hreury

Áp dụng bất đẳng thức Bunhiacopxki cho cặp 3 số ta có:

\(\left[\left(\dfrac{x}{\sqrt{a}}\right)^2+\left(\dfrac{y}{\sqrt{b}}\right)^2+\left(\dfrac{z}{\sqrt{c}}\right)^2\right]\left[\sqrt{a}^2+\sqrt{b}^2+\sqrt{c}^2\right]\ge\left[\dfrac{x}{\sqrt{a}}\cdot\sqrt{a}+\dfrac{y}{\sqrt{b}}\cdot\sqrt{b}+\dfrac{z}{\sqrt{c}}\cdot\sqrt{c}\right]^2=\left(x+y+z\right)^2\)

Dấu = xảy ra khi x/a=y/b=z/c

\(\dfrac{x^2}{a}\) + \(\dfrac{y^2}{b}\) + \(\dfrac{z^2}{c}\)≥ \(\dfrac{\left(x+y+z\right)^2}{a+b+c}\)

\(a+b+c=0\Rightarrow\left\{{}\begin{matrix}a^2=\left(b+c\right)^2\\b^2=\left(c+a\right)^2\\c^2=\left(a+b\right)^2\end{matrix}\right.\)

\(P=a^2x+b^2y+c^2z=\left(b+c\right)^2x+\left(c+a\right)^2y+\left(a+b\right)^2z\)\(=\left(b^2x+c^2x+c^2y+a^2y+a^2z+b^2z\right)+2\left(bcx+acy+abz\right)\)\(=a^2\left(y+z\right)+b^2\left(z+x\right)+c^2\left(x+y\right)+2\left(bcx+acy+abz\right)=0\)ta có: \(\frac{x}{a}+\frac{y}{b}+\frac{z}{c}=0\Leftrightarrow xbc+ayc+abz=0\)

\(\Rightarrow P=-a^2x-b^2y-c^2z\)

\(\Rightarrow a^2x+b^2y+c^2z=-\left(a^2x+b^2y+c^2z\right)\Rightarrow2\left(a^2x+b^2y+c^2z\right)=0\Rightarrow P=0\)

Đáp án đúng : D

ta có: a+b+c=1 
<=>(a+b+c)^2=1 
<=>ab+bc+ca=0 (1) 
mặt khác: áp dụng tính chất dãy tỉ số bằng nhau ta có: 
x/a=y/b=z/c=(x+y+z)/(a+b+c)=x+y+z 
<=> x=a(x+y+z) ; y=b(x+y+z) ; z=c(x+y+z) 
=>xy+yz+zx=ab(x+y+z)^2+bc(x+y+z)^2+ca(x... 
<=>xy+yz+zx=(ab+bc+ca)(x+y+z)^2 (2) 
từ (1) và (2) ta có đpcm 
Chúc bạn học giỏi!

:3

Ta có \(\dfrac{a}{x}+\dfrac{b}{y}+\dfrac{c}{z}=0\Leftrightarrow ayz+bzx+cxy=0\).

Do đó: \(ax^2+by^2+cz^2=\left(ax+by+cz\right)\left(x+y+z\right)-axy-axz-byz-byx-czx-czy=0-xy\left(a+b\right)-yz\left(b+c\right)-zx\left(c+a\right)=0+xyc+yza+zxb=0\).