Chứng minh: (a-b)(b-c)(c-a)(a^2+b^2+c^2+ab+bc+ca) khác 0 khi a;b;c là 3 số phân biệt
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Cho a+ b + c =0 (a,b,c khác 0). Chứng minh rằng a^2/bc+b^2/ca+c^2/ab-3=0
1) Phân tích nhân tử
a) a(b^2+c^2)+b(c^2+a^2)+c(a^2+b^2)+2abc
b) ab(a+b)-bc(b+c)-ca(c-a)
c) (x^2+x)^2+2(x^2+x)-3
2) Cho 3 số a,b,c khác 0 biết
ab(a-b)+bc(b-c)+ca(c-a)=0.Chứng minh a=b=c
Bài 1:
a ) a.( b2 + c2 ) + b.( a2 + c2 ) + c.( a2 + b2 ) + 2abc
= ab2 + ac2 + a2b + bc2 + a2c + b2c + 2abc
= ( ab2 + a2b ) + ( ac2 + bc2 ) + ( a2c + 2abc + b2c )
= ab.( a + b ) + c2.( a + b ) + c.( a2 + 2ab + b2 )
= ab.( a + b ) + c2.( a + b )v + c.( a + b)2
= ( a + b ).[ ( ab + c2 + c. ( a + b ) ]
= ( a + b ).( ab + c2 + ac + bc )
= ( a + b ).[ ( ab + ac ) + ( c2 + bc) ]
= ( a + b ).[ a.( b + c ) + c.( b + c ) ]
= ( a + b ).( b + c ).( a + c )
b) ab.( a + b ) - bc.( b + c ) + ac.( a - c )
= ab.( a + b ) - bc.( b + c ) + ac.[ ( a + b ) - ( b + c ) ]
= ab.( a + b ) - bc. ( b + c ) + ac.( a + b ) - ac.( b + c )
= ab.( a + b ) + ac.( a + b ) - bc.( b + c ) - ac.( b + c )
= ( a + b ).( ab + ac ) + ( b + c ).( -bc - ac )
= ( a + b ).a.( b + c ) - ( b + c ).c.( a + b )
= ( a + b ).( b + c ).( a - c )
c) ( x2 + x )2 + 2.( x2 + x ) - 3
Đặt x2 + x = a
Khi đó đa thức trở thành:
a2 + 2a - 3
= a2 + 3a - a - 3
= a.( a + 3 ) - ( a + 3 )
= ( a - 1 ).( a - 3 )
\(\Rightarrow\) ( x2 + x - 1 ).( x2 + x - 3 )
B2
ab.( a - b ) + bc.( b - c ) + ca.( c - a ) = 0
\(\Leftrightarrow\)ab.( a - b ) + bc.( b - c ) - ca.[ ( a - b ) + ( b - c ) ] = 0
\(\Leftrightarrow\)ab.( a - b ) + bc.( b - c ) - ca.( a - b ) - ca.( b - c ) = 0
\(\Leftrightarrow\)ab.( a - b ) - ca.( a - b ) + bc.( b - c ) - ca.( b - c ) = 0
\(\Leftrightarrow\) ( a - b ).( ab - ca ) + ( b - c ).( bc - ca ) = 0
\(\Leftrightarrow\) ( a - b ).a.( b - c ) - ( b - c ).c.( a - b ) = 0
\(\Leftrightarrow\) ( a - b ).( b - c ).( a - c ) = 0
\(\Leftrightarrow\) ( a - b ).( b - c ).( a - c ) = 0
\(\Leftrightarrow\) a = b , b = c , a = c
\(\Rightarrow\) a = b = c
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cho a,b,c>0, chứng minh:
1)ab+bc+ca >= a√ab+b√ca+c√ab
2)a^2+b^2+c^2 >= a√ab+b√ca+c√ab
1, Áp dụng BĐT cosi cho a,b,c>0
\(ab+bc\ge2\sqrt{ab^2c}=2b\sqrt{ac}\\ bc+ca\ge2\sqrt{abc^2}=2c\sqrt{ab}\\ ca+ab\ge2\sqrt{a^2bc}=2a\sqrt{bc}\)
Cộng VTV 3 BĐT trên:
\(\Leftrightarrow2\left(ab+bc+ac\right)\ge2\left(b\sqrt{ac}+a\sqrt{bc}+c\sqrt{ab}\right)\\ \Leftrightarrow ab+bc+ca\ge a\sqrt{bc}+b\sqrt{ac}+c\sqrt{ab}\)
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\(2,\)
Ta có
\(\left(a-b\right)^2+\left(b-c\right)^2+\left(c-a\right)^2\ge0\\ \Leftrightarrow2a^2+2b^2+2c^2-2ab-2ac-2bc\ge0\\ \Leftrightarrow a^2+b^2+c^2-ab-ac-bc\ge0\\ \Leftrightarrow a^2+b^2+c^2\ge ab+bc+ca\)
Áp dụng BĐT cm ở câu 1
Suy ra đpcm
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Cho các số thực a,b,c khác 0 thỏa mãn ab+bc+ca=1 và a2b+c=b2c+a=c2a+b. Chứng minh rằng a=b=c
Cho a,b,c khác 0 . Chứng minh: (ab+ac)/2=(ba+bc)/3=(cb+ca)/4 thì a/3=b/5=c/15
cho c^2 +2(ab -ac -bc ) =0 và b khác c, a+b khác 0. Chứng minh a^2 +(a-c)^2 /b^2+(b-c)^2 = a-c / b-c
\(a^2+b^2+c^2+2ab-2ac-2bc=a^2+b^2\)
\(\Rightarrow\left(a+b-c\right)^2=a^2+b^2\)
