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Ta CM BĐT phụ sau: \(\frac{1}{a}+\frac{1}{b}\ge\frac{4}{a+b}\)
Ta có: \(\frac{1}{a}+\frac{1}{b}\ge\frac{2}{\sqrt{ab}},a+b\ge2\sqrt{ab}\)( co si với a,b>0)
Suy ra \(\left(\frac{1}{a}+\frac{1}{b}\right)\left(a+b\right)\ge4\RightarrowĐPCM\)\(\Rightarrow\frac{1}{a+b}\le\frac{1}{4}\left(\frac{1}{a}+\frac{1}{b}\right)\left(1\right)\)
a/Áp dụng (1) có
\(\frac{1}{a+b+2c}\le\frac{1}{4}\left(\frac{1}{a+c}+\frac{1}{b+c}\right)\left(2\right)\).Tương tự ta cũng có:
\(\frac{1}{b+c+2a}\le\frac{1}{4}\left(\frac{1}{a+b}+\frac{1}{a+c}\right)\left(3\right),\frac{1}{c+a+2b}\le\frac{1}{4}\left(\frac{1}{b+c}+\frac{1}{a+b}\right)\left(4\right)\)
Cộng (2),(3) và (4) có \(VT\le\frac{1}{4}.\left(6+6\right)=3\left(ĐPCM\right)\)
b/Áp dụng (1) có:
\(\frac{1}{3a+3b+2c}=\frac{1}{\left(a+b+2c\right)+2\left(a+b\right)}\le\frac{1}{4}\left(\frac{1}{a+b+2c}+\frac{1}{2\left(a+b\right)}\right)\left(5\right)\)
Tương tự có: \(\frac{1}{3a+2b+3c}\le\frac{1}{4}\left(\frac{1}{a+c+2b}+\frac{1}{2\left(a+c\right)}\right)\left(6\right)\)
\(\frac{1}{2a+3b+3c}\le\frac{1}{4}\left(\frac{1}{2a+b+c}+\frac{1}{2\left(b+c\right)}\right)\left(7\right)\)
Cộng (5),(6) và (7) có:
\(VT\le\frac{1}{4}\left(\frac{1}{a+b+2c}+\frac{1}{a+c+2b}+\frac{1}{2a+b+c}+\frac{1}{2}\left(\frac{1}{a+b}+\frac{1}{b+c}+\frac{1}{a+c}\right)\right)\le\frac{1}{4}.9=\frac{3}{2}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
Ta có: \(a^2+b^2\ge2ab;b^2+1\ge2b\Rightarrow a^2+2b^2+3\ge2\left(ab+b+1\right)\)
\(\Rightarrow\frac{1}{a^3+2b^2+3}\le\frac{1}{2\left(ab+b+1\right)}\)
Tương tự ta cũng có:
\(\frac{1}{a^2+2b^2+3}+\frac{1}{b^2+2c^2+3}+\frac{1}{c^2+2a^2+3}\le\)\(\frac{1}{2}\left(\frac{1}{ab+b+1}+\frac{1}{bc+b+1}+\frac{1}{ca+a+1}\right)\)
Mà: \(\frac{1}{ab+b+1}+\frac{1}{bc+c+1}+\frac{1}{ca+a+1}=\frac{1}{ab+b+1}+\)\(\frac{ab}{ab^2+abc+ab}+\frac{b}{bca+ab+b}=1\)
\(\Rightarrow\frac{1}{a^2+2b^2+3}+\frac{1}{b^2+2c^2+3}+\frac{1}{c^2+2a^2+3}\le\frac{1}{2}\left(đpcm\right)\)\(\Leftrightarrow a=b=c=1\)
![](https://rs.olm.vn/images/avt/0.png?1311)
Ta có :\(a^2+2b^2+3=\left(a^2+b^2\right)+\left(b^2+1\right)+2\)\(>=2ab+2b+2=2\left(ab+b+1\right)\)
tương tự ta được \(b^2+2c^2+3>=2\left(bc+c+1\right)\)
\(c^2+2a^2+3>=2\left(ac+a+1\right)\)
theo đề bài abc=1
=> \(\frac{1}{ab+b+1}+\frac{1}{bc+c+1}+\frac{1}{ca+a+1}\)=\(\frac{1}{ab+b+1}+\frac{ab}{b+ab+1}+\frac{b}{ab+b+1}\)=1
=> VT<=1/2
Dấu bằng khi a=b=c=1
Ta có :$a^2+2b^2+3=\left(a^2+b^2\right)+\left(b^2+1\right)+2$a2+2b2+3=(a2+b2)+(b2+1)+2$>=2ab+2b+2=2\left(ab+b+1\right)$>=2ab+2b+2=2(ab+b+1)
tương tự ta được $b^2+2c^2+3>=2\left(bc+c+1\right)$b2+2c2+3>=2(bc+c+1)
