a>b>c
chứng minh (a-b)^3 + (b-c)^3 + (c-a)^3 < 0
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CH
1
S
14 tháng 7 2018
\(\left(a-b\right)^3+\left(b-c\right)^3+\left(c-a\right)^3\)
\(=\left(a-b+b-c\right)\left[\left(a-b\right)^2-\left(a-b\right)\left(b-c\right)+\left(b-c\right)^2\right]+\left(c-a\right)^3\)
\(=\left(a-c\right)\left[\left(a-b\right)^2-\left(a-b\right)\left(b-c\right)+\left(b-c\right)^2\right]-\left(a-c\right)^3\)
\(=\left(a-c\right)\left[\left(a-b\right)^2-\left(a-b\right)\left(b-c\right)+\left(b-c\right)^2-\left(a-c\right)^2\right]\)
\(=\left(a-c\right)\left[\left(a-b\right)\left(a-b-b+c\right)+\left(b-c+a-c\right)\left(b-c-a+c\right)\right]\)
\(=\left(a-c\right)\left[\left(a-b\right)\left(a-2b+c\right)+\left(a+b-2c\right)\left(b-a\right)\right]\)
\(=\left(a-c\right)\left[\left(a-b\right)\left(a-2b+c\right)-\left(a+b-2c\right)\left(a-b\right)\right]\)
\(=\left(a-c\right)\left(a-b\right)\left(a-2b+c-a-b+2c\right)\)
\(=-\left(c-a\right)\left(a-b\right)\left(-3b+3c\right)\)
\(=3\left(a-b\right)\left(b-c\right)\left(c-a\right)\)
Vì a > b > c nên a - b > 0 ; b - c > 0 ; c - a < 0
Do đó \(3\left(a-b\right)\left(b-c\right)\left(c-a\right)< 0\) hay \(\left(a-b\right)^3+\left(b-c\right)^3+\left(c-a\right)^3< 0\) (đpcm)
HH
0
LA
3 tháng 7 2017
Ta có a² + \(\sqrt{a}\) + \(\sqrt{a}\) ≥ 3a ( 1 )
b² + \(\sqrt{b}\) + \(\sqrt{b}\) ≥ 3b ( 2 )
c² + \(\sqrt{c}\) + \(\sqrt{c}\) ≥ 3c ( 3 )
Cộng từng vế ( 1 ) ( 2 ) ( 3 ) cho ta
a² + b² + c² + 2 ( \(\sqrt{a}+\sqrt{b}+\sqrt{c}\) ) ≥ 3 ( a + b + c ) = 9
2 ( \(\sqrt{a}+\sqrt{b}+\sqrt{c}\)) ≥ 9 - ( a² + b² + c² )
2 ( \(\sqrt{a}+\sqrt{b}+\sqrt{c}\) ) ≥ 9 - ( a + b + c )² + 2 (ab + bc + ca) = 2 (ab + bc + ca)
Vậy\(\sqrt{a}+\sqrt{b}+\sqrt{c}\) ≥ ab + bc + ca
Dấu bằng xãy ra khi và chỉ khi a = b = c = 1
Vậy......
4 tháng 12 2018
Ta có \(x^3+y^3=\left(x+y\right)^3-3xy\left(x+y\right)\)
\(x^3-y^3=\left(x-y\right)^3+3xy\left(x-y\right)\)
Áp dụng, ta có \(\left(a+b+c\right)^3-\left(a+b-c\right)^3=\left(a+b+c-a-b+c\right)^3+3\left(a+b+c\right)\left(a+b-c\right)\left(a+b+c-a-b+c\right)=\left(2c\right)^3+3\left(a+b+c\right)\left(a+b-c\right).2c=\left(2c\right)^3+6c\left(a+b+c\right)\left(a+b-c\right)\left(1\right)\)\(\left(b+c-a\right)^3+\left(a+c-b\right)^3=\left(b+c-a+a+c-b\right)^3-3\left(b+c-a\right)\left(a+c-b\right)\left(b+c-a+a+c-b\right)=\left(2c\right)^3-3\left(b+c-a\right)\left(a+c-b\right).2c=\left(2c\right)^3-6c\left(b+c-a\right)\left(a+c-b\right)\left(2\right)\)Từ (1),(2)\(\Rightarrow\left(a+b+c\right)^3-\left(a+b-c\right)^3-\left(b+c-a\right)^3-\left(a+c-b\right)^3=\left(2c\right)^3+6c\left(a+b+c\right)\left(a+b-c\right)-\left[\left(2c\right)^3-6c\left(b+c-a\right)\left(a+c-b\right)\right]=\left(2c\right)^3+6c\left(a+b+c\right)\left(a+b-c\right)-\left(2c\right)^3+6c\left(b+c-a\right)\left(a+c-b\right)=6c\left(a+b+c\right)\left(a+b-c\right)+6c\left(b+c-a\right)\left(a+c-b\right)=6c\left(a^2+2ab+b^2-c^2+ab+bc-b^2+ac+c^2-bc-a^2-ac+ab\right)=6c\left(4ab\right)=24abc\)Vậy \(\left(a+b+c\right)^3-\left(a+b-c\right)^3-\left(b+c-a\right)^3-\left(a+c-b\right)^3=24abc\)(3)
Ta có a,b,c sẽ có một số lẻ và 2 số chẵn nên \(abc⋮4\Rightarrow24abc⋮96\left(4\right)\)
Từ (3),(4)\(\Rightarrow\left(a+b+c\right)^3-\left(a+b-c\right)^3-\left(b+c-a\right)^3-\left(a+c-b\right)^3⋮96\)
15 tháng 4 2019
1. (a+b)^2 ≥ 4ab
<=> a2+2ab+b2≥ 4ab
<=> a2+2ab+b2-4ab≥ 0
<=> a2-2ab+b2≥ 0
<=> (a-b)^2 ≥ 0 ( luôn đúng )
15 tháng 4 2019
2. a^2 + b^2 + c^2 ≥ ab + bc + ca
<=> 2a^2 + 2b^2 + 2c^2 ≥ 2ab + 2bc + 2ca
<=> 2a^2 + 2b^2 + 2c^2 - 2ab - 2bc - 2ca ≥ 0
<=> (a^2- 2ab+b^2) + (b^2-2bc+c^2) + (c^2-2ca+a^2) ≥ 0
<=> (a-b)^2 + (b-c)^2 + (c-a)^2 ≥ 0 ( luôn đúng)
Đặt: a-b=x, b-c=y, c-a=z
=> x+y+z = (a-b)+(b-c)+(c-a)=0
Ta có: Nếu x+y+z=0 thì x3+y3+z3=3xyz
=> (a-b)3+(b-c)3+(c-a)3=3(a-b)(b-c)(c-a)
Từ a>b>c => a-b>0,b-c>0, c-a<0
=> 3(a-b)(b-c)(c-a)<0
=> (a-b)3+(b-c)3+(c-a)3<0 (đpcm)