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1/1.2 + 1/2.3 + 1/3.4 + .......................+ 1/99.100
= 1 - 1/2 + 1/2 - 1/3 +1/3 - 1/4 +..................+ 1/99 - 1/100
= 1 - 1/100
= 99/100
1/1.2 + 1/2.3 + 1/3.4 + ... + 1/99.100 = 1 - 1/2 + 1/2 - 1/3 + 1/3 - 1/4 + ... + 1/99 - 1/100
= 1 - 1/100
= 99/100
Ma 99/100 < 1.
=> 1/1.2 + 1/2.3 + 1/3.4 + ... + 1/99.100 < 1 (dccm)
Đặt P = ...
* Chứng minh P > 1/2 :
\(P\ge\frac{\left(1+1+1+...+1\right)^2}{n+1+n+2+n+3+...+n+n}\)
Từ \(n+1\) đến \(n+n\) có n số => tổng \(\left(n+1\right)+\left(n+2\right)+\left(n+3\right)+...+\left(n+n\right)\) là:
\(\frac{n\left(n+n+n+1\right)}{2}=\frac{n\left(3n+1\right)}{2}\)
\(\Rightarrow\)\(P\ge\frac{n^2}{\frac{n\left(3n+1\right)}{2}}=\frac{2n}{3n+1}\)
Mà \(n>1\)\(\Leftrightarrow\)\(4n>3n+1\)\(\Leftrightarrow\)\(\frac{n}{3n+1}>\frac{1}{2}\)
\(\Rightarrow\)\(P>\frac{1}{2}\)
* Chứng minh P < 3/4 :
Có: \(\frac{1}{n+1}\le\frac{1}{4}\left(\frac{1}{n}+1\right)\)
\(\frac{1}{n+2}\le\frac{1}{4}\left(\frac{1}{n}+\frac{1}{2}\right)\)
\(\frac{1}{n+3}\le\frac{1}{4}\left(\frac{1}{n}+\frac{1}{3}\right)\)
...
\(\frac{1}{n+n}=\frac{1}{2n}=\frac{1}{4}\left(\frac{1}{n}+\frac{1}{n}\right)\)
\(\Rightarrow\)\(P\le\frac{1}{4}\left(\frac{1}{n}+1+\frac{1}{n}+\frac{1}{2}+\frac{1}{n}+\frac{1}{3}+...+\frac{1}{n}+\frac{1}{n}\right)\)
\(\Leftrightarrow\)\(P\le\frac{1}{4}\left(\frac{1}{n}+\frac{1}{n}+\frac{1}{n}+...+\frac{1}{n}\right)+\frac{1}{4}\left(1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{n}\right)\)
\(\Leftrightarrow\)\(P\le\frac{1}{4}\left(n.\frac{1}{n}\right)+\frac{1}{4}\left(1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{n}\right)< \frac{1}{4}+\frac{1}{4}=\frac{2}{4}< \frac{3}{4}\) ( do n>1 )
\(\Rightarrow\)\(P< \frac{3}{4}\)
Ta có:\(\frac{1}{2^2}\)+\(\frac{1}{3^2}\)+\(\frac{1}{4^2}\)+...+\(\frac{1}{50^2}\)<\(\frac{1}{1\cdot2}\)+\(\frac{1}{2\cdot3}\)+\(\frac{1}{3\cdot4}\)+...+\(\frac{1}{49\cdot50}\)
<1-\(\frac{1}{2}\)+\(\frac{1}{2}\)-\(\frac{1}{3}\)+\(\frac{1}{3}\)-\(\frac{1}{4}\)+...+\(\frac{1}{49}\)-\(\frac{1}{50}\)
<1-\(\frac{1}{50}\)<1
Nên \(\frac{1}{2^2}\)+\(\frac{1}{3^2}\)+\(\frac{1}{4^2}\)+...+\(\frac{1}{50^2}\)<1
Ta có: \(\frac{1}{2^2}< \frac{1}{1.2};\frac{1}{3^2}< \frac{1}{2.3};.....;\frac{1}{50^2}< \frac{1}{49.50}\)
\(\rightarrow\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+....+\frac{1}{50^2}< \frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+....+\frac{1}{49.50}=S\)
Đặt S = \(\frac{1}{1.2}+\frac{1}{2.3}+....+\frac{1}{49.50}\)