$c^2+2a^2+3>=2\left(ac+a+1\right)$c2+2a2+3>=2(ac+a+1)
theo đề bài abc=1
=> $\frac{1}{ab+b+1}+\frac{1}{bc+c+1}+\frac{1}{ca+a+1}$1ab+b+1 +1bc+c+1 +1ca+a+1 =$\frac{1}{ab+b+1}+\frac{ab}{b+ab+1}+\frac{b}{ab+b+1}$1ab+b+1 +abb+ab+1 +bab+b+1 =1
=> VT<=1/2
Dấu bằng khi a=b=c=1
![](https://rs.olm.vn/images/avt/0.png?1311)
Bài làm:
Ta có: \(a^2+2b^2+3=\left(a^2+b^2\right)+\left(b^2+1\right)+2\ge2ab+2b+2=2\left(ab+b+1\right)\)
\(\Rightarrow\frac{1}{a^2+2b^2+3}\le\frac{1}{2\left(ab+b+1\right)}\)
Tương tự ta CM được:
\(\frac{1}{b^2+2c^2+3}\le\frac{1}{2\left(bc+c+1\right)}\)
\(\frac{1}{c^2+2a^2+3}\le\frac{1}{2\left(ca+a+1\right)}\)
Cộng vế 3 BĐT trên ta được:
\(VP\le\frac{1}{2}\left(\frac{1}{ab+b+1}+\frac{1}{bc+c+1}+\frac{1}{ca+a+1}\right)\)
\(=\frac{1}{2}\left(\frac{1}{ab+b+1}+\frac{ab}{ab^2c+abc+ab}+\frac{b}{abc+ab+b}\right)\)
\(=\frac{1}{2}\left(\frac{1}{ab+b+1}+\frac{ab}{b+1+ab}+\frac{b}{1+ab+b}\right)\)
\(=\frac{1}{2}.\frac{ab+b+1}{ab+b+1}=\frac{1}{2}.1=\frac{1}{2}\)
Dấu "=" xảy ra khi: \(a=b=c=1\)
p/s : đéo biết làm thì câm mẹ mồm lại , loại súc vật như bạn ý thì cút khỏi olm cho sạch ạ !
Theo Cauchy ta dễ có : \(b^2+1\ge2\sqrt{b^2}=2b\)
\(a^2+b^2\ge2\sqrt{a^2b^2}=2ab\)
Khi đó : \(\frac{1}{a^2+2b^2+3}\le\frac{1}{2+2b+2ab}=\frac{1}{2\left(ab+b+1\right)}\)
Bằng cách chứng minh tương tự rồi cộng theo vế các bđt cùng chiều thì ta được :
\(VT\le\frac{1}{2}.\frac{1}{ab+b+1}+\frac{1}{2}.\frac{1}{bc+c+1}+\frac{1}{2}.\frac{1}{ca+a+1}=\frac{1}{2}.\left(\frac{1}{ab+b+1}+\frac{1}{bc+c+1}+\frac{1}{ca+a+1}\right)\)
Đặt \(A=\frac{1}{ab+b+1}+\frac{1}{bc+c+1}+\frac{1}{ca+a+1}=\frac{ac}{abc.c+abc+ac}+\frac{a}{abc+ca+1}+\frac{1}{ca+a+1}=1\)
Từ đó ta thu được \(VT\le\frac{1}{2}.1=\frac{1}{2}\)hay \(\frac{1}{a^2+2b^2+3}+\frac{1}{b^2+2c^2+3}+\frac{1}{c^2+2a^2+3}\le1\)
Đẳng thức xảy ra khi và chỉ khi \(a=b=c=1\)
Vậy ta có điều phải chứng minh
![](https://rs.olm.vn/images/avt/0.png?1311)
Bạn tham khảo lời giải tại đây:
Câu hỏi của Ngo Hiệu - Toán lớp 9 | Học trực tuyến
![](https://rs.olm.vn/images/avt/0.png?1311)
Bài này chả khó với lại đầy người đăng rồi
Ta có: \(a^2+b^2\ge2ab\) và \(b^2+1\ge2b\)
\(\Rightarrow a^2+b^2+b^2+1+2\ge2ab+2b+2=2\left(ab+b+1\right)\)
\(\Rightarrow\frac{1}{a^2+2b^2+3}\le\frac{1}{2ab+2b+2}=\frac{1}{2\left(ab+b+1\right)}\left(1\right)\)
Tương tự ta có: \(\frac{1}{b^2+2c^2+3}\le\frac{1}{2\left(bc+c+1\right)}\left(2\right);\frac{1}{c^2+2a^2+3}\le\frac{1}{2\left(ac+a+1\right)}\left(3\right)\)
Cộng theo vế của \(\left(1\right);\left(2\right);\left(3\right)\) ta có:
\(VT\le\frac{1}{2\left(ab+b+1\right)}+\frac{1}{2\left(bc+c+1\right)}+\frac{1}{2\left(ac+a+1\right)}\)
