Монгол Бодлогын Сан

Заавар:

Бодолт: Тэнцэтгэл бишийг $B^2\leq A\cdot C$ хэлбэрт бичье. $$\begin{aligned} B^2&=\Big(\mathop{\int\limits_{0}^{1}}(apx^2+(bp+aq)x+bq)\,\mathrm{d}x\Big)^2=\\ &=\left(\dfrac{ap}{3}+\dfrac{bp+aq}{2}+bq\right)^2=\left(\left(\dfrac p3+\dfrac q2\right)a+\left(\dfrac p2+q\right)b\right)^2\\ &=\left(\dfrac p3+\dfrac q2\right)^2a^2+2ab \left(\dfrac p3+\dfrac q2\right)\left(\dfrac p2+q\right)+\left(\dfrac p2+q^2\right)b^2\nonumber \end{aligned}$$ болно. $$A=\mathop{\int\limits_{0}^{1}}(ax^2+2ab+b^2)\,\mathrm{d}x=\dfrac{a}{3}+ab+b^2$$

ба $$A\cdot C=\left(\dfrac a3+ab+b^2\right)\left(\dfrac p3+pq+q^2\right)$$

тул $AC-B^2=\dfrac{1}{12}a^2q^2-\dfrac16abpq+\dfrac{1}{12}b^2p^2=\dfrac{1}{12}(aq-bp)^2\geq 0$ гэдгээс $AC\geq B^2$ болж батлагдав.