Ta lại có: \(\frac{1}{1.2}=\frac{1}{1}-\frac{1}{2};\frac{1}{2.3}=\frac{1}{2}-\frac{1}{3};....;\frac{1}{49.50}=\frac{1}{49}-\frac{1}{50}\)
\(S=\frac{1}{1.2}+\frac{1}{2.3}+....+\frac{1}{49.50}=\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-....-\frac{1}{50}=\frac{49}{50}\)
\(\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+....+\frac{1}{50^2}< S=\frac{49}{50}< 1\)
Vậy \(\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{50^2}< 1\) (đpcm)
Ta có:
\(A=\frac{1^2}{1.2}.\frac{2^2}{2.3}.\frac{3^2}{3.4}.\frac{4^2}{4.5}=\frac{1.1}{1.2}.\frac{2.2}{2.3}.\frac{3.3}{3.4}.\frac{4.4}{4.5}=\frac{1.1.2.2.3.3.4.4}{1.2.2.3.3.4.4.5}=\frac{1}{5}\)
Ta có :\(\dfrac{1}{x}+\dfrac{1}{y}+\dfrac{1}{z}=1\Leftrightarrow\dfrac{1}{x}=\dfrac{1}{2}-\dfrac{1}{y}+\dfrac{1}{2}-\dfrac{1}{z}\Leftrightarrow\dfrac{1}{x}=\dfrac{y-2}{2y}+\dfrac{z-2}{2z}\)
Áp dụng bất đẳng thức cô si ta có :\(\dfrac{y-2}{2y}+\dfrac{z-2}{2z}\ge2\sqrt{\dfrac{\left(y-2\right)\left(z-2\right)}{4yz}}=\dfrac{\sqrt{\left(y-2\right)\left(z-2\right)}}{\sqrt{yz}}\)
\(\Rightarrow\)\(\dfrac{1}{x}\ge\dfrac{\sqrt{\left(y-2\right)\left(z-2\right)}}{\sqrt{yz}}\) (1)
Chứng minh tương tự :\(\dfrac{1}{y}\ge\dfrac{\sqrt{\left(x-2\right)\left(z-2\right)}}{\sqrt{xz}}\) (2)
\(\dfrac{1}{z}\ge\dfrac{\sqrt{\left(x-2\right)\left(y-2\right)}}{\sqrt{xy}}\) (3)
Nhân 3 bất đẳng thức (1),(2) và (3) vế theo vế ta được :
\(\dfrac{1}{xyz}\ge\dfrac{\left(x-2\right)\left(y-2\right)\left(z-2\right)}{xyz}\)
\(\Rightarrow\left(x-2\right)\left(y-2\right)\left(z-2\right)\le1\)
Dấu "=" xảy ra khi :\(x=y=z=3\)
Bổ đề 1: Với m, n < 1 ta có bất đẳng thức:
\(\frac{1}{1+m^2}+\frac{1}{1+n^2}\le\frac{2}{1+mn}\).
Thật vậy, bất đẳng thức trên tương đương với: \(\left(mn-1\right)\left(m-n\right)^2\le0\) (luôn đúng).
Bổ đề 2: Với m, n, p < 1 ta có bất đẳng thức:
\(\frac{1}{1+m^3}+\frac{1}{1+n^3}+\frac{1}{1+p^3}\le\frac{3}{1+mnp}\left(2\right)\).
Thật vậy, áp dụng bổ đề (1) ta có:
\(VT_{\left(2\right)}=\left(\frac{1}{1+m^3}+\frac{1}{1+n^3}\right)+\left(\frac{1}{1+p^3}+\frac{1}{1+mnp}\right)-\frac{1}{1+mnp}\le\frac{2}{1+\sqrt{m^3n^3}}+\frac{2}{1+\sqrt{mnp^4}}-\frac{1}{1+mnp}\le\frac{4}{1+\sqrt[4]{m^3n^3.mnp^4}}-\frac{1}{1+mnp}=\frac{4}{1+mnp}-\frac{1}{1+mnp}=\frac{3}{1+mnp}\left(đpcm\right)\).
Quay trở lại bài toán.
Đặt \(\left(\sqrt[3]{a},\sqrt[3]{b},\sqrt[3]{c}\right)=\left(x,y,z\right)\). Ta có: \(0< x,y,z< 1\).
BĐT cần chứng minh trở thành:
\(\frac{1}{1+x^3+y^3}+\frac{1}{1+y^3+z^3}+\frac{1}{1+z^3+x^3}\le\frac{3}{1+2xyz}\left(1\right)\).
