Điều phải chứng minh tương đương với:
\(\left(ax+by\right)^2\le\left(a^2+b^2\right)\left(x^2+y^2\right)\\ \Leftrightarrow a^2x^2+b^2y^2+2axby\le a^2x^2+b^2x^2+a^2y^2+b^2y^2\Leftrightarrow2axby\le a^2y^2+b^2x^2\\ \Leftrightarrow a^2y^2+b^2x^2-2\cdot ay\cdot bx\ge0\\ \Leftrightarrow\left(ay-bx\right)^2\ge0\left(luon.dung\right)\)
Dấu = xảy ra <=> ay=bx
Cho a+b+c=0, x+y+z=0, a/x+b/y+c/z=0. CMR: \(ax^2+by^2+cz^2=0\)
CMR:4 số a,b,x,y có
\(\left(a^2+b^2\right)\left(x^2+y^2\right)\ge\left(ax+by\right)^2\)
\(Cho\) \(ax+by+cz=0;a+b+c=\dfrac{1}{2018}\) . CMR: \(\dfrac{ax^{2\:}+by^2+cz^2}{bc\left(y-z\right)^2+ac\left(x-z\right)^2+ab\left(x-y\right)^2}=2018\)
CMR: nếu: x/a =y/b = z/c thì ( x2 +y2 + z2).( a2 +b2 +c2) = (ax+by + cz)2
Cho \(a+b+c=x+y+z=\dfrac{a}{x}+\dfrac{b}{y}+\dfrac{c}{z}\)
CMR: \(ax^2+by^2+cz^2=0\)
Cho: \(\dfrac{a}{x}=\dfrac{b}{y}=\dfrac{c}{z}\)và x, y, z khác 0
CMR: \(\dfrac{x^2+y^2+z^2}{\left(ax+by+cz\right)^2}=\dfrac{1}{a^2+b^2+c^2}\)
CMR:
a,\(a^2+b^2+2\ge2\left(a+b\right)\)
b,\(\left(a^2+b^2\right)\left(x^2+y^2\right)\ge\left(ax+by\right)^2\)
Cho (ax+by+cz)2 = (a2+b2+c2)(x2+y2+z2)
CMR:\(\dfrac{x}{a}=\dfrac{y}{b}=\dfrac{z}{c}\)
cho: a + b + c = x + y + z = \(\frac{a}{x}+\frac{b}{y}+\frac{c}{z}=0\)
CMR: ax2 + by2 + cz2 = 0