\(=\frac{1}{2}\left(\frac{ac}{a^2bc+abc+ac}+\frac{a}{abc+ac+a}+\frac{1}{ac+a+1}\right)\)
\(=\frac{1}{2}\left(\frac{ac}{ac+a+1}+\frac{a}{ac+a+1}+\frac{1}{ac+a+1}\right)\left(abc=1\right)\)
\(=\frac{1}{2}\left(\frac{ac+a+1}{ac+a+1}\right)=\frac{1}{2}=VP\) (ĐPCM)
Đẳng thức xảy ra khi \(a=b=c=\frac{1}{3}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
Ta có:
\(\frac{1}{a^2+2b^2+3}=\frac{1}{\left(a^2+b^2\right)+\left(b^2+1\right)+2}\le\frac{1}{2ab+2b+2}=\frac{1}{2}\cdot\frac{1}{ab+b+1}\)
Tương tự CM được:
\(\frac{1}{b^2+2c^2+3}\le\frac{1}{2}\cdot\frac{1}{bc+c+1}\) và \(\frac{1}{c^2+2a^2+3}\le\frac{1}{2}\cdot\frac{1}{ca+a+1}\)
\(\Rightarrow VT\le\frac{1}{2}\left(\frac{1}{ab+b+1}+\frac{1}{bc+c+1}+\frac{1}{ca+a+1}\right)\)
\(=\frac{1}{2}\left(\frac{1}{ab+b+1}+\frac{ab}{ab^2c+abc+ab}+\frac{b}{abc+ab+b}\right)\)
\(=\frac{1}{2}\left(\frac{1}{ab+b+1}+\frac{ab}{b+1+ab}+\frac{b}{1+ab+b}\right)=\frac{1}{2}\cdot1=\frac{1}{2}\)
Dấu "=" xảy ra khi: a = b = c = 1
A=\(\frac{1}{a^2+2b^2+3}\)+\(\frac{1}{b^2+2c^2+3}\)+\(\frac{1}{c^2+2a^2+3}\)
ta có: \(\frac{1}{a^2+2b^2+3}\)=\(\frac{1}{\left(a^2+b^2\right)+\left(b^2+1\right)+2}\)\(\le\)\(\frac{1}{2\left(ab+b+1\right)}\)
vì : a2+b2\(\ge\)2\(\sqrt{a^2b^2}\)=2ab
b2+1\(\ge\)2\(\sqrt{b^2x1}\)=2b
cmtt => A\(\le\)\(\frac{1}{2}\)x(\(\frac{1}{ab+b+1}\)+\(\frac{1}{bc+c+1}\)+\(\frac{1}{ca+a+1}\))
=\(\frac{1}{2}\)x(\(\frac{1}{ab+b+1}\)+\(\frac{ab}{ab^2c+abc+ab}\)+\(\frac{b}{cba+ab+b}\))
=\(\frac{1}{2}\)x (\(\frac{1}{ab+b+1}\)+\(\frac{ab}{ab+b+1}\)+\(\frac{b}{ab+b+1}\))=\(\frac{1}{2}\)x\(\frac{ab+b+1}{ab+b+1}\)=\(\frac{1}{2}\)
dấu "=" xảy ra <=> a=b=c=1
![](https://rs.olm.vn/images/avt/0.png?1311)
Áp dụng BĐT Cô-si ta có:
\(a^2+b^2\ge2ab;b^2+1^2\ge2b\)
\(\Rightarrow a^2+b^2+b^2+1+2\ge2ab+2b+2\)
\(\Rightarrow a^2+2b^2+3\ge2\left(ab+b+1\right)\)
\(\Rightarrow\frac{1}{a^2+2b^2+3}\le\frac{1}{2\left(ab+b+1\right)}=\frac{1}{2}.\frac{1}{ab+b+1}\)
chứng minh tương tự
\(\Rightarrow\frac{1}{b^2+2c^2+3}\le\frac{1}{2}.\frac{1}{bc+c+1};\frac{1}{c^2+2a^2+3}\le\frac{1}{2}.\frac{1}{ac+a+1}\)
\(\Rightarrow\frac{1}{a^2+2b^2+3}+\frac{1}{b^2+2c^2+3}+\frac{1}{c^2+2a^2+3}\le\frac{1}{2}.\frac{1}{ab+b+1}+\frac{1}{2}.\frac{1}{bc+c+1}+\frac{1}{2}.\frac{1}{ac+a+1}\)
\(\Rightarrow\frac{1}{a^2+2b^2+3}+\frac{1}{b^2+2c^2+3}+\frac{1}{c^2+2a^2+3}\le\frac{1}{2}.\left(\frac{1}{ab+b+1}+\frac{1}{bc+c+1}+\frac{1}{ac+a+1}\right)\)
đặt \(A=\frac{1}{ab+b+1}+\frac{1}{bc+c+1}+\frac{1}{ac+a+1}\)
\(=\frac{ac}{a^2bc+abc+ac}+\frac{a}{abc+ac+a}+\frac{1}{ac+a+1}\)
\(=\frac{ac}{ac+a+1}+\frac{a}{ac+a+1}+\frac{1}{ac+a+1}=\frac{ac+a+1}{ac+a+1}=1\)
\(\Rightarrow\frac{1}{a^2+2b^2+3}+\frac{1}{b^2+2c^2+3}+\frac{1}{c^2+2a^2+3}\le\frac{1}{2}.1=2\)
=>đpcm