương tự \(\hept{\begin{cases}\frac{1}{y}\le\frac{1}{\sqrt{2y-1}}\\\frac{1}{z}\le\frac{1}{\sqrt{2z-1}}\end{cases}}\)
Do đó \(A\le\frac{1}{\sqrt{2x-1}}+\frac{1}{\sqrt{2y-1}}+\frac{1}{\sqrt{2z-1}}=3\)(dấu "=" xảy ra <=> x=y=z=1)
Vậy MaxA=3 đạt được khi x=y=z=1
ĐKXĐ : \(x>\frac{1}{2};y>\frac{1}{2};z>\frac{1}{2}\)
Áp dụng ( a+b)2 \(\ge4ab\)ta có :
( x+ 2y)2 = \(\left(\frac{2x+y}{2}+\frac{3y}{2}\right)^2\ge4.\left(\frac{2x+y}{2}\right).\frac{3y}{2}\)
\(\Rightarrow\left(x+2y\right)^2\ge3y\left(2x+y\right)\)
\(\Rightarrow\frac{2x+y}{x+2y}\le\frac{x+2y}{3y}\)
\(\Rightarrow\frac{2x+y}{x\left(x+2y\right)}\le\frac{1}{3}\left(\frac{2}{x}+\frac{1}{y}\right)\)
Tương tự : \(\frac{2y+z}{y\left(y+2\right)}\le\frac{1}{3}\left(\frac{2}{y}+\frac{1}{z}\right)\)
\(\frac{2z+x}{z.\left(z+2x\right)}\le\frac{1}{3}\left(\frac{2}{z}+\frac{1}{x}\right)\)
=> \(A\le\frac{1}{x}+\frac{1}{y}+\frac{1}{z}\)
Ta có : \(\sqrt{\left(2x-1\right)1}\le\frac{2x-1+1}{2}\)
\(\Rightarrow\sqrt{2x-1}\le x\)
\(\Rightarrow\frac{1}{x}\le\frac{1}{\sqrt{2x-1}}\)
\(\frac{1}{y}\le\frac{1}{\sqrt{2y-1}}\)
\(\frac{1}{z}\le\frac{1}{\sqrt{2z-1}}\)
Do đó
A \(\le\frac{1}{\sqrt{2x-1}}+\frac{1}{\sqrt{2y-1}}+\frac{1}{\sqrt{2z-1}}\)
Vậy Max A = 3 khi x = y = z = 1
Theo Cô-si ta có:
\(3=\frac{1}{\sqrt{2x-1}}+\frac{1}{\sqrt{2y-1}}+\frac{1}{\sqrt{2z-1}}\ge\frac{1}{x}+\frac{1}{y}+\frac{1}{z}\)
\(\Rightarrow\frac{1}{x}+\frac{1}{y}+\frac{1}{z}\le3\)
Xét:
\(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}-\Sigma_{cyc}\frac{2x+y}{x\left(x+2y\right)}=\frac{1}{3}\left[\frac{\left(x-y\right)^2}{xy\left(x+2y\right)}+\frac{\left(y-z\right)^2}{yz\left(y+2z\right)}+\frac{\left(z-x\right)^2}{zx\left(z+2x\right)}\right]\ge0\)
\(\Rightarrow\Sigma_{cyc}\frac{2x+y}{x\left(x+2y\right)}\le3\)
\(2\sqrt{xy}+\sqrt{2x}+\sqrt{2y}\ge8\)
Mà \(\left\{{}\begin{matrix}2\sqrt{xy}\le x+y\\\sqrt{2x}+\sqrt{2y}\le2\sqrt{x+y}\end{matrix}\right.\)
\(\Rightarrow x+y+2\sqrt{x+y}\ge8\)
\(\Leftrightarrow\left(\sqrt{x+y}-2\right)\left(\sqrt{x+y}+4\right)\ge0\)
\(\Rightarrow x+y\ge4\)
\(P=\frac{x^2}{y}+\frac{y^2}{x}+\frac{1}{x}+\frac{1}{y}\ge x+y+\frac{4}{x+y}\)
\(P\ge\frac{x+y}{4}+\frac{4}{x+y}+\frac{3\left(x+y\right)}{4}\ge2\sqrt{\frac{4\left(x+y\right)}{4\left(x+y\right)}}+\frac{3.4}{4}=5\)
Dấu "=" xảy ra khi \(x=y=2\)