CMR: \(\left(1+\dfrac{1}{m}\right)^m\)<\(\left(1+\dfrac{1}{n}\right)^n\) với m<n và m,nϵN*
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20 tháng 2 2018
M=\(\dfrac{1}{2}+\left(\dfrac{1}{2}\right)^2+\left(\dfrac{1}{2}\right)^3+.....+\left(\dfrac{1}{2}\right)^{100}\)
ta có:
2M=\(1+\dfrac{1}{2}+\dfrac{1}{2^2}+...+\dfrac{1}{2^{99}}\)
=>2M-M=1-\(\dfrac{1}{2^{100}}\)
=>M=1-\(\dfrac{1}{2^{100}}\)<1(đpcm)
chúc bạn học tốt ^ ^
AH
Akai Haruma
Giáo viên
23 tháng 5 2021
Lời giải:
\(M=\frac{1.2.3.4.5.6.7...(2n-1)}{2.4.6...(2n-2).(n+1)(n+2)....2n}=\frac{(2n-1)!}{2.1.2.2.2.3...2(n-1).(n+1).(n+2)...2n}\)
\(=\frac{(2n-1)!}{2^{n-1}.1.2...(n-1).(n+1).(n+2)....2n}=\frac{(2n-1)!}{2^{n-1}.1.2...(n-1).n(n+1)..(2n-1).2}\)
\(=\frac{(2n-1)!}{2^{n-1}.(2n-1)!.2}=\frac{1}{2^{n-1}.2}<\frac{1}{2^{n-1}}\)
Ta có đpcm.
NV
Nguyễn Việt Lâm
Giáo viên
25 tháng 3 2023
\(1-\dfrac{3}{n\left(n+2\right)}=\dfrac{n\left(n+2\right)-3}{n\left(n+2\right)}=\dfrac{\left(n-1\right)\left(n+3\right)}{n\left(n+2\right)}\)
\(\Rightarrow M=\dfrac{1.5}{2.4}.\dfrac{2.6}{3.5}.\dfrac{3.7}{4.6}...\dfrac{\left(n-1\right)\left(n+3\right)}{n\left(n+2\right)}\)
\(=\dfrac{1.2.3...\left(n-1\right)}{2.3.4...n}.\dfrac{5.6.7...\left(n+3\right)}{4.5.6...\left(n+2\right)}\)
\(=\dfrac{1}{n}.\dfrac{n+3}{4}=\dfrac{n+3}{4n}=\dfrac{1}{4}+\dfrac{3}{4n}>\dfrac{1}{4}\) (đpcm)
16 tháng 5 2018
1)\(2a^4+1\ge2a^3+a^2\)
\(\Leftrightarrow2a^4-2a^3-a^2+1\ge0\)
\(\Leftrightarrow\left(a^4-2a^3+a^2\right)+\left(a^4-2a^2+1\right)\ge0\)
\(\Leftrightarrow\left(a^2-a\right)^2+\left(a^2-1\right)^2\ge0\)(luôn đúng)
"="<=>a=1
16 tháng 5 2018
Ta có:\(2A=\dfrac{2}{1\cdot3}+\dfrac{2}{3\cdot5}+...+\dfrac{2}{9\cdot11}\)
\(2A=1-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{5}+...+\dfrac{1}{9}-\dfrac{1}{11}\)
\(2A=1-\dfrac{1}{11}=\dfrac{10}{11}\)
\(B=\left(1+\dfrac{1}{1\cdot3}\right)\left(1+\dfrac{1}{2\cdot4}\right)\cdot...\cdot\left(1+\dfrac{1}{9\cdot11}\right)\)
\(B=\dfrac{4}{1\cdot3}\cdot\dfrac{9}{2\cdot4}\cdot...\cdot\dfrac{100}{9\cdot11}\)
\(B=\dfrac{2\cdot2\cdot3\cdot3\cdot...\cdot10\cdot10}{1\cdot3\cdot2\cdot4\cdot...\cdot9\cdot11}\)
\(B=\dfrac{20}{11}\)
\(\Rightarrow11< 2x< 20\)
\(\Rightarrow x\in\left\{6;7;8;9\right\}\)
11 tháng 4 2017
Áp dụng BĐT AM-GM ta có:
\(\dfrac{a^3}{b\left(c+1\right)}+\dfrac{c+1}{4}+\dfrac{b}{2}\ge3\sqrt[3]{\dfrac{a^3}{b\left(c+1\right)}\cdot\dfrac{c+1}{4}\cdot\dfrac{b}{2}}\)
\(=3\sqrt[3]{\dfrac{a^3}{4\cdot2}\cdot\dfrac{c+1}{c+1}\cdot\dfrac{b}{b}}=3\sqrt[3]{\dfrac{a^3}{8}}=\dfrac{3a}{2}\)
Tương tự cho 2 BĐT còn lại ta cũng có:
\(\dfrac{b^3}{c\left(a+1\right)}\ge\dfrac{3b}{2};\dfrac{c^3}{a\left(b+1\right)}\ge\dfrac{3c}{2}\)
