CMR:\(\forall a,b,c\)ta luôn có \(a^4+b^4+c^4\ge abc\left(a+b+c\right)\)
Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
![](https://rs.olm.vn/images/avt/0.png?1311)
![](https://rs.olm.vn/images/avt/0.png?1311)
Sử dụng BĐT: \(\left(x+y+z\right)^3\ge27xyz\Rightarrow\left(\frac{x+y+z}{3}\right)^3\ge xyz\)
\(\Rightarrow\left(\frac{1+a+1+b+1+c}{3}\right)^3\ge\left(1+a\right)\left(1+b\right)\left(1+c\right)\)
Ta có: \(\frac{1}{1+a}+\frac{1}{1+b}+\frac{1}{1+c}\ge3\sqrt[3]{\frac{1}{\left(1+a\right)\left(1+b\right)\left(1+c\right)}}\)
\(\frac{a}{1+a}+\frac{b}{1+b}+\frac{c}{1+c}\ge3\sqrt[3]{\frac{abc}{\left(1+a\right)\left(1+b\right)\left(1+c\right)}}\)
Cộng vế với vế:
\(1\ge\frac{1+\sqrt[3]{abc}}{\sqrt[3]{\left(1+a\right)\left(1+b\right)\left(1+c\right)}}\Rightarrow\left(1+a\right)\left(1+b\right)\left(1+c\right)\ge\left(1+\sqrt[3]{abc}\right)^3\)
Dấu "=" 3 BĐT trên xảy ra khi \(a=b=c\)
Lại có:
\(1+\sqrt[3]{abc}\ge2\sqrt{\sqrt[3]{abc}}\Rightarrow\left(1+\sqrt[3]{abc}\right)^3\ge\left(2\sqrt{\sqrt[3]{abc}}\right)^3=8\sqrt{abc}\)Dấu "=" xảy ra khi \(a=b=c=1\)
![](https://rs.olm.vn/images/avt/0.png?1311)
Bất đẳng thức cần chứng minh tương đương:
\(a^{10}b^2+b^{10}a^2\ge a^8b^4+b^8a^4\)
\(\Leftrightarrow a^8+b^8\ge a^6b^2+b^6a^2\) (Do \(a^2b^2\ge0\))
\(\Leftrightarrow\left(a^6-b^6\right)\left(a^2-b^2\right)\ge0\)
\(\Leftrightarrow\left(a^2-b^2\right)^2\left(a^4+a^2b^2+b^4\right)\ge0\) (luôn đúng).
Vậy ta có đpcm.
![](https://rs.olm.vn/images/avt/0.png?1311)
Nhận xét thấy : \(x^4+y^4+z^4+t^4\ge2x^2y^2+2z^2t^2\ge4xyzt\)
Dấu " =" xảy ra khi \(x=y=z=t\)
Áp dụng :
\(a^4+a^4+b^4+c^4\ge4a^2bc\)
\(a^4+b^4+b^4+c^4\ge4ab^2c\)
\(a^4+b^4+c^4+c^4\ge4abc^2\)
\(\Rightarrow4\left(a^4+b^4+c^4\right)\ge4abc\left(a+b+c\right)\)
\(\Leftrightarrowđpcm\)
Dấu " = " xảy ra khi \(a=b=c\)
![](https://rs.olm.vn/images/avt/0.png?1311)
a ) CM : \(a^4+b^4\ge a^3b+b^3a\)
Giả sử điều cần c/m là đúng
\(\Rightarrow a^4+b^4-a^3b-b^3a\ge0\)
\(\Rightarrow a^3\left(a-b\right)-b^3\left(a-b\right)\ge0\)
\(\Rightarrow\left(a^3-b^3\right)\left(a-b\right)\ge0\)
\(\Rightarrow\left(a-b\right)^2\left(a^2+ab+b^2\right)\ge0\)
