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Câu a/ Thì chứng minh ở dưới rồi nhé e
b/ Ta cần chứng minh
\(2\left(a^2b^2+b^2c^2+c^2a^2\right)=2\left(ab+bc+ca\right)^2\)
\(\Leftrightarrow a^2b^2+b^2c^2+c^2a^2=a^2b^2+b^2c^2+c^2a^2+2a^2bc+2ab^2c+2abc^2\)
\(\Leftrightarrow2abc\left(a+b+c\right)=0\)(đúng)
=> ĐPCM
c/ Ta có
\(\frac{\left(a^2+b^2+c^2\right)^2}{2}=\frac{a^4+b^4+c^4+2\left(a^2b^2+b^2c^2+c^2a^2\right)}{2}=a^4+b^4+c^4\)
Cái này là áp dụng câu a vô nhé e
a)\(\left(ab+bc+ca\right)^2=a^2b^2+b^2c^2+c^2a^2\)\(+2\left(ab^2c+abc^2+a^2bc\right)\)
=\(a^2b^2+b^2c^2+c^2a^2+2abc\left(a+b+c\right)\)
=\(a^2b^2+b^{2^2}c^2+c^2a^2+2abc.0\)
=\(a^2b^2+b^2c^2+c^2a^2\)
b) \(a+b+c=0\)=>\(\left(a+b+c\right)^2=0\)
<=>\(a^2+b^2+c^2+2\left(ab+bc+xa\right)=0\)
<=>\(a^2+b^2+c^2=-2\left(ab+bc+ca\right)\)
=>\(\left(a^2+b^2+c^2\right)^2=4\left(ab+bc+ca\right)^2\)
<=>\(a^4+b^4+c^4+2\left(a^2b^2+b^2c^2+c^2a^2\right)\)\(=4\left(ab+bc+ca\right)^2\)
Do \(\left(ab+bc+ca\right)^2=a^2b^2+b^2c^2+c^2a^2\)
=>\(a^4+b^4+c^4+2\left(ab+bc+ca\right)^2\)\(=4\left(ab+bc+ca\right)^2\)
=>\(a^4+b^4+c^4=2\left(ab+bc+ca\right)^2\)
Ta có:
a)
\(a^4+b^4+c^4=\left(a^2+b^2+c^2\right)^2-2a^2b^2-2a^2c^2-2b^2c^2\)
\(=\left[\left(a+b+c\right)^2-2ab-2ac-2bc\right]^2-2a^2b^2-2b^2c^2-2a^2c^2\)
\(=4\left[ab+ac+bc\right]^2-2a^2b^2-2b^2c^2-2a^2c^2\)
\(=4\left(ab\right)^2+4\left(ac\right)^2+4\left(bc\right)^2-8abc\left(a+b+c\right)-2a^2b^2-2b^2c^2-2a^2c^2\)
\(=2\left(a^2b^2+b^2c^2+c^2a^2\right)\)
b)\(=2\left(ab+bc+ac\right)^2-4\left(abbc+abca+bcca\right)\)
\(=2\left(ab+bc+ac\right)^2-4abc\left(a+b+c\right)=2\left(ab+bc+ac\right)^2\)
c) \(\frac{\left(a^2+b^2+c^2\right)^2}{2}=\frac{a^4+b^4+c^4+2\left(a^2b^2+b^2c^2+c^2a^2\right)}{2}=\frac{a^4+b^4+c^4+a^4+b^4+c^4}{2}\)
\(=a^4+b^4+c^4\)
1a)\(a^2+b^2+1\ge ab+a+b\)
\(\Leftrightarrow2\left(a^2+b^2+1\right)\ge2\left(ab+b+a\right)\)
\(\Leftrightarrow\left(a^2-2ab+b^2\right)+\left(a^2-2a+1\right)+\left(b^2-2b+1\right)\ge0\)
\(\Leftrightarrow\left(a-b\right)^2+\left(a-1\right)^2+\left(b-1\right)^2\ge0\)(luôn đúng)
Dấu "=" xảy ra khi x=y=1
b)\(a^2+b^2+c^2\ge a\left(b+c\right)\)
\(\Leftrightarrow2a^2+2b^2+2c^2\ge2ab+2ac\)
\(\Leftrightarrow\left(a^2-2ab+b^2\right)+\left(a^2-2ac+c^2\right)+b^2+c^2\ge0\)
\(\Leftrightarrow\left(a-b\right)^2+\left(a-c\right)^2+b^2+c^2\ge0\)(luôn đúng)
Dấu "=" xảy ra khi a=b=c=0
=a, a(b2+c2)+b(a2+c2)+c(a2+b2)+2abc
= ab2+ac2+ba2+bc2+ca2+cb2+2abc
= c2(a+b)+ab(a+b)+c(a2+b2+2ab)
= c2(a+b)+ab(a+b)+c(a+b)2
= (a+b)\(\left[c^2+ab+c\left(a+b\right)\right]\)
