Chứng minh: \(\frac{1}{a+b}+\frac{1}{b+c}+\frac{1}{c+a}+\frac{\left(a+b+c\sqrt[3]{abc}\right)^2}{\left(a+b\right)\left(b+c\right)\left(c+a\right)}\) với mọi a,b,c >0
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![](https://rs.olm.vn/images/avt/0.png?1311)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(\frac{1}{a+b}+\frac{1}{b+c}+\frac{1}{c+a}+\frac{1}{2\sqrt[3]{abc}}\ge\frac{\left(a+b+c+\sqrt[3]{abc}\right)^2}{\left(a+b\right)\left(b+c\right)\left(c+a\right)}\)
cauchy-schwarz:
\(VT=\frac{c^2}{ac^2+bc^2}+\frac{a^2}{a^2b+a^2c}+\frac{b^2}{b^2c+b^2a}+\frac{\sqrt[3]{a^2b^2c^2}}{2abc}\ge\frac{\left(a+b+c+\sqrt[3]{abc}\right)^2}{\left(a+b\right)\left(b+c\right)\left(c+a\right)}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(VT=\frac{\left(\sqrt[3]{abc}\right)^2}{2abc}+\Sigma\frac{a^2}{a^2\left(b+c\right)}\ge\frac{\left(a+b+c+\sqrt[3]{abc}\right)^2}{\Sigma a^2\left(b+c\right)+2abc}=\frac{\left(a+b+c+\sqrt[3]{abc}\right)^2}{\left(a+b\right)\left(b+c\right)\left(c+a\right)}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
Áp dụng BĐT AM-GM ta có \(\frac{1^2}{a\left(a+b\right)}+\frac{1^2}{b\left(b+c\right)}+\frac{1^2}{c\left(c+a\right)}\ge\)
\(\ge\frac{\left(1+1+1\right)^2}{a\left(a+b\right)+b\left(b+c\right)+c\left(c+a\right)}=\frac{9}{a\left(a+b\right)+b\left(b+c\right)+c\left(c+a\right)}\ge\)
\(\ge\frac{9}{3.\sqrt[3]{abc\left(a+b\right)\left(b+c\right)\left(c+a\right)}}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
Áp dụng BĐT Cô-si cho 3 số dương, ta có :
\(\frac{1}{a\left(a+b\right)}+\frac{1}{b\left(b+c\right)}+\frac{1}{c\left(a+c\right)}\ge3\sqrt[3]{\frac{1}{abc\left(a+b\right)\left(b+c\right)\left(a+c\right)}}\)
Cần chứng minh : \(\sqrt[3]{\frac{1}{abc\left(a+b\right)\left(b+c\right)\left(a+c\right)}}\ge\frac{9}{2\left(a+b+c\right)^2}\)
hay \(8\left(a+b+c\right)^6\ge729abc\left(a+b\right)\left(b+c\right)\left(a+c\right)\)
Thật vậy, ta có : \(\left(a+b+c\right)^3\ge\left(3\sqrt[3]{abc}\right)^3=27abc\)
\(8\left(a+b+c\right)^3=\left(2\left(a+b+c\right)\right)^3=\left(a+b+b+c+a+c\right)^3\)
\(\ge\left(3\sqrt[3]{\left(a+b\right)\left(b+c\right)\left(a+c\right)}\right)^3=27\left(a+b\right)\left(b+c\right)\left(a+c\right)\)
Nhân từng vế 2 bất đẳng thức trên, ta được đpcm
Dấu "=" xảy ra khi a = b = c
Vậy ...
2. Áp dụng BĐT Cô-si cho 3 số không âm, ta có :
\(B\ge3\sqrt[3]{\sqrt{\left(a^3+b^3+1\right)\left(b^3+c^3+1\right)\left(a^3+c^3+1\right)}}\)
Ta có : \(a^3+b^3+1\ge3\sqrt[3]{a^3b^3}=3ab\Rightarrow\sqrt{a^3+b^3+1}\ge\sqrt{3ab}\)
Tương tự : ....
\(\Rightarrow\sqrt{\left(a^3+b^3+1\right)\left(b^3+c^3+1\right)\left(c^3+a^3+1\right)}\ge\sqrt{27a^2b^2c^2}=\sqrt{27}\)
\(\Rightarrow B\ge3\sqrt[3]{\sqrt{27}}=3\sqrt{3}\)
Vậy GTNN của B là \(3\sqrt{3}\)khi a = b = c = 1
![](https://rs.olm.vn/images/avt/0.png?1311)
Áp dụng bđt Cauchy cho 2 số không âm :
\(x^2+\frac{1}{x}\ge2\sqrt[2]{\frac{x^2}{x}}=2.\sqrt{x}\)
\(y^2+\frac{1}{y}\ge2\sqrt[2]{\frac{y^2}{y}}=2.\sqrt{y}\)
Cộng vế với vế ta được :
\(x^2+y^2+\frac{1}{x}+\frac{1}{y}\ge2.\sqrt{x}+2.\sqrt{y}=2\left(\sqrt{x}+\sqrt{y}\right)\)
Vậy ta có điều phải chứng mình
Ta đi chứng minh:\(a^3+b^3\ge ab\left(a+b\right)\)
\(\Leftrightarrow\left(a-b\right)^2\left(a+b\right)\ge0\)* đúng *
Khi đó:
\(\frac{1}{a^3+b^3+abc}\le\frac{1}{ab\left(a+b\right)+abc}=\frac{1}{ab\left(a+b+c\right)}=\frac{c}{abc\left(a+b+c\right)}\)
Tương tự:
\(\frac{1}{b^3+c^3+abc}\le\frac{a}{abc\left(a+b+c\right)};\frac{1}{c^3+a^3+abc}\le\frac{b}{abc\left(a+b+c\right)}\)
\(\Rightarrow LHS\le\frac{a+b+c}{abc\left(a+b+c\right)}=\frac{1}{abc}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
Lời giải:
Áp dụng BĐT Cauchy-Schwarz ta có:
\(\frac{1}{a+b}+\frac{1}{b+c}+\frac{1}{c+a}+\frac{1}{2\sqrt[3]{abc}}=\frac{c^2}{c^2(a+b)}+\frac{a^2}{a^2(b+c)}+\frac{b^2}{b^2(c+a)}+\frac{(\sqrt[3]{abc})^2}{2abc}\)
\(\geq \frac{(c+a+b+\sqrt[3]{abc})^2}{c^2(a+b)+a^2(b+c)+b^2(c+a)+2abc}=\frac{(a+b+c+\sqrt[3]{abc})^2}{(a+b)(b+c)(c+a)}\)
Ta có đpcm
Dấu "=" xảy ra khi $a=b=c$