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Hãy chứng minh \(a^4+b^4+c^4=\frac{\left(a^2+b^2+c^2\right)^2}{2}\)
Ta có: \(a+b+c=0\)
⇒\(\left(a+b+c\right)^2=0\)
\(\Rightarrow a^2+b^2+c^2+2ab+2ac+2bc=0\)
mà \(a^2+b^2+c^2=1\)
nên \(2ab+2ac+2bc=-1\)
\(\Leftrightarrow2\cdot\left(ab+ac+bc\right)=-1\)
\(\Leftrightarrow\left(ab+ac+bc\right)^2=\frac{1}{4}\)
\(\Leftrightarrow a^2b^2+a^2c^2+b^2c^2+2abc\left(a+b+c\right)=\frac{1}{4}\)
\(\Leftrightarrow a^2b^2+b^2c^2+c^2a^2=\frac{1}{4}\)
Ta có: \(a^2+b^2+c^2=1\)
\(\Leftrightarrow\left(a^2+b^2+c^2\right)^2=1\)
\(\Leftrightarrow a^4+b^4+c^4+2\left(a^2b^2+b^2c^2+c^2a^2\right)=1\)
\(\Leftrightarrow a^4+b^4+c^4+\frac{1}{2}=1\)
hay \(a^4+b^4+c^4=1-\frac{1}{2}=\frac{1}{2}\)(đpcm)
Ta có: a+b+c=0
=> (a+b+c)2 = \(a^2+b^2+c^2+2\left(ab+bc+ac\right)=0\)
mà \(a^2+b^2+c^2=1\) => 1 + 2(ab + bc + ac) = 0
=> 2(ab + bc + ac) = -1 => ab + bc + ac = \(\frac{-1}{2}\)
=> (ab + bc + ac)2 = \(\left(\frac{-1}{2}\right)^2\)
=> a2b2 + b2c2 + a2c2 + 2(ab2c+abc2+a2bc) = \(\frac{1}{4}\)
=> a2b2 + b2c2 + a2c2 + 2abc(a+b+c) = \(\frac{1}{4}\)
mà a+b+c = 0 => a2b2 + b2c2 + a2c2 = \(\frac{1}{4}\)
Do a2 + b2 + c2 =1
=> (a2 + b2 + c2)2 = a4 + b4 + c4 + 2(a2b2 + b2c2 + a2c2)=1
=> a4 + b4 + c4 + 2.\(\frac{1}{4}\) = 1
=> a4 + b4 + c4 = 1 - 2.\(\frac{1}{4}\) =\(\frac{1}{2}\)
Bạn không hiểu chỗ nào thì hỏi lại mình nhé
Từ a+b+c=6 \(\Rightarrow\)a+b=6-c
Ta có: ab+bc+ac=9\(\Leftrightarrow\)ab+c(a+b)=9
\(\Leftrightarrow\)ab=9-c(a+b)
Mà a+b=6-c (cmt)
\(\Rightarrow\)ab=9-c(6-c)
\(\Rightarrow\)ab=9-6c+c2
Ta có: (b-a)2\(\ge\)0 \(\forall\)b, c
\(\Rightarrow\)b2+a2-2ab\(\ge\)0
\(\Rightarrow\)(b+a)2-4ab\(\ge\)0
\(\Rightarrow\)(a+b)2\(\ge\)4ab
Mà a+b=6-c (cmt)
ab= 9-6c+c2 (cmt)
\(\Rightarrow\)(6-c)2\(\ge\)4(9-6c+c2)
\(\Rightarrow\)36+c2-12c\(\ge\)36-24c+4c2
\(\Rightarrow\)36+c2-12c-36+24c-4c2\(\ge\)0
\(\Rightarrow\)-3c2+12c\(\ge\)0
\(\Rightarrow\)3c2-12c\(\le\)0
\(\Rightarrow\)3c(c-4)\(\le\)0
\(\Rightarrow\)c(c-4)\(\le\)0
\(\Rightarrow\hept{\begin{cases}c\ge0\\c-4\le0\end{cases}}\)hoặc\(\hept{\begin{cases}c\le0\\c-4\ge0\end{cases}}\)
*\(\hept{\begin{cases}c\ge0\\c-4\le0\end{cases}\Leftrightarrow\hept{\begin{cases}c\ge0\\c\le4\end{cases}\Leftrightarrow}0\le c\le4}\)
*
câu 2
a^4 + b^4 + c^4 + d^4 = 4abcd
<=> \(a^4-2a^2b^2+b^4+c^4-2c^2d^2+d^4+2a^2b^2-4abcd+2b^2d^2=0\)
<=> \(\left(a^2-b^2\right)^2+\left(c^2-d^2\right)^2+2\left(ab-cd\right)^2=0\)
<=> \(\left\{{}\begin{matrix}a^2=b^2\\c^2=d^2\\ab=cd\end{matrix}\right.\Leftrightarrow a=b=c=d\)
Bài 1:
Áp dụng BĐt cauchy dạng phân thức:
\(\dfrac{1}{2x+y}+\dfrac{1}{x+2y}\ge\dfrac{4}{3\left(x+y\right)}\)
\(\Rightarrow\left(3x+3y\right)\left(\dfrac{1}{2x+y}+\dfrac{1}{x+2y}\right)\ge\left(3x+3y\right).\dfrac{4}{3x+3y}=4\)
dấu = xảy ra khi 2x+y=x+2y <=> x=y
Bài 2:
ta có: \(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}+\dfrac{1}{d}\ge\dfrac{4^2}{a+b+c+d}=\dfrac{16}{a+b+c+d}\)(theo BĐt cauchy-schwarz)
\(\Rightarrow\dfrac{1}{a+b+c+d}\le\dfrac{1}{16}\left(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}+\dfrac{1}{d}\right)\)
Áp dụng BĐT trên vào bài toán ta có:
\(A=\dfrac{1}{2a+b+c}+\dfrac{1}{a+2b+c}+\dfrac{1}{a+b+2c}\le\dfrac{1}{16}\left(\dfrac{2}{a}+\dfrac{1}{b}+\dfrac{1}{c}+\dfrac{1}{a}+\dfrac{2}{b}+\dfrac{1}{c}+\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{2}{c}\right)\)\(A\le\dfrac{1}{16}.4\left(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}\right)=\dfrac{1}{4}\left(\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}\right)\)
......
