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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 :\(\dfrac{1}{x}+\dfrac{1}{y}+\dfrac{1}{z}=1\Leftrightarrow\dfrac{1}{x}=\dfrac{1}{2}-\dfrac{1}{y}+\dfrac{1}{2}-\dfrac{1}{z}\Leftrightarrow\dfrac{1}{x}=\dfrac{y-2}{2y}+\dfrac{z-2}{2z}\)
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 :\(x=y=z=3\)
Chỉ đúng trong trường hợp các số thực dương (kì lạ là các bạn rất thích quên điều kiện này khi đăng đề lên)
a/ \(\frac{a^3}{b^2}+a\ge2\sqrt{\frac{a^4}{b^2}}=\frac{2a^2}{b}\) ; \(\frac{b^3}{c^2}+b\ge\frac{2b^2}{c}\); \(\frac{c^3}{a^2}+c\ge\frac{2c^2}{a}\)
Cộng vế với vế:
\(VT+a+b+c\ge2VP\Rightarrow VT\ge2VP-\left(a+b+c\right)\)
Mà \(2VP=\frac{a^2}{b}+\frac{b^2}{c}+\frac{c^2}{a}+\frac{a^2}{b}+\frac{b^2}{c}+\frac{c^2}{a}\ge\frac{a^2}{b}+\frac{b^2}{c}+\frac{c^2}{a}+\frac{\left(a+b+c\right)^2}{a+b+c}\)
\(\Rightarrow2VP\ge VP+a+b+c\)
\(\Rightarrow2VP-\left(a+b+c\right)\ge VP\)
\(\Rightarrow VT\ge VP\)
Dấu "=" xảy ra khi \(a=b=c\)
Câu dưới tương tự:
\(\frac{a^5}{b^3}+a^2+a^2\ge\frac{3a^3}{b}\) , làm tương tự với 2 cái còn lại và cộng lại:
\(\Rightarrow VT+2\left(a^2+b^2+c^2\right)\ge3\left(\frac{a^3}{b}+\frac{b^3}{c}+\frac{c^3}{a}\right)=3\left(\frac{a^4}{ab}+\frac{b^4}{ca}+\frac{c^4}{ab}\right)\ge\frac{3\left(a^2+b^2+c^2\right)^2}{ab+bc+ca}\ge3\left(a^2+b^2+c^2\right)\)
\(\Rightarrow VT\ge a^2+b^2+c^2\)
Dấu "=" xảy ra khi \(a=b=c\)
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 )
\(\Rightarrow\)\(P< \frac{3}{4}\)
\(\pi< a< \frac{3\pi}{2}\Rightarrow\left\{{}\begin{matrix}sina< 0\\cosa< 0\end{matrix}\right.\) \(\Rightarrow sin2a=2sina.cosa>0\)
\(\Rightarrow sin2a=\sqrt{1-cos^22a}=\frac{3\sqrt{7}}{8}\)
\(cos2a=1-2sin^2a=\frac{1}{8}\)
\(\Leftrightarrow sin^2a=\frac{7}{16}\Rightarrow sina=-\frac{\sqrt{7}}{4}\)
\(\Rightarrow M=\frac{-\frac{\sqrt{7}}{4}-\frac{3\sqrt{7}}{8}}{-\frac{\sqrt{7}}{4}+\frac{3\sqrt{7}}{8}}=...\)
\(sinx\left(1-tan^2\frac{x}{2}\right)=sinx\left(1-\frac{sin^2\frac{x}{2}}{cos^2\frac{x}{2}}\right)=sinx\left(1-\frac{1-cosx}{1+cosx}\right)\)
\(=sinx\left(\frac{1+cosx-\left(1-cosx\right)}{1+cosx}\right)=\frac{2sinx.cosx}{1+cosx}\)
\(1-sin2x.sin3x-cos2x.cos3x=1-\left(cos3x.cos2x+sin3x.sin2x\right)=1-cos\left(3x-2x\right)=1-cosx\)
\(\Rightarrow\frac{1-sin2x.sin3x-cos2x.cos3x}{sinx\left(1-tan^2\frac{x}{2}\right)}=\frac{1-cosx}{\frac{2sinx.cosx}{1+cosx}}=\frac{\left(1-cosx\right)\left(1+cosx\right)}{2sinx.cosx}\)
\(=\frac{1-cos^2x}{2sinx.cosx}=\frac{sin^2x}{2sinx.cosx}=\frac{sinx}{2cosx}=\frac{1}{2}tanx\)
\(\frac{3\pi}{2}< a< 2\pi\Rightarrow sina< 0\) \(\Rightarrow sina=-\sqrt{1-cos^2a}=\frac{-2\sqrt{2}}{3}\)
\(tan\left(a-\frac{\pi}{3}\right)=\frac{sin\left(a-\frac{\pi}{3}\right)}{cos\left(a-\frac{\pi}{3}\right)}=\frac{sina.cos\frac{\pi}{3}-cosa.sin\frac{\pi}{3}}{cosa.cos\frac{\pi}{3}+sina.sin\frac{\pi}{3}}=\frac{\frac{-2\sqrt{2}}{3}.\frac{1}{2}-\frac{1}{3}.\frac{\sqrt{3}}{2}}{\frac{1}{3}.\frac{1}{2}-\frac{2\sqrt{2}}{3}.\frac{\sqrt{3}}{2}}=\frac{9\sqrt{3}+8\sqrt{2}}{23}\)
\(\frac{1}{25}\)<1