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7 tháng 12 2021
Áp dụng BĐT cosi:
\(\left(a+b+b+c+c+a\right)\left(\dfrac{1}{a+b}+\dfrac{1}{b+c}+\dfrac{1}{c+a}\right)\\ \ge3\sqrt[3]{\left(a+b\right)\left(b+c\right)\left(c+a\right)}\cdot3\sqrt[3]{\dfrac{1}{\left(a+b\right)\left(b+c\right)\left(c+a\right)}}=9\\ \Leftrightarrow2\left(a+b+c\right)\left(\dfrac{1}{a+b}+\dfrac{1}{b+c}+\dfrac{1}{c+a}\right)\ge9\\ \Leftrightarrow\left(a+b+c\right)\left(\dfrac{1}{a+b}+\dfrac{1}{b+c}+\dfrac{1}{c+a}\right)\ge\dfrac{9}{2}\left(đpcm\right)\)
Dấu \("="\Leftrightarrow a=b=c\)
22 tháng 9 2020
\(\left(a+b+c\right)^2+a^2+b^2+c^2=\left(a+b\right)^2+\left(b+c\right)^2+\left(c+a\right)^2\)
VT : (a + b + c)2 + a2 + b2 + c2
= a2 + b2 + c2 + 2ab +2bc + 2ac + a2 + b2 + c2
= ( a2 + 2ab + b2 ) + (b2 + 2bc + c2) + ( a2 + 2ac + c2)
= (a + b)2 + (b + c)2 + (a + c)2 = VP
Vậy \(\left(a+b+c\right)^2+a^2+b^2+c^2=\left(a+b\right)^2+\left(b+c\right)^2+\left(c+a\right)^2\)(đpcm)
QN
24 tháng 7 2017
a) Ta có: \(\left(a+b+c\right)^2+\left(b+c-a\right)^2+\left(a+c-b\right)^2+\left(a+b-c\right)^2\)
\(=a^2+b^2+c^2+2ab+2bc+2ac+a^2+b^2+c^2+2bc-2ab-2ac+a^2+b^2+c^2-2ab-2bc+2ac+a^2+b^2+c^2+2ab-2bc-2ca\)
\(=a^2+b^2+c^2+a^2+b^2+c^2+a^2+b^2+c^2+a^2+b^2+c^2\)
\(=4a^2+4b^2+4c^2\)
\(=4\left(a^2+b^2+c^2\right)\)
QN
24 tháng 7 2017
b) Đặt x = b + c - a
y = c + a - b
z = a + b - c
\(\Rightarrow\left\{{}\begin{matrix}c=\dfrac{x+y}{2}\\a=\dfrac{y+z}{2}\\b=\dfrac{x+z}{2}\end{matrix}\right.\)
\(\Rightarrow a+b+c=x+y+z\)
Ta có: \(\left(a+b+c\right)^3-x^3-y^3-z^3\)
\(=\left(x+y+z\right)^3-x^3-y^3-z^3\)
\(=\left[\left(x+y\right)+z\right]^3-x^3-y^3-z^3\)
\(=\left(x+y\right)^3+3\left(x+y\right)z+3\left(x+y\right)z^2+z^3-x^3-y^3-z^2\)
\(=3x^2y+3xy^2+3\left(x+y\right)^2z+3\left(x+y\right)z^2\)
\(=3xy\left(x+y\right)+3\left(x+y\right)^2z+3\left(x+y\right)z^2\)
\(=3\left(x+y\right)\left[xy+\left(x+y\right)z+z^2\right]\)
\(=3\left(x+y\right)\left[z^2+xy+xz+yz\right]\)
\(=3\left(x+y\right)\left[z\left(x+y\right)+y\left(x+y\right)\right]\)
\(=3\left(x+y\right)\left(x+z\right)\left(y+z\right)\)
\(=3.2a.2b.2c\)
\(=24abc\) (đpcm)
TN
5 tháng 8 2018
\(\left(a+b\right)-\left(-a+b-c\right)+\left(c-a-b\right)\)
\(=a+b+a-b+c+c-a-b\)
\(=\)\(a-b+2c\)( đpcm )
\(a\left(b-c\right)-a\left(b+d\right)\)
\(=a\left(b-c-b-d\right)\)
\(=\)\(a\left(-c-d\right)\)
\(=-a\left(c+d\right)\)( đpcm )
học tốt
S
27 tháng 3 2020
Ta có :\(\frac{b+c}{\left(a-b\right)\left(a-c\right)}+\frac{c+a}{\left(b-c\right)\left(b-a\right)}+\frac{a+b}{\left(c-a\right)\left(c-b\right)}\)
\(=\frac{\left(b+c\right)\left(b-c\right)}{\left(a-b\right)\left(a-c\right)\left(b-c\right)}+\frac{\left(c+a\right)\left(c-a\right)}{\left(b-c\right)\left(b-a\right)\left(c-a\right)}+\frac{\left(a+b\right)\left(a-b\right)}{\left(c-a\right)\left(c-b\right)\left(a-b\right)}\)
\(=\frac{b^2-c^2}{\left(a-b\right)\left(a-c\right)\left(b-c\right)}+\frac{c^2-a^2}{\left(a-b\right)\left(a-c\right)\left(b-c\right)}+\frac{a^2-b^2}{\left(a-b\right)\left(a-c\right)\left(b-c\right)}\)
\(=\frac{b^2-c^2+c^2-a^2+a^2-b^2}{\left(a-b\right)\left(a-c\right)\left(b-c\right)}=0\left(ĐPCM\right)\)
25 tháng 3 2020
Ta có: VP = \(a\left(b^2-2bc+c^2\right)+b\left(c^2-2ac+a^2\right)+c\left(a^2-2ab+b^2\right)\)
= \(ab^2+ac^2+bc^2+ba^2+ca^2+cb^2-6abc\)(1)
\(VT=\left(ab+b^2+ac+bc\right)\left(c+a\right)-8abc\)
\(=abc+b^2c+ac^2+bc^2+a^2b+b^2a+a^2c+abc-8abc\)
= \(ab^2+ac^2+bc^2+ba^2+ca^2+cb^2-6abc\)(2)
Từ (1) ; (2) => VT = VP
Vậy đẳng thức luôn đúng.
28 tháng 6 2018
VT =\(\left(a+b+c\right)^2+a^2+b^2+c^2=a^2+b^2+c^2+2ab+2ac+2bc+a^2+b^2+c^2\)
=\(\left(a^2+2ab+b^2\right)+\left(b^2+2bc+c^2\right)+\left(c^2+2ca+a^2\right)\)
=\(\left(a+b\right)^2+\left(b+c\right)^2+\left(c+a\right)^2=VP\)
=> đpcm
Ta gọi vế bên trái là A vế bên phải là B
Ta có:
A = - ( - a + b + c ) + ( b + c - 1 )
A = a - b - c + b + c - 1
A = a - 1
B = ( b - c + 6 ) - ( 7 - a + b ) + c
B = b - c + 6 - 7 + a - b + c
B = a - 1
Ta thấy A = B = a - 1
Vậy - ( - a + b + c ) + ( b + c - 1 ) = ( b - c + 6 ) - ( 7 - a + b ) + c