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Áp dụng bất đẳng thức Cô - si với n số dương ta được
\(a_1+a_2+...+a_n\ge n\sqrt[n]{a_1.a_2....a_n}\)
\(\frac{1}{a_1}+\frac{1}{a_2}+...+\frac{1}{a_n}\ge n\sqrt[n]{\frac{1}{a_1}.\frac{1}{a_2}....\frac{1}{a_n}}\)
Suy ra \(\left(a_1+a_2+...+a_n\right)\left(\frac{1}{a_1}+\frac{1}{a_2}+...+\frac{1}{a_n}\right)\ge n^2.\sqrt[n]{1}=n^2\)
(dấu "=" xẩy ra <=> a1=a2 =...=an)
Theo bat dang thuc cauchy ta co
a1+a2+...+an lon hon hoc bang n.can bac n cua (a1.a2....an) (1)
1/a1+1/a2...1/an lon hon hoac bang n.1/can bac n cua (a1.a2...an) (2)
Nhan 2 ve (1) va (2) ta duoc
(a1+a2+...+an).(1/a1+1/a2+...1/an) lon hon hoac bang n tren 2
=>1/a1+1/a2+...1/an lon hon hoac bang n tren 2/a1+a2+...+an
Dau bang xay ra khi a1=a2=...=an
Mk giai co hieu ko
ÁP DỤNG BĐT Cauchy ta có :
\(\text{a}_1+\text{a}_2+...+\text{a}_n\ge n^n\sqrt{\text{a}_1.\text{a}_2....\text{a}_n}\) (1)
\(\frac{1}{\text{a}_1}+\frac{1}{\text{a}_2}+...+\frac{1}{\text{a}_n}\ge n^n\sqrt{\frac{1}{\text{a}_1}\cdot\frac{1}{\text{a}_2}\cdot...\cdot\frac{1}{\text{a}_n}}\)(2)
Nhân (1) và (2) vế với vế tương ứng ta có được BĐT (*)
Đẳng thức xảy ra \(\Leftrightarrow\hept{\begin{cases}\text{a}_1=\text{a}_2=...=\text{a}_n\\\frac{1}{\text{a}_1}=\frac{1}{\text{a}_2}=...=\frac{1}{\text{a}_n}\end{cases}}\)
\(\Leftrightarrow\text{a}_1=\text{a}_2=...=\text{a}_n\)
\(a_n=\frac{2}{\left(2n+1\right)\left(\sqrt{n}+\sqrt{n+1}\right)}=\frac{2\left(\sqrt{n+1}-\sqrt{n}\right)}{\left(2n+1\right)\left(n+1-n\right)}=\frac{2\left(\sqrt{n+1}-\sqrt{n}\right)}{n+n+1}\)
\(< \frac{\sqrt{n+1}-\sqrt{n}}{\sqrt{n\left(n+1\right)}}=\frac{1}{\sqrt{n}}-\frac{1}{\sqrt{n+1}}\)
\(a_1+a_2+a_3+...+a_{2009}< 1-\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{2}}-\frac{1}{\sqrt{3}}+...-\frac{1}{\sqrt{2010}}=1-\frac{1}{\sqrt{2010}}< \frac{2008}{2010}\)
Ta có:
\(1-a_1\ge a_2+a_3+...+a_n\ge\left(n-1\right)\sqrt[n-1]{a_2a_3...a_n}\)
\(1-a_2\ge a_1+a_3+...+a_n\ge\left(n-1\right)\sqrt[n-1]{a_1a_3...a_n}\)
....
