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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 ) : \(\left\{{}\begin{matrix}a^4+b^4\ge a^3b+b^3a\\b^4+c^4\ge b^3c+c^3b\\a^4+c^4\ge a^3c+c^3a\end{matrix}\right.\)
\(\Rightarrow2\left(a^4+b^4+c^4\right)\ge a^3b+b^3a+b^3c+c^3b+a^3c+c^3a\)
\(\Rightarrow3\left(a^4+b^4+c^4\right)\ge a^4+b^4+c^4+a^3b+b^3a+b^3c+c^3b+a^3c+c^3a\)
\(\Rightarrow3\left(a^4+b^4+c^4\right)\ge\left(a+b+c\right)\left(a^3+b^3+c^3\right)\left(đpcm\right)\)
\(\left(a+b+c\right)^3\)
\(=\left(a+b\right)^3+3\left(a+b\right)^2c+3\left(a+b\right)c^2+c^3\)
\(=a^3+b^3+3a^2b+3b^2a+3\left(a^2+2ab+b^2\right)c+3ac^2+3bc^2+c^3\)
\(=a^3+b^3+c^3+3a^2b+3b^2a+3a^2c+6abc+3b^2c+3ac^2+3bc^2\)
\(=a^3+b^3+c^3+3\left(a^2b+b^2a+a^2c+2abc+b^2c+ac^2+bc^2\right)\)
\(=a^3+b^3+c^3+3\left[ab\left(a+b\right)+ac\left(a+b\right)+bc\left(a+b\right)+c^2\left(a+b\right)\right]\)
\(=a^3+b^3+c^3+3\left(a+b\right)\left(ab+ac+bc+c^2\right)\)
\(=a^3+b^3+c^3+3\left(a+b\right)\left[a\left(b+c\right)+c\left(b+c\right)\right]\)
\(=a^3+b^3+c^3+3\left(a+b\right)\left(b+c\right)\left(a+c\right)\)
\(\left(đpcm\right)\)
\(\left(a+b+c\right)^3=\left(a+b\right)^3+3\left(a+b\right)^2.c+3\left(a+b\right).c^2+c^3\)
..........................\(=a^3+b^3+c^3+\left[3a^2b+3ab^2+3\left(a+b\right)^2.c+3\left(a+b\right).c^2\right]\)
..........................\(=a^3+b^3+c^3\left[3ab\left(a+b\right)+3\left(a+b\right)^2.c+3\left(a+b\right).c^2\right]\)
...........................\(=a^3+b^3+c^3+3\left(a+b\right)\left[ab+\left(a+b\right)c+c^2\right]\)
...........................\(=a^3+b^3+c^3+3\left(a+b\right)\left(ab+ac+bc+c^2\right)\)
...........................\(=a^3+b^3+c^3+3\left(a+b\right)\left[a\left(b+c\right)+c\left(b+c\right)\right]\)
...........................\(=a^3+b^3+c^3+3\left(a+b\right)\left(b+c\right)\left(c+a\right)\)
4) Ta có : A=(a+b+c+d)(a-b-c+d)=(a-b+c-d)(a+b-c-d)
=> (a+d)2 - (b+c)2= (a-d)2 - (c-b)2
=> a2+ d2+ 2ad - b2- c2- 2bc=a2 + d2 - 2ad - c2-b2+2bc
Rút gọn ta được: 4ad = 4bc => ad = bc =>\(\dfrac{a}{c}=\dfrac{b}{d}\)
1) a2+b2+c2+3=2(a+b+c) =>(a-1)2+(b-1)2+(c-1)2=0
=> a-1=b-1=c-1=0 => a=b=c=1 =>đpcm
Xét : a^3+b^3-ab.(a+b)
= (a+b).(a^2-ab+b^2)-ab.(a+b)
= (a+b).(a^2-2ab+b^2)
= (a+b).(a-b)^2 >= 0 ( vì a;b > 0 )
=> a^3+b^3 >= ab.(a+b)
<=> (a+b)^3 = a^3+b^3+3ab.(a+b) < = a^3+b^3+3a^3+3b^3 = 4a^3+4b^3
Tương tự ........
=> (a+b)^3 + (b+c)^3 + (c+a)^3 < = 8a^3+8b^3+8c^3 = 8.(a^3+b^3+c^3)
=> ĐPCM
Tk mk nha
Lời giải:
Áp dụng BĐT AM-GM:
\(a^3+b^3+b^3\geq 3ab^2\)
\(a^3+a^3+b^3\geq 3a^2b\)
\(\Rightarrow 3(a^3+b^3)\geq 3ab(a+b)\)
\(\Leftrightarrow 4(a^3+b^3)\geq a^3+b^3+3ab(a+b)=(a+b)^3\)
Tương tự:
\(\left\{\begin{matrix} 4(b^3+c^3)\geq (b+c)^3\\ 4(c^3+a^3)\geq (c+a)^3\end{matrix}\right.\)
Cộng theo vế:
\(8(a^3+b^3+c^3)\geq (a+b)^3+(b+c)^3+(c+a)^3\)
Do đó ta có đpcm
Dấu bằng xảy ra khi a=b=c
\(VT=\left(a+b+c\right)^2=\left(a+b\right)^3+3\left(a+b\right)^2c+3\left(a+b\right)c^2+c^3\)
\(=a^3+b^3+3a^2b+3ab^2+3\left(a+b\right)c.\left(a+b+c\right)+c^3\)
\(=a^3+b^3+c^3+3ab\left(a+b\right)+3\left(a+b\right)c.\left(a+b+c\right)\)
\(=a^3+b^3+c^3+3\left(a+b\right).\left[ab+c\left(a+b+c\right)\right]\)
\(=a^3+b^3+c^3+3\left(a+b\right)\left(ab+ac+bc+c^2\right)\)
\(=a^3+b^3+c^3+3\left(a+b\right)\left[a.\left(b+c\right)+c.\left(b+c\right)\right]\)
\(=a^3+b^3+c^3+3\left(a+b\right)\left(b+c\right)\left(c+a\right)\)
\(=VP\)
\(\Rightarrow\text{Điều phải chứng minh}\)