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\(ĐKXĐ:x,y,z\ge1\left(x,y,z\inℤ\right)\)
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:
\(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}-\Sigma_{cyc}\frac{2x+y}{x\left(x+2y\right)}=\frac{1}{3}\left[\frac{\left(x-y\right)^2}{xy\left(x+2y\right)}+\frac{\left(y-z\right)^2}{yz\left(y+2z\right)}+\frac{\left(z-x\right)^2}{zx\left(z+2x\right)}\right]\ge0\)
\(\Rightarrow\Sigma_{cyc}\frac{2x+y}{x\left(x+2y\right)}\le3\)
\(ĐK:x,y,z>\frac{1}{2}\)
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
\(P=\frac{1}{x+x+y+z}+\frac{1}{x+y+y+z}+\frac{1}{x+y+z+z}\)
\(P\le\frac{1}{16}\left(\frac{1}{x}+\frac{1}{x}+\frac{1}{y}+\frac{1}{z}+\frac{1}{x}+\frac{1}{y}+\frac{1}{y}+\frac{1}{z}+\frac{1}{x}+\frac{1}{y}+\frac{1}{z}+\frac{1}{z}\right)\)
\(P\le\frac{1}{4}\left(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}\right)=\frac{1007}{2}\)
\(P_{max}=\frac{1007}{2}\) khi \(x=y=z=\frac{3}{2014}\)
\(\frac{16}{2x+y+z}=\frac{16}{x+x+y+z}\le\frac{1}{x}+\frac{1}{x}+\frac{1}{y}+\frac{1}{z}=\frac{2}{x}+\frac{1}{y}+\frac{1}{z}\)
Tương tự:
\(\frac{16}{x+2y+z}\le\frac{1}{x}+\frac{2}{y}+\frac{1}{z};\frac{16}{x+y+2z}\le\frac{1}{x}+\frac{1}{y}+\frac{2}{z}\)
Cộng lại:
\(16P\le4\left(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}\right)=16\Rightarrow P\le1\)
dấu "=" xảy ra tại \(x=y=z=\frac{3}{4}\)
\(\frac{1}{x}+\frac{1}{y}\ge\frac{4}{x+y}\Rightarrow2\left(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}\right)\ge4\left(\frac{1}{x+y}+\frac{1}{y+z}+\frac{1}{z+x}\right)\)
\(\Leftrightarrow2\left(\frac{1}{x+y}+\frac{1}{y+z}+\frac{1}{z+x}\right)\le1\)
\(\frac{1}{x+y}+\frac{1}{y+z}\ge\frac{4}{2x+y+z}\Rightarrow2\left(\frac{1}{x+y}+\frac{1}{y+z}+\frac{1}{z+x}\right)\ge4\left(\frac{1}{2x+y+z}+\frac{1}{2y+z+x}+\frac{1}{2z+x+y}\right)\)
\(4M\le1\Leftrightarrow M\le\frac{1}{4}\) \(M=\frac{1}{4}\Leftrightarrow x=y=z=3\)
a/ \(\frac{2x+1}{\sqrt{x^2+2}}+\left(x+1\right)\left(\sqrt{1+\frac{2x+1}{x^2+2}}-1\right)+2x+1=0\)
\(\Leftrightarrow\frac{2x+1}{\sqrt{x^2+2}}+\frac{\left(x+1\right)\left(2x+1\right)}{\sqrt{1+\frac{2x+1}{x^2+2}}+1}+2x+1=0\)
\(\Leftrightarrow\left(2x+1\right)\left(\frac{1}{\sqrt{x^2+2}}+\frac{x+1}{\sqrt{1+\frac{2x+1}{x^2+2}}+1}+1\right)=0\)
\(\Rightarrow x=-\frac{1}{2}\)
b/ \(Q\ge\frac{\left(x+y+z\right)^2}{xyz\left(x+y+z\right)}+\frac{\left(x^3+y^3+z^3\right)^2}{xy+yz+zx}\ge\frac{x+y+z}{xyz}+\frac{\left(x^2+y^2+z^2\right)^3}{\left(x+y+z\right)^2}\)
\(Q\ge\frac{27\left(x+y+z\right)}{\left(x+y+z\right)^3}+\frac{\left(x+y+z\right)^6}{27\left(x+y+z\right)^2}=\frac{27}{\left(x+y+z\right)^2}+\frac{\left(x+y+z\right)^4}{27}\)
\(Q\ge\frac{27}{64\left(x+y+z\right)^2}+\frac{27}{64\left(x+y+z\right)^2}+\frac{\left(x+y+z\right)^4}{27}+\frac{837}{32\left(x+y+z\right)^2}\)
\(Q\ge3\sqrt[3]{\frac{27^2\left(x+y+z\right)^4}{64^2.27\left(x+y+z\right)^4}}+\frac{837}{32.\left(\frac{3}{2}\right)^2}=\frac{195}{16}\)
"=" \(\Leftrightarrow x=y=z=\frac{1}{2}\)
Nguyễn Trúc Giang, Duy Khang, Vũ Minh Tuấn, Võ Hồng Phúc, tth, No choice teen, Phạm Lan Hương,
Nguyễn Lê Phước Thịnh, @Nguyễn Việt Lâm, @Akai Haruma
giúp em vs ạ! Cần trước 5h chiều nay ạ
Thanks nhiều
\(\left(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}\right)^2\le3\left(\frac{1}{x^2}+\frac{1}{y^2}+\frac{1}{z^2}\right)\Rightarrow\frac{1}{x}+\frac{1}{y}+\frac{1}{z}\le\sqrt{3}\)
\(P=\frac{1}{x+x+y}+\frac{1}{y+y+z}+\frac{1}{x+z+z}\)
\(P\le\frac{1}{9}\left(\frac{1}{x}+\frac{1}{x}+\frac{1}{y}+\frac{1}{y}+\frac{1}{y}+\frac{1}{z}+\frac{1}{x}+\frac{1}{z}+\frac{1}{z}\right)=\frac{1}{3}\left(\frac{1}{x}+\frac{1}{y}+\frac{1}{z}\right)=\frac{\sqrt{3}}{3}\)
\(\Rightarrow P_{max}=\frac{\sqrt{3}}{3}\) khi \(x=y=z=\sqrt{3}\)