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HN
30 tháng 6 2017
Không mất tính tổng quát ta giả sử \(a\ge b\ge c\)
Đặt \(\left\{{}\begin{matrix}a-b=x\\b-c=y\\a-c=z\end{matrix}\right.\) \(\Rightarrow\left\{{}\begin{matrix}z\ge x\ge0\\z\ge y\ge0\end{matrix}\right.\)
Ta có:
\(x^2+y^2+z^2=\left(x-y\right)^2+\left(x+z\right)^2+\left(y+z\right)^2\)
\(\Leftrightarrow x^2+y^2+z^2+2xz+2yz-2xy=0\)
\(\Leftrightarrow z^2+2xz+2yz+\left(x-y\right)^2=0\)
Vì \(\Rightarrow\left\{{}\begin{matrix}z\ge x\ge0\\z\ge y\ge0\end{matrix}\right.\)
\(\Rightarrow z^2+2xz+2yz+\left(x-y\right)^2\ge0\)
Dấu = xảy ra khi \(x=y=z=0\)
Hay \(a=b=c\)
30 tháng 6 2017
\(VT=\left(a-b\right)^2+\left(b-c\right)^2+\left(c-a\right)^2\)
\(=\left(a+b\right)^2+\left(b+c\right)^2+\left(c+a\right)^2-4ab-4bc-4ca\)
\(VP=\left[\left(a+b\right)-2c\right]^2+\left[\left(b+c\right)-2a\right]^2+\left[\left(c+a\right)-2b\right]^2\)
\(=\left(a+b\right)^2-4\left(a+b\right)c+4c^2+\left(b+c\right)^2-4\left(b+c\right)a+4a^2+\left(a+c\right)^2-4\left(a+c\right)b+4b^2\)
\(=\left(a+b\right)^2+\left(b+c\right)^2+\left(c+a\right)^2-4\left(a+b\right)c+4c^2-4\left(b+c\right)a+4a^2-4\left(a+c\right)b+4b^2\)
Nhìn vào thấy 2 vế có \(\left(a+b\right)^2+\left(b+c\right)^2+\left(c+a\right)^2\) rút gọn luôn thì được
\(-4ab-4bc-4ca=-4\left(a+b\right)c+4c^2-4\left(b+c\right)a+4a^2-4\left(a+c\right)b+4b^2\)
\(\Rightarrow ab-\left(a+b\right)c+c^2+bc-\left(b+c\right)a+a^2+ac-\left(a+c\right)c+b^2=0\)
\(\Rightarrow ab-ac-bc+c^2+bc-ab-ac+a^2+ac-ab-bc+b^2=0\)
\(\Rightarrow a^2+b^2+c^2-ab-bc-ca=0\)
\(\Rightarrow\left(a-b\right)^2+\left(b-c\right)^2+\left(c-a\right)^2=0\)
Xảy ra khi \(\left\{{}\begin{matrix}a-b=0\\b-c=0\\c-a=0\end{matrix}\right.\Rightarrow a=b=c\)
KN
2
1 tháng 6 2019
\(a^3+b^3+c^3-3abc\)
\(=\left(a^3+3a^2b+3ab^2+b^3\right)+c^3-\left(3a^2b+3ab^2+3abc\right)\)
\(=\left(a+b\right)^3+c^3-3ab\left(a+b+c\right)\)
\(=\left(a+b+c\right)\left[\left(a+b\right)^2-\left(a+b\right)c+c^2\right]-3ab\left(a+b+c\right)\)
\(=\left(a+b+c\right)\left(a^2+b^2+2ab-ac-bc+c^2-3ab\right)\)
\(=\left(a+b+c\right)\left(a^2+b^2+c^2-ab-ac-bc\right)\)\(\left(đpcm\right)\)
1 tháng 6 2019
\(a^3+b^3+c^3-3abc\)
\(=\left(a+b\right)^3+c^3-3a^2b-3ab^2-3abc\)
\(=\left(a+b\right)^3+c^3-3ab\left(a+b+c\right)\)
\(=\left(a+b+c\right)\left(\left(a+b\right)^2-c\left(a+b\right)+c^2\right)-3ab\left(a+b+c\right)\)
\(=\left(a+b+c\right)\left(\left(a+b\right)^2-c\left(a+b\right)+c^2-3ab\right)\)
\(=\left(a+b+c\right)\left(a^2+b^2+2ab-ac-bc+c^2-3ab\right)\)
\(=\left(a+b+c\right)\left(a^2+b^2+c^2-ab-bc-ca\right)\)
HH
2 tháng 2 2018
Giải:
Ta có:
\(\left(a-b\right)^2+\left(b-c\right)^2+\left(c-a\right)^2=4\left(a^2+b^2+c^2-ab-bc-ac\right)\)
\(\Leftrightarrow a^2-2ab+b^2+b^2-2bc+c^2+c^2-2ca+a^2=4a^2+4b^2+4c^2-4ab-4bc-4ac\)
\(\Leftrightarrow2a^2+2b^2+2c^2-2ab-2bc-2ca=4a^2+4b^2+4c^2-4ab-4bc-4ac\)
\(\Leftrightarrow2\left(a^2+b^2+c^2-ab-bc-ca\right)=4\left(a^2+b^2+c^2-ab-bc-ac\right)\)
\(\Leftrightarrow\left(a^2+b^2+c^2-ab-bc-ca\right)=2\left(a^2+b^2+c^2-ab-bc-ac\right)\)
\(\Rightarrowđpcm\)
ND
17 tháng 4 2018
2)
Xét hiệu:
