Question

[KHANH QUYNH MAI PHAM]

\(\left(a^2+b^2+c^2\right)\left(x^2+y^2+z^2\right)\) \(\ge\left(ax+by+cz\right)^2\)  

Cm bt trên đúng

Answer 1

[Kiệt Nguyễn]

\(\left(a^2+b^2+c^2\right)\left(x^2+y^2+z^2\right)\)

\(=a^2x^2+a^2y^2+a^2z^2+b^2x^2+b^2y^2+b^2z^2+c^2x^2+c^2y^2+c^2z^2\)

\(\left(ax+by+cz\right)^2\)

\(=c^2z^2+2bcyz+2acxz+b^2y^2+2abxy+a^2x^2\)

\(\left(a^2+b^2+c^2\right)\left(a^2+b^2+c^2\right)\)\(\ge\left(ax+by+cz\right)^2\)

\(\Leftrightarrow a^2x^2+a^2y^2+a^2z^2+b^2x^2+b^2y^2+b^2z^2+c^2x^2+c^2y^2+c^2z^2\)

\(\ge c^2z^2+2bcyz+2acxz+b^2y^2+2abxy+a^2x^2\)

\(\Leftrightarrow a^2y^2+a^2z^2+b^2x^2+b^2z^2+c^2x^2+c^2y^2\)

\(\ge2bcyz+2acxz+2abxy\)

\(\Leftrightarrow a^2y^2+a^2z^2+b^2x^2+b^2z^2+c^2x^2+c^2y^2\)\(-2bcyz-2acxz-2abxy\ge0\)

\(\Leftrightarrow\left(a^2y^2-2abxy+b^2x^2\right)+\left(a^2z^2-2acxz+c^2x^2\right)\)

\(+\left(b^2z^2-2bcyz+c^2y^2\right)\ge0\)

\(\Leftrightarrow\left(ay-bx\right)^2+\left(az-cx\right)^2+\left(bz-cy\right)^2\ge0\)

(Điều trên đúng vì \(\hept{\begin{cases}\left(ay-bx\right)^2\ge0\\\left(az-cx\right)^2\ge0\\\left(bz-cy\right)^2\ge0\end{cases}}\))

Vậy\(\left(a^2+b^2+c^2\right)\left(x^2+y^2+z^2\right)\) \(\ge\left(ax+by+cz\right)^2\)