toán nâng cao 8

V

vipboycodon

$m^2+n^2+p^2+q^2+1 \ge m(n+p+q+1)$
<=> $m^2+n^2+p^2+q^2+1-mn-mp-mp-m \ge 0$
<=> $4m^2+4n^2+4p^2+4q^2+4-4mn-4mp-4mq-4m \ge 0$
<=> $(m^2-4mn+4n^2)+(m^2-4mp+4p^2)+(m^2-4mq+4q^2)+(m^2-4m+4) \ge 0$
<=> $(m-2n)^2+(m-2p)^2+(m-2q)^2+(m-2)^2 \ge 0$ (đúng)
Dấu " = " xảy ra khi { $\begin{matrix} m = 2n \\ m = 2p \\ m = 2q \\ m= 2 \end{matrix}$
<=> $m = 2n = 2p = 2q = 2$
<=> $n = p = q = 1$
 
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V

vipboycodon

$m^2+n^2+p^2+q^2+1 \ge m(n+p+q+1)$
<=> $m^2+n^2+p^2+q^2+1-mn-mp-mp-m \ge 0$

<=> $(\dfrac{m^2}{4}-mn+n^2)+(\dfrac{m^2}{4}-mp+p^2)+(\dfrac{m^2}{4}-mq+q^2)+(\dfrac{m^2}{4}-m+1) \ge 0$

<=> $(\dfrac{m}{2}-n)^2+(\dfrac{m}{2}-p)^2+(\dfrac{m}{2}-q)^2+(\dfrac{m}{2}-1)^2 \ge 0$ (đúng)
Chém chơi cho giảm bài tồn đọng.
 
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