Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
\(a^3b^2-a^3c^2+b^3c^2-b^3a^2+c^3a^2-c^3b^2\)
\(=a^2b^2\left(a-b\right)-c^2\left(a^3-b^3\right)+c^3\left(a^2-b^2\right)\)
\(=a^2b^2\left(a-b\right)-c^2\left(a-b\right)\left(a^2+ab+b^2\right)+c^3\left(a+b\right)\left(a-b\right)\)
\(=\left(a-b\right)\left(a^2b^2-c^2a^2-c^2ab-c^2b^2+c^3a+c^3b\right)\)
\(=\left(a-b\right)\left[\left(a^2b^2-c^2b^2\right)-\left(c^2a^2-c^3a\right)-\left(c^2ab-c^3b\right)\right]\)
\(=\left(a-b\right)\left[b^2\left(a-c\right)\left(a+c\right)-c^2a\left(a-c\right)-c^2b\left(a-c\right)\right]\)
\(=\left(a-b\right)\left[\left(a-c\right)\left(b^2a+b^2c-c^2a-c^2b\right)\right]\)
\(=\left(a-b\right)\left(a-c\right)\left[\left(b^2a-c^2a\right)+\left(b^2c-c^2b\right)\right]\)
\(=\left(a-b\right)\left(a-c\right)\left[a\left(b-c\right)\left(b+c\right)+bc\left(b-c\right)\right]\)
\(=\left(a-b\right)\left(a-c\right)\left(b-c\right)\left(ab+ac+bc\right)\)
\(a< b< c\Leftrightarrow\left\{{}\begin{matrix}a-b< 0\\a-c< 0\\b-c< 0\end{matrix}\right.\)
ab+ac+bc hiển nhiên lớn hơn 0 suy ra tích nhỏ hơn 0 => đpcm
Vì \(0\le a,b,c\le2\)nên:
\(abc+\left(2-a\right)\left(2-b\right)\left(2-c\right)\ge0\)
\(\Leftrightarrow abc+2bc-abc+2ac-4c+2ab-4b-4a+8\ge0\)
\(\Leftrightarrow2bc+2ac+2ab-4\left(a+b+c\right)+8\ge0\)
\(\Leftrightarrow2\left(ab+bc+ac\right)-12+8\ge0\)
\(\Leftrightarrow2\left(ab+bc+ac\right)\ge4\)
Do đó: \(a^2+b^2+c^2=\left(a+b+c\right)^2-2\left(ab+bc+ac\right)\le3^2-4=5\)
(Dấu "="\(\Leftrightarrow\)(a,b,c) là các hoán vị của (0,1,2))
Cho a = 1; b =0,5; c = 0,5
1^2+0,5^2+0,5^2=1+0,25+0,25=1,5
\(VT=a^3\left(b^2-c^2\right)+b^3\left(b^2-a^2\right)+b^3\left(c^2-b^2\right)+c^3\left(a^2-b^2\right)\)
\(=\left(b^2-c^2\right)\left(a^3-b^3\right)-\left(a^2-b^2\right)\left(b^3-c^3\right)\)
\(=\left(a-b\right)\left(b-c\right)\left(b+c\right)\left(a^2+b^2+ab\right)-\left(a-b\right)\left(b-c\right)\left(a+b\right)\left(b^2+c^2+bc\right)\)
\(=\left(a-b\right)\left(b-c\right)\left(a^2b+a^2c-ac^2-bc^2\right)\)
\(=\left(a-b\right)\left(b-c\right)\left[b\left(a-c\right)\left(a+c\right)+ac\left(a-c\right)\right]\)
\(=\left(a-b\right)\left(b-c\right)\left(a-c\right)\left(ab+bc+ca\right)\)
Do \(a< b< c\Rightarrow\left\{{}\begin{matrix}a-b< 0\\b-c< 0\\a-c< 0\end{matrix}\right.\) \(\Rightarrow\left(a-b\right)\left(b-c\right)\left(a-c\right)< 0\)
\(\Rightarrow\left(a-b\right)\left(b-c\right)\left(a-c\right)\left(ab+bc+ca\right)< 0\) (đpcm)
Vì \(0< a,b,c< 1\) nên
\(\Rightarrow\left\{{}\begin{matrix}a^2< a\\b^2< b\\c^2< c\end{matrix}\right.\)
\(\Rightarrow a^2+b^2+c^2< a+b+c=2\)
vì 0<a<1 ;0<b<2 ;0<c<3
=> 1-a > 0 <=> 0<\(\sqrt{1-a}\) < 1
=> 0 <\(\dfrac{\sqrt{1-a}}{a}\) ≤ 1 (1)
c/m tương tự với b,c
=> 0 < \(\dfrac{\sqrt{2-b}}{b}\) ≤ 2 (2)
và 0 < \(\dfrac{\sqrt{3-c}}{c}\) ≤ 3 (3)
Cộng các vế của bđt với nhau
=> 0 < \(\dfrac{\sqrt{1-a}}{a}+\dfrac{\sqrt{2-b}}{b}+\dfrac{\sqrt{3-c}}{c}\) ≤ 6
Vậy GTLN của A là 6
Ta có: a+b+c=1 <=>(a+b+c)2 = 1 <=> ab+bc+ca=0 (1)
Theo dãy tỉ số bằng nhau ta có:
xa=yb=zc=x+y+za+b+c=x+y+z1=x+y+zxa=yb=zc=x+y+za+b+c=x+y+z1=x+y+z
<=> x = a(x+y+z) ; y = b(x+y+z) ; z = c(x+y+z)
=> xy+yz+zx= ab(x+y+z)2+bc(x+y+z)2+ca(x + y + z)2
<=> xy+yz+zx =(ab+bc+ca)(x+y+z)2 (2)
từ (1) và (2) => xy + yz + zx = 0
W.L.O.G: \(a\ge b\ge c\Rightarrow2\ge a\ge\frac{a+b+c}{3}=1\Rightarrow\left(a-2\right)\left(a-1\right)\le0\)
\(\therefore a^2+b^2+c^2\le a^2+\left(b+c\right)^2=2\left(a-1\right)\left(a-2\right)+5\le5\)
Equality holds when \(\left(a;b;c\right)=\left(2;1;0\right)\) and ..
Ta có: a2 + b2 > 2ab, b2 + c2 > 2bc, c2 + a2 > 2ca
=> 2(a2 + b2 + c2) >= 2(ab + bc + ca)
=>3(a2 + b2 + c2) >= (a + b + c)2
=> a2 + b2 + c2 >= \(\frac{\text{(a + b + c)}^2}{3}\)
=> a2 + b2 + c2 >= 3
Dâu = xảy ra khi: a = b = c = 1