dấu = xảy ra khi a=b=c
Bài 2:
Áp dụng BĐT cauchy cho 2 số dương:
\(a^2+1\ge2a\)
\(\Leftrightarrow\dfrac{a}{a^2+1}\le\dfrac{a}{2a}=\dfrac{1}{2}\)
thiết lập tương tự:\(\dfrac{b}{b^2+1}\le\dfrac{1}{2};\dfrac{c}{c^2+1}\le\dfrac{1}{2}\)
cả 2 vế các BĐT đều dương ,cộng vế với vế,ta có dpcm
dấu = xảy ra khi a=b=c=1
Lời giải:
a)
$a+b+c=0\Leftrightarrow (a+b+c)^2=0$
$\Leftrightarrow a^2+b^2+c^2+2(ab+bc+ac)=0$
$\Rightarrow ab+bc+ac=-\frac{a^2+b^2+c^2}{2}\leq 0$
Mà $a^2\geq 0$
Do đó: $a^2(ab+bc+ac)\leq 0$
$\Leftrightarrow a^3b+a^2bc+a^3c\leq 0$ (đpcm)
Dấu "=" xảy ra khi $a=0$
b)
Từ ĐKĐB \(\Rightarrow \left\{\begin{matrix} a+b=(3c+3)\\ 4ab=9c^2\end{matrix}\right.\)
Ta biết rằng $(a+b)^2=(a-b)^2+4ab\geq 4ab$
$\Leftrightarrow (3c+3)^2\geq 9c^2$
$\Leftrightarrow (c+1)^2\geq c^2$
$\Leftrightarrow 2c+1\geq 0\Leftrightarrow c\geq \frac{-1}{2}$ (đpcm)
Vậy.......
Ta có: a+b+c=0
=> \(\left(a+b+c\right)^2=0\)
=> \(a^2+b^2+c^2+2ab+2bc+2ac=0\)
=> 2ab + 2bc + 2ac = -1 (do \(a^2+b^2+c^2=1\) )
=> \(\left(2ab+2bc+2ac\right)^2=\left(-1\right)^2\)
=> \(4a^2b^2+4b^2c^2+4a^2c^2+8ab^2c+8abc^2+8a^2bc=1\)
=>\(4a^2b^2+4b^2c^2+4a^2c^2+8abc\left(a+b+c\right)=1\)
=>\(2\left(2a^2b^2+2b^2c^2+2a^2c^2\right)=1\) (do a+b+c=0)
=>\(2a^2b^2+2b^2c^2+2a^2c^2=\frac{1}{2}\)
Lại có: \(a^2+b^2+c^2=1\)
=> \(\left(a^2+b^2+c^2\right)^2=1\) = 1
=> \(a^4+b^4+c^4+2a^2b^2+2b^2c^2+2a^2c^2=1\)
=> \(a^4+b^4+c^4+\frac{1}{2}=1\)
=> \(a^4+b^4+c^4=\frac{1}{2}\)
=> ĐPCM
Ta có a+b+c=0=>\(\left(a+b+c\right)^2=0\)
=>\(a^2+b^2+c^2+2\left(ab+bc+ca\right)=0\)(1)
Vì \(a^2+b^2+c^2=1\)
Thay vào (1) có ab+bc+ca=\(-\frac{1}{2}\)
Ta có\(a^4+b^4+c^4=\left(a^2+b^2+c^2\right)^2-2\left(a^2b^2+b^2c^2+c^2a^2\right)\)
=1-2\(\left[\left(ab+bc+ca\right)^2-2a^2bc-2ab^2c-2abc^2\right]\)
=1-2\(\left[\frac{1}{4}-2abc\left(a+b+c\right)\right]\)
=1-2\(\left(\frac{1}{4}-0\right)\)
=1-\(\frac{1}{2}\)=\(\frac{1}{2}\)(đpcm