Cộng theo vế 3 BĐT trên ta có:
\(VT+\dfrac{a+b+c+3}{4}+\dfrac{a+b+c}{2}\ge\dfrac{3a+3b+3c}{2}\)
\(\Leftrightarrow VT+\dfrac{3\left(a+b+c\right)}{4}+\dfrac{3}{4}\ge\dfrac{3\left(a+b+c\right)}{2}\)
\(\Leftrightarrow VT+\dfrac{3}{4}\ge\dfrac{3\left(a+b+c\right)}{4}\). Mà theo AM-GM ta có:
\(a+b+c\ge3\sqrt[3]{abc}=3\)\(\Rightarrow VT+\dfrac{3}{4}\ge\dfrac{9}{4}\Rightarrow VT\ge\dfrac{3}{2}=VP\)
Đẳng thức xảy ra khi \(a=b=c=1\)
AT
20 tháng 7 2018
fix đề: CMR:\(\dfrac{x^3}{\left(1+y\right)\left(1+z\right)}+\dfrac{y^3}{\left(1+z\right)\left(1+x\right)}+\dfrac{z^3}{\left(1+y\right)\left(1+x\right)}\)
Áp dụng AM-GM có:
\(\dfrac{x^3}{\left(1+y\right)\left(1+z\right)}+\dfrac{1+y}{8}+\dfrac{1+z}{8}\ge3\sqrt[3]{\dfrac{x^3\left(1+y\right)\left(1+z\right)}{8\cdot8\cdot\left(1+y\right)\left(1+z\right)}}=3\sqrt[3]{\dfrac{x^3}{64}}=\dfrac{3x}{4}\)
Tương tự ta có: \(\left\{{}\begin{matrix}\dfrac{y^3}{\left(1+z\right)\left(1+x\right)}+\dfrac{1+z}{8}+\dfrac{1+x}{8}\ge\dfrac{3y}{4}\\\dfrac{z^3}{\left(1+y\right)\left(1+x\right)}+\dfrac{1+y}{8}+\dfrac{1+x}{8}\ge\dfrac{3z}{4}\end{matrix}\right.\)
Cộng theo về các BĐT trên ta được:
\(\dfrac{x^3}{\left(1+y\right)\left(1+z\right)}+\dfrac{y^3}{\left(1+z\right)\left(1+x\right)}+\dfrac{z^3}{\left(1+y\right)\left(1+x\right)}+\dfrac{3+x+y+z}{4}\ge\dfrac{3\left(x+y+z\right)}{4}\)
\(\Rightarrow\dfrac{x^3}{\left(1+y\right)\left(1+z\right)}+\dfrac{y^3}{\left(1+z\right)\left(1+x\right)}+\dfrac{z^3}{\left(1+y\right)\left(1+x\right)}\ge\dfrac{3x+3y+3z-x-y-z-3}{4}=\dfrac{2\left(x+y+z\right)-3}{4}\)
\(\Rightarrow\dfrac{x^3}{\left(1+y\right)\left(1+z\right)}+\dfrac{y^3}{\left(1+z\right)\left(1+x\right)}+\dfrac{z^3}{\left(1+y\right)\left(1+x\right)}\ge\dfrac{2\cdot3\sqrt[3]{xyz}-3}{4}=\dfrac{2\cdot3-3}{4}=\dfrac{3}{4}\)
-> Đpcm
Dấu ''='' xảy ra khi x = y = z = 1
AH
Akai Haruma
Giáo viên
10 tháng 3 2018
Lời giải:
Áp dụng BĐT AM-GM cho các số dương ta có:
\(\frac{a^3}{(a+1)(b+1)}+\frac{a+1}{8}+\frac{b+1}{8}\geq 3\sqrt[3]{\frac{a^3}{64}}=\frac{3a}{4}\)
\(\frac{b^3}{(b+1)(c+1)}+\frac{b+1}{8}+\frac{c+1}{8}\geq 3\sqrt[3]{\frac{b^3}{64}}=\frac{3b}{4}\)
\(\frac{c^3}{(c+1)(a+1)}+\frac{c+1}{8}+\frac{a+1}{8}\geq 3\sqrt[3]{\frac{c^3}{64}}=\frac{3c}{4}\)
Cộng theo vế:
\(\Rightarrow \frac{a^3}{(a+1)(b+1)}+\frac{b^3}{(b+1)(c+1)}+\frac{c^3}{(c+1)(a+1)}+\frac{a+b+c+3}{4}\geq \frac{3}{4}(a+b+c)\)
\(\Leftrightarrow \frac{a^3}{(a+1)(b+1)}+\frac{b^3}{(b+1)(c+1)}+\frac{c^3}{(c+1)(a+1)}+\frac{3}{2}\geq \frac{9}{4}\)
\(\Leftrightarrow \frac{a^3}{(a+1)(b+1)}+\frac{b^3}{(b+1)(c+1)}+\frac{c^3}{(c+1)(a+1)}\geq \frac{3}{4}\) (đpcm)
Dấu bằng xảy ra khi \(a=b=c=1\)