\(1-a_n\ge a_1+a_2+...+a_{n-1}\ge\left(n-1\right)\sqrt[n-1]{a_1a_2...a_{n-1}}\)
Nhân vế với vế:
\(\left(1-a_1\right)\left(1-a_2\right)...\left(1-a_n\right)\ge\left(n-1\right)^n.a_1a_2...a_n\)
\(\Leftrightarrow\frac{a_1a_2...a_n}{\left(1-a_1\right)\left(1-a_2\right)...\left(1-a_n\right)}\le\frac{1}{\left(n-1\right)^n}\)
Dấu "=" xảy ra khi \(a_1=a_2=...=a_n=\frac{1}{n}\)
\(\frac{x^3}{\left(1+y\right)\left(1+z\right)}+\frac{y^3}{\left(1+z\right)\left(1+x\right)}+\frac{z^3}{\left(1+x\right)\left(1+y\right)}\ge\frac{3}{4}\)
\(=\frac{x^3}{1+z+y+yz}+\frac{y^3}{1+x+z+xz}+\frac{z^3}{1+y+x+xy}\)
\(=\frac{x^3}{1+x+y+2y}\ge\frac{x}{2}\Rightarrow TổngBPT\ge\frac{x}{2}+\frac{y}{2}+\frac{z}{2}\ge\frac{2}{3}\left(đpcm\right)\)
(Không chắc à nha)
Ta có : \(\frac{x^3}{\left(1+y\right)\left(1+z\right)}+\frac{1+y}{8}+\frac{1+z}{8}\ge\frac{3x}{4}\)
\(\Rightarrow\frac{x^3}{\left(1+y\right)\left(1+z\right)}\ge\frac{6x-y-z-2}{8}\left(1\right)\)
Tương tự ta có : \(\hept{\begin{cases}\frac{y^3}{\left(1+z\right)\left(1+x\right)}\ge\frac{6y-z-x-2}{8}\left(2\right)\\\frac{z^3}{\left(1+x\right)\left(1+y\right)}\ge\frac{6z-x-y-2}{8}\left(3\right)\end{cases}}\)
Từ (1) , (2) và (3)
\(\Rightarrow\frac{x^3}{\left(1+y\right)\left(1+z\right)}+\frac{y^3}{\left(1+z\right)\left(1+x\right)}+\frac{z^3}{\left(1+x\right)\left(1+y\right)}\)
\(\ge\frac{6x-y-z-2}{8}+\frac{6y-z-x-2}{8}+\frac{6z-x-y-2}{8}\)
\(=\frac{1}{2}\left(x+y+z\right)-\frac{3}{4}\ge\frac{3}{2}-\frac{3}{4}=\frac{3}{4}\)
Chúc bạn học tốt !!!
Ta có : \(\frac{x^3}{\left(1+y\right)\left(1+z\right)}+\frac{1+y}{8}+\frac{1+z}{8}\ge\frac{3x}{4}\)
\(\Rightarrow\frac{x^3}{\left(1+y\right)\left(1+z\right)}\ge\frac{6x-y-z-2}{8}\left(1\right)\)
Tương tự ta có : \(\hept{\begin{cases}\frac{y^3}{\left(1+z\right)\left(1+x\right)}\ge\frac{6y-z-x-2}{8}\left(2\right)\\\frac{z^3}{\left(1+x\right)\left(1+y\right)}\ge\frac{6z-x-y-2}{8}\left(3\right)\end{cases}}\)
Từ (1) , (2) , (3)
\(\Rightarrow\frac{x^3}{\left(1+y\right)\left(1+z\right)}+\frac{y^3}{\left(1+z\right)\left(1+x\right)}+\frac{z^3}{\left(1+x\right)\left(1+y\right)}\)
\(\ge\frac{6x-y-z-2}{8}+\frac{6y-z-x-2}{8}+\frac{6z-x-y-2}{8}\)
\(=\frac{1}{2}\left(x+y+z\right)-\frac{3}{4}\ge\frac{3}{2}-\frac{3}{4}=\frac{3}{4}\)
Chúc bạn học tốt !!!
Áp dụng bđt AM-GM ta có:
\(\hept{\begin{cases}\frac{x^3}{\left(1+y\right)\left(1+z\right)}+\frac{1+y}{8}+\frac{1+z}{8}\ge3\sqrt[3]{\frac{x^3}{\left(1+y\right)\left(1+z\right)}.\frac{1+y}{8}.\frac{1+z}{8}}=\frac{3x}{4}\left(1\right)\\\frac{y^3}{\left(1+z\right)\left(1+x\right)}+\frac{1+z}{8}+\frac{1+x}{8}\ge3\sqrt[3]{\frac{y^3}{\left(1+z\right)\left(1+x\right)}.\frac{1+z}{8}.\frac{1+x}{8}}=\frac{3y}{4}\left(2\right)\\\frac{z^3}{\left(1+x\right)\left(1+y\right)}+\frac{1+x}{8}+\frac{1+y}{8}\ge3\sqrt[3]{\frac{z^3}{\left(1+x\right)\left(1+y\right)}.\frac{1+x}{8}.\frac{1+y}{8}}=\frac{3z}{4}\left(3\right)\end{cases}}\)
Lấy \(\left(1\right)+\left(2\right)+\left(3\right)\)ta được:
\(P+\frac{3+x+y+z}{4}\ge\frac{3\left(x+y+z\right)}{4}\)
\(\Leftrightarrow P\ge\frac{3\left(x+y+z\right)}{4}-\frac{3+x+y+z}{4}\)
\(\Leftrightarrow P\ge\frac{2\left(x+y+z\right)-3}{4}\left(1\right)\)
Áp dụng bdt AM-GM ta có:
\(x+y+z\ge3\sqrt[3]{xyz}=3\)Thay vào (1) ta được:
\(P\ge\frac{2.3-3}{4}\)
\(\Rightarrow P\ge\frac{3}{4}\)Dấu"="xảy ra \(\Leftrightarrow x=y=z\)