Ta có : \(\left\{{}\begin{matrix}\left(a-b\right)^2\ge0\\a^2+ab+b^2=\left(a+\dfrac{b}{2}\right)^2+\dfrac{3b^2}{4}\ge0\end{matrix}\right.\)
\(\Rightarrow\left(a-b\right)^2\left(a^2+ab+b^2\right)\ge0\)
\(\Rightarrow a^4+b^4-a^3b-b^3a\ge0\)
\(\Rightarrow a^4+b^4\ge a^3b+b^3a\)
\(\Rightarrow2\left(a^4+b^4\right)\ge a^4+a^3b+b^4+b^3a\)
\(\Rightarrow2\left(a^4+b^4\right)\ge\left(a+b\right)\left(a^3+b^3\right)\)
\(\left(đpcm\right)\)
b ) \(\left(a+b+c\right)\left(a^3+b^3+c^3\right)\)
\(=a^4+a^3b+a^3c+b^3a+b^4+b^3c+c^3a+c^3b+c^4\)
\(=\left(a^4+b^4+c^4\right)+\left(a^3b+b^3a\right)+\left(b^3c+c^3b\right)+\left(a^3c+c^3a\right)\)
CMTT như a ) : \(\left\{{}\begin{matrix}a^4+b^4\ge a^3b+b^3a\\b^4+c^4\ge b^3c+c^3b\\a^4+c^4\ge a^3c+c^3a\end{matrix}\right.\)
\(\Rightarrow2\left(a^4+b^4+c^4\right)\ge a^3b+b^3a+b^3c+c^3b+a^3c+c^3a\)
\(\Rightarrow3\left(a^4+b^4+c^4\right)\ge a^4+b^4+c^4+a^3b+b^3a+b^3c+c^3b+a^3c+c^3a\)
\(\Rightarrow3\left(a^4+b^4+c^4\right)\ge\left(a+b+c\right)\left(a^3+b^3+c^3\right)\left(đpcm\right)\)
Áp dụng BĐT Cô si, ta có :
\(a^4+b^4\ge2a^2b^2\)
\(b^4+c^4\ge2b^2c^2\)
\(c^4+a^4\ge2c^2a^2\)
\(\Rightarrow a^4+b^4+b^4+c^4+c^4+a^4\ge2a^2b^2+2b^2c^2+2c^2a^2\)
\(\Rightarrow a^4+b^4+c^4\ge a^2b^2+b^2c^2+c^2a^2\)( 1 )
Ta lại có :
\(a^2b^2+b^2c^2\ge2ab^2c\)
\(b^2c^2+c^2a^2\ge2bc^2a\)
\(c^2a^2+a^2b^2\ge2ca^2b\)
\(\Rightarrow a^2b^2+b^2c^2+c^2a^2\ge ab^2c+bc^2a+ca^2b=abc\left(a+b+c\right)\)( 2 )
Từ ( 1 ) và ( 2 ) \(\Rightarrow a^4+b^4+c^4\ge abc\left(a+b+c\right)\forall a;b;c\)( Đpcm )
Ta có \(a^4+b^4+c^4\ge abc\left(a+b+c\right)\forall a;b;c>0\)
\(\Leftrightarrow a^4+b^4+c^4-a^2bc-b^2ac-c^2ab\ge0\)
\(\Leftrightarrow2a^4+2b^4+2c^4-2a^2bc-2b^2ac-2c^2ab\ge0\)
\(\Leftrightarrow\left(a^2-b^2\right)^2+2a^2b^2+\left(b^2-c^2\right)^2+2b^2c^2+\left(c^2-a^2\right)^2-2a^2c^2-2b^2ac-2c^2ab\ge0\)
\(\Leftrightarrow\left(a^2-b^2\right)^2+\left(b^2-c^2\right)^2-\left(c^2-a^2\right)^2+\left(a^2b^2+b^2c^2-2b^2ac\right)\)\(+\left(b^2c^2+c^2a^2-2c^2ab\right)+\left(a^2b^2+c^2a^2-2a^2bc\right)\ge0\)
\(\Leftrightarrow\left(a^2-b^2\right)^2+\left(b^2-c^2\right)^2+\left(c^2-a^2\right)^2+\left(ab-bc\right)^2+\left(bc-ca\right)^2+\left(ab-ac\right)^2\ge0\)
Luôn đúng với mọi a,b,c