\(A^2+B^2+C^2+D^2+4-2A-2B-2C-2D\)
\(=\left(A^2-2A+1\right)+\left(B^2-2B+1\right)+\left(C^2-2C+1\right)+\left(D^2-2D+1\right)\)
\(=\left(A-1\right)^2+\left(B-1\right)^2+\left(C-1\right)^2+\left(D-1\right)^2\ge0\)
=> BĐT luôn đúng
Vậy \(A^2+B^2+C^2+D^2+4\ge2\left(A+B+C+D\right)\)
ND
17 tháng 4 2018
1)
Áp dụng BĐT Cauchy cho 2 số không âm, ta có:
\(\dfrac{AB}{C}+\dfrac{BC}{A}\ge2\sqrt{\dfrac{AB}{C}.\dfrac{BC}{A}}=2B\) (1)
\(\dfrac{BC}{A}+\dfrac{AC}{B}\ge2\sqrt{\dfrac{BC}{A}.\dfrac{AC}{B}}=2C\) (2)
\(\dfrac{AB}{C}+\dfrac{AC}{B}\ge2\sqrt{\dfrac{AB}{C}.\dfrac{AC}{B}}=2A\) (3)
Từ (1)(2)(3) cộng vế theo vế:
\(2\left(\dfrac{AB}{C}+\dfrac{AC}{B}+\dfrac{BC}{A}\right)\ge2\left(A+B+C\right)\)
\(\Rightarrow\dfrac{AB}{C}+\dfrac{AC}{B}+\dfrac{BC}{A}\ge A+B+C\)
13 tháng 8 2017
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 =>\(\dfrac{a}{c}=\dfrac{b}{d}\)
13 tháng 8 2017
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
26 tháng 8 2017
a) \(A=\left(3x+1\right)^2-2\left(3x+1\right)\left(3x+5\right)+\left(5x+5\right)^2\)
\(A=\left[\left(3x+1\right)-\left(5x+5\right)\right]^2\)
\(A=\left(-2x-4\right)^2\)
KN
30 tháng 9 2017
A = (3x + 1)2 - 2(3x + 1)(5x + 5) + (5x + 5)2
= [(3x + 1)-(5x + 5)]2
= (3x + 1 - 5x - 5)2
= [(-2x) - 4]2
B = (3 + 1)(32 + 1)(34 + 1)(38 + 1)(316 +1)(332 + 1)
=> (3 - 1)B = (3 - 1)(3 + 1)(32 + 1)(34 + 1)(38 + 1)(316 +1)(332 + 1)
=>2B = (32 - 1)(32 + 1)(34 + 1)(38 + 1)(316 +1)(332 + 1)
= (34 - 1)(34 + 1)(38 + 1)(316 +1)(332 + 1)
= (38 - 1)(38 + 1)(316 +1)(332 + 1)
= (316 - 1)316 +1)(332 + 1)
= (332 - 1)(332 + 1)
= 364 - 1
vì 2B = 364 - 1
=> B = \(\dfrac{3^{64}-1}{2}\)
C = a2 + b2 + c2 + 2ab - 2ac - 2bc + a2 + b2 + c2 - 2ab + 2ac - 2bc - 2( b2 - 2bc + c2)
= 2a2 + 2b2 + 2c2 - 4bc - 2b2 + 4bc - 2c2
= 2a2
2 tháng 6 2020
Chứng minh cái này đi: \(\frac{a^3+a^2+a+1}{a^2+a+1}\ge\frac{2}{3}a+\frac{2}{3}\) ( gợi ý: bđt \(\Leftrightarrow\)\(\left(a-1\right)^2\left(a+1\right)\ge0\))
Tương tự với 2 ẩn kia \(\Rightarrow\)\(\Sigma\frac{a^3+a^2+a+1}{a^2+a+1}\ge\frac{8}{27}\Pi\left(a+1\right)\ge\frac{64}{27}\sqrt{abc}\ge\frac{64}{27}\)
dấu "=" xảy ra khi \(a=b=c=1\)
\(a^2+b^2+c^2+3=2\left(a+b+c\right)\)
\(\Leftrightarrow a^2+b^2+c^2+3=2a+2b+2c\)
\(\Leftrightarrow a^2-2a+1+b^2-2b+1+c^2-2c+1=0\) \(\Leftrightarrow\left(a-1\right)^2+\left(b-1\right)^2+\left(c-1\right)^2=0\)
\(\Rightarrow\left[{}\begin{matrix}a-1=0\\b-1=0\\c-1=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}a=1\\b=1\\c=1\end{matrix}\right.\Rightarrow a=b=c=1\Rightarrowđpcm\)
\(a^2+b^2+c^2+3=2\left(a+b+c\right)\Leftrightarrow a^2+b^2+c^2+3-2\left(a+b+c\right)=0\)
\(\Leftrightarrow a^2+b^2+c^2+3-2a-2b-2c=0\)
\(\Leftrightarrow\left(a^2-2a+1\right)+\left(b^2-2b+1\right)+\left(c^2-2c+1\right)=0\)
\(\Leftrightarrow\left(a-1\right)^2+\left(b-1\right)^2+\left(c-1\right)^2=0\)
Vì \(\left\{{}\begin{matrix}\left(a-1\right)^2\ge0\\\left(b-1\right)^2\ge0\\\left(c-1\right)^2\ge0\end{matrix}\right.\)\(\Rightarrow\left(a-1\right)^2+\left(b-1\right)^2+\left(c-1\right)^2\ge0\)
Dấu "=" xảy ra khi \(\left(a-1\right)^2=\left(b-1\right)^2=\left(c-1\right)^2=0\)
<=>a-1=b-1=c-1=0<=>a=b=c=1(đpcm)