\(a+b+c=0\Rightarrow\left(a+b+c\right)^2=0\)
hay \(a^2+b^2+c^2+2\left(ab+bc+ca\right)=0\Leftrightarrow a^2+b^2+c^2=-2\left(ab+bc+ca\right)\)Ta có: \(a^2+b^2+c^2\ge0\) .Dấu "=" xảy ra \(\Leftrightarrow a=b=c=0\)
Suy ra \(ab+bc+ca=-\dfrac{a^2+b^2+c^2}{2}\le-\dfrac{0}{2}=0\)
Dấu "=" xảy ra \(\Leftrightarrow a^2=b^2=c^2=0\Leftrightarrow a=b=c=0\)
cho a, b, c >0. c/m rằng: bc/a+ca/b+ab/c>= a+b+c
Cho a;b;c khác0 và thỏa mãn:ab+bc+ca=0
Tính \(B=\dfrac{bc}{a^2}+\dfrac{ca}{b^2}+\dfrac{ab}{c^2}\)
Hung nguyen
\(A=\dfrac{4bc-a^2}{bc+2a^2}\\ B=\dfrac{4ca-b^2}{ca+2b^2}\\ C=\dfrac{4ab-c^2}{ab+2c^2}\\ \)
CMR: nếu a+b+c=0 thì A.B.C=1
Cho a,b,c >0 và a2 + b2 + c2 = 1 . CMR
\(\dfrac{1}{1-ab}+\dfrac{1}{1-bc}+\dfrac{1}{1-ca}\le\dfrac{9}{2}\)
Cho a,bc thuộc R . Cm
\(\dfrac{ab}{c}+\dfrac{bc}{a}+\dfrac{ca}{b}\ge a+b+c\left(vớia,b,c>0\right)\)
1. Cho a + b + c = 0. CM:
a/ a3 + b3 + c3 = 3abc.
b/ (ab + bc + ca)2 = a2b2 + b2c2 + c2a2.
c/ a4 + b4 + c4 = 2(ab + bc +ca)2.
2. Cho a + b + c + d = 0. CM:
a3 + b3 + c3 + d3 = 3(b + c)(ad - bc)
cho 1/(a^2-bc)+1/(b^2-ca)+1/(c^2-ca)=0
Tinh A= a/(a^2-bc)^2+b/(b^2-ca)^2+c/(c^2-ca)^2
cho ba số a, b, c thỏa mãn a+b+c=0
CMR a2+a2c-abc+b2c+b2=0
Cho ab,c thuộc R, CM:
\(\dfrac{a}{bc}+\dfrac{b}{ca}+\dfrac{c}{ab}\ge\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}\left(vớia,b,c>0\right)\)