\(\Rightarrow\hept{\begin{cases}a^2=\left(a+b-c\right)^2-b^2=\left(a+b-c-b\right)\left(a+b-c+b\right)=\left(a-c\right)\left(a+2b-c\right)\\b^2=\left(a+b-c\right)^2-a^2=\left(a+b-c-a\right)\left(a+b-c+a\right)=\left(b-c\right)\left(2a+b-c\right)\end{cases}}\)
\(a^2+\left(a-c\right)^2=\left(a-c\right)\left(a+2b-c\right)+\left(a-c\right)^2\)
\(=\left(a-c\right)\left(a+2b-c+a-c\right)=2\left(a-c\right)\left(a+b-c\right)\)
\(b^2+\left(b-c\right)^2=\left(b-c\right)\left(2a+b-c\right)+\left(b-c\right)^2\)
\(=\left(b-c\right)\left(2a+b-c+b-c\right)=2\left(b-c\right)\left(a+b-c\right)\)
Vậy \(\frac{a^2+\left(a-c\right)^2}{b^2+\left(b-c\right)^2}=\frac{2\left(a-c\right)\left(a+b+c\right)}{2\left(b-c\right)\left(a+b+c\right)}=\frac{a-c}{b-c}\)
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cho a,b,c>=0, a+b+c=1. chứng minh rằng (a-bc)/(a+bc)+(b-ca)/(b+ca)+(c-ab)/(c+ab)<=3/2
\(\frac{a-bc}{a+bc}=\frac{a-bc}{a\left(a+b+c\right)+bc}=\frac{a-bc}{a^2+ab+bc+ca}=\frac{a-bc}{\left(a+b\right)\left(c+a\right)}\)
\(=\left(a-bc\right)\sqrt{\frac{1}{\left(a+b\right)^2\left(c+a\right)^2}}\le\frac{\frac{a-bc}{\left(a+b\right)^2}+\frac{a-bc}{\left(c+a\right)^2}}{2}=\frac{a-bc}{2\left(a+b\right)^2}+\frac{a-bc}{2\left(c+a\right)^2}\)
Tương tự, ta có: \(\frac{b-ca}{b+ca}\le\frac{b-ca}{2\left(b+c\right)^2}+\frac{b-ca}{2\left(a+b\right)^2}\)\(;\)\(\frac{c-ab}{c+ab}\le\frac{c-ab}{2\left(c+a\right)^2}+\frac{c-ab}{2\left(b+c\right)^2}\)
=> \(\frac{a-bc}{a+bc}+\frac{b-ca}{b+ca}+\frac{c-ab}{c+ab}\le\frac{a-bc+b-ca}{2\left(a+b\right)^2}+\frac{b-ca+c-ab}{2\left(b+c\right)^2}+\frac{a-bc+c-ab}{2\left(c+a\right)^2}\)
\(\frac{\left(a+b\right)\left(1-c\right)}{2\left(a+b\right)\left(1-c\right)}+\frac{\left(b+c\right)\left(1-a\right)}{2\left(b+c\right)\left(1-a\right)}+\frac{\left(c+a\right)\left(1-b\right)}{2\left(c+a\right)\left(1-b\right)}=\frac{3}{2}\)
Dấu "=" xảy ra \(\Leftrightarrow\)\(a=b=c=\frac{1}{3}\)
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Bài 1: Choa;b;c là các số khác 0 và a^2= bc; b^2= ab; c^2=ac.Cmr a=b=c
Bài2: Cho a;b;c là các số khác 0 thỏa mãn ab+ac/2=bc+ba/3=ca+cb/4. Chứng tỏ : a/3= b/5=c/15
Cho a, b, c>0
Chứng minh rằng: (a+b+c)^2\(\ge\) 3(ab+bc+ca) và ((a+b+c)^2/ab+bc+ca)+(ab+bc+ca/(a+b+c)^2)\(\ge\) 10/3
Biến đổi tương đương:
\(\left(a+b+c\right)^2\ge3\left(ab+ac+bc\right)\)
\(\Leftrightarrow a^2+b^2+c^2+2ab+2ac+2bc\ge3\left(ab+ac+bc\right)\)
\(\Leftrightarrow a^2+b^2+c^2-ab-ac-bc\ge0\)
\(\Leftrightarrow2a^2+2b^2+2c^2-2ab-2ac-2bc\ge0\)
\(\Leftrightarrow\left(a-b\right)^2+\left(a-c\right)^2+\left(b-c\right)^2\ge0\) (luôn đúng)
Dấu "=" xảy ra khi \(a=b=c\)
\(\Rightarrow\frac{\left(a+b+c\right)^2}{ab+ac+bc}\ge3\)
b/ \(VT=\frac{\left(a+b+c\right)^2}{ab+ac+bc}+\frac{ab+ac+bc}{\left(a+b+c\right)^2}=\frac{8\left(a+b+c\right)^2}{9\left(ab+ac+bc\right)}+\frac{\left(a+b+c\right)^2}{9\left(ab+ac+bc\right)}+\frac{ab+ac+bc}{\left(a+b+c\right)^2}\)
\(\Rightarrow VT\ge\frac{8\left(a+b+c\right)^2}{9\left(ab+ac+bc\right)}+2\sqrt{\frac{\left(a+b+c\right)^2\left(ab+ac+bc\right)}{9\left(ab+ac+bc\right)\left(a+b+c\right)^2}}\ge\frac{8.3}{9}+\frac{2}{3}=\frac{10}{3}\) (đpcm)
Dấu "=" xảy ra khi \(a=b=c\)
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