Áp dụng BĐT AM - GM và bổ đề 2 ta có: \(VT_{\left(1\right)}\le\frac{1}{1+\left(\sqrt[3]{2}\sqrt{xy}\right)^3}+\frac{1}{1+\left(\sqrt[3]{2}\sqrt{yz}\right)^3}+\frac{1}{1+\left(\sqrt[3]{2}\sqrt{zx}\right)^3}\le\frac{3}{1+\sqrt[3]{2}\sqrt[3]{2}\sqrt[3]{2}\sqrt{xy.yz.zx}}=\frac{3}{1+2xyz}=VP_{\left(1\right)}\left(đpcm\right)\)
Bạn bổ sung cho mình thêm điều kiện ở hai bổ đề:
Bổ đề 1: Thêm m, n > 0.
Bổ đề 2: Thêm m, n, p > 0.
bđt \(\Leftrightarrow\)\(\Sigma_{cyc}\frac{a^2}{2}+\Sigma_{cyc}\frac{a}{bc}\ge\frac{9}{2}\)
mặt khác: \(\Sigma_{cyc}\frac{a}{bc}=\frac{1}{2}\Sigma_{cyc}\left(\frac{b}{ca}+\frac{c}{ab}\right)\ge\Sigma\frac{1}{a}\)\(\Rightarrow\)\(\Sigma_{cyc}\frac{a}{bc}\ge\Sigma_{cyc}\frac{1}{a}\)
do đó cần CM: \(\Sigma_{cyc}\frac{a^2}{2}+\Sigma_{cyc}\frac{1}{a}\ge\frac{9}{2}\) (1)
\(VT_{\left(1\right)}=\Sigma_{cyc}\left(\frac{a^2}{2}+\frac{1}{2a}+\frac{1}{2a}\right)\ge3.\frac{3}{2}=\frac{9}{2}\)
"=" \(\Leftrightarrow\)\(a=b=c=1\)
sửa: chứng minh \(\frac{1}{1+ab}+\frac{1}{1+bc}+\frac{1}{1+ca}\ge\frac{3}{2}\)
áp dụng bđt Cauchy ta có
\(\frac{1}{1+ab}=1-\frac{1}{1+ab}\ge1-\frac{ab}{2\sqrt{ab}}=1-\frac{\sqrt{ab}}{2}\)
tương tự ta có \(\hept{\begin{cases}\frac{1}{1+bc}\ge1-\frac{\sqrt{bc}}{2}\\\frac{1}{1+ca}\ge1-\frac{\sqrt{ca}}{2}\end{cases}}\)
cộng theo vế các bđt trên và áp dụng bđt Cauchy ta được
\(\frac{1}{1+ab}+\frac{1}{1+bc}+\frac{1}{1+ac}\ge3-\frac{1}{2}\left(\sqrt{ab}+\sqrt{bc}+\sqrt{ca}\right)\)
\(\ge3-\frac{1}{2}\left(\frac{a+b}{2}+\frac{b+c}{2}+\frac{c+a}{2}\right)=3-\frac{a+b+c}{2}\ge3-\frac{3}{2}=\frac{3}{2}\)
dấu "=" xảy ra khi \(\hept{\begin{cases}1+ab=1+bc=1+ca\\a=b=c\\a+b+c=3\end{cases}\Leftrightarrow a=b=c=1}\)
Đặt \(A=\frac{1}{2.3}+\frac{1}{4.5}+\frac{1}{6.7}+...+\frac{1}{100.101}\)
\(\Rightarrow A< \left(\frac{1}{2.3}+\frac{1}{4.5}+\frac{1}{6.7}+...+\frac{1}{100.101}\right)+\left(\frac{1}{3.4}+\frac{1}{5.6}+\frac{1}{7.8}+...+\frac{1}{99.100}\right)\)
\(\Rightarrow A< \frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+\frac{1}{5.6}+\frac{1}{6.7}+...+\frac{1}{99.100}+\frac{1}{100.101}\)
\(\Rightarrow A< \frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+\frac{1}{5}-\frac{1}{6}+\frac{1}{6}-\frac{1}{7}+...+\frac{1}{99}-\frac{1}{100}+\frac{1}{100}-\frac{1}{101}\)
\(\Rightarrow A< \frac{1}{2}-\frac{1}{101}< \frac{1}{2}\)
\(\Rightarrow A< \frac{1}{2}\)
Vậy \(\frac{1}{2.3}+\frac{1}{4.5}+\frac{1}{6.7}+...+\frac{1}{100.101}< 2\) (đpcm)
Mai ơi, bài này thầy dạy hôm chiều cậu nghỉ đó