= (a+b)(c2+ab+ca+cb)
= (a+b)\(\left[c\left(a+c\right)+b\left(a+c\right)\right]\)
=(a+b)(a+c)(b+c)
b, a(b-c)3+b(c-a)3+c(a-b)3
= a(b-c)3-b\(\left[\left(b-c\right)+\left(a-b\right)\right]\)3+c(a-b)3
= a(b-c)3-b(b-c)3-3b(b-c)2(a-b)-3b(b-c)(a-b)2-b(a-b)3+c(a-b)3
= a(b-c)3-b(b-c)3-3b(b-c)(a-b)(b-c+a-b)-b(a-b)3+c(a-b)3
= a(b-c)3-b(b-c)3-3b(b-c)(a-b)(a-c)-b(a-b)3+c(a-b)3
= (b-c)3(a-b)-3b(b-c)(a-b)(a-c)-(a-b)3(b-c)
= (b-c)(a-b)\(\left[\left(b-c\right)^2-3b\left(a-c\right)-\left(a-b\right)^2\right]\)
=(b-c)(a-b)(b2-2bc+c2-3ab+3bc-a2+2ab-b2)
= (b-c)(a-b)(c2-a2+bc-ab)
= (b-c)(a-b)\(\left[\left(c-a\right)\left(c+a\right)+b\left(c-a\right)\right]\)
= (b-c)(a-b)(c-a)(c+a+b)
c, a2b2(a-b)+b2c2(b-c)+c2a2(c-a)
= a2b2(a-b)-b2c2\(\left[\left(a-b\right)+\left(c-a\right)\right]\)+c2a2(c-a)
= a2b2(a-b)-b2c2(a-b)-b2c2(c-a)+c2a2(c-a)
= b2(a-b)(a2-c2)+c2(c-a)(a2-b2)
= b2(a-b)(a-c)(a+c)-c2(a-c)(a-b)(a+b)
= (a-c)(a-b)\(\left[b^2\left(a+c\right)-c^2\left(a+b\right)\right]\)
= (a-c)(a-b)(b2a+b2c-c2a-c2b)
= (a-c)(a-b)\(\left[a\left(b^2-c^2\right)+bc\left(b-c\right)\right]\)
= (a-c)(a-b)\(\left[a\left(b-c\right)\left(b+c\right)+bc\left(b-c\right)\right]\)
= (a-c)(a-b)(b-c)\(\left[a\left(b+c\right)+bc\right]\)
= (a-c)(a-b)(b-c)(ab+ac+bc)
d, a4(b-c)+b4(c-a)+c4(a-b)
= a4(b-c)-b4[(b-c)+(a-b)]+c4(a-b)
= (b-c)(a4-b4)+(a-b)(c4-b4)
= (b-c)(a2-b2)(a2+b2)+(a-b)(c2-b2)(c2+b2)
= (b-c)(a-b)(a+b)(a^2+b^2)-(a-b)(b-c)(b+c)(b2+c2)
= (b-c)(a-b)(a3+ab2+ba2+b3-bc2-b3-cb2-c3)
= (b-c)(a-b)(a3+ab2+ba2-bc2-c3-cb2)
= (b-c)(a-b)(a3-c3)+b2(a-c)+b(a2-c2)
= (b-c)(a-b)(a-c)(a2+ac+c2)+b2(a-c)+b(a-c)(a+c)
= (b-c)(a-b)(a-c)(a2+ac+c2+b2+ab+ac)
= (a-b)(b-c)(c-a)(a2+b2+c2+ab+bc+ca)
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
1. (a2+b2+ab)2-a2b2-b2c2-c2a2
=a4+b4+a2b2+2(a2b2+ab3+a3b)-a2b2-b2c2-c2a2
=a4+b4+2a2b2+2ab3+2a3b-b2c2-c2a2
=(a2+b2)2+2ab(a2+b2)-c2(a2+b2)
=(a2+b2)[(a+b)2-c2]
=(a2+b2)(a+b+c)(a+b-c)
2. a4+b4+c4-2a2b2-2b2c2-2a2c2=(a2-b2-c2)2
3. a(b3-c3)+b(c3-a3)+c(a3-b3)
=ab3-ac3+bc3-ba3+ca3-cb3
=a3(c-b)+b3(a-c)+c3(b-a)
=a3(c-b)-b3(c-a)+c3(b-a)
=a3(c-b)-b3(c-b+b-a)+c3(b-a)
=a3(c-b)-b3(c-b)-b3(b-a)+c3(b-a)
=(c-b)(a-b)(a2+ab+b2)-(b-a)(b-c)(b2+bc+c2)
=(a-b)(c-b)(a2+ab+2b2+bc+c2)
4. a6-a4+2a3+2a2=a4(a+1)(a-1)+2a2(a+1)=(a+1)(a5-a4+2a2)=a2(a+1)(a3-a2+2)
5. (a+b)3-(a-b)3=(a+b-a+b)[(a+b)2+(a+b)(a-b)+(a-b)2]
=2b(3a2+b2)
6. x3-3x2+3x-1-y3=(x-1)3-y3=(x-1-y)[(x-1)2+(x-1)y+y2]
=(x-y-1)(x2+y2+xy-2x-y+1)
7. xm+4+xm+3-x-1=xm+3(x+1)-(x+1)=(x+1)(xm+3-1)
(Đúng nhớ like nhá !)
Minh Hải,Lê Thiên Anh,Nguyễn Huy Tú,Ace Legona,...giúp mk vs mai mk đi hk rùi