toán nâng cao 8

V

vipboycodon

m2+n2+p2+q2+1m(n+p+q+1)m^2+n^2+p^2+q^2+1 \ge m(n+p+q+1)
<=> m2+n2+p2+q2+1mnmpmpm0m^2+n^2+p^2+q^2+1-mn-mp-mp-m \ge 0
<=> 4m2+4n2+4p2+4q2+44mn4mp4mq4m04m^2+4n^2+4p^2+4q^2+4-4mn-4mp-4mq-4m \ge 0
<=> (m24mn+4n2)+(m24mp+4p2)+(m24mq+4q2)+(m24m+4)0(m^2-4mn+4n^2)+(m^2-4mp+4p^2)+(m^2-4mq+4q^2)+(m^2-4m+4) \ge 0
<=> (m2n)2+(m2p)2+(m2q)2+(m2)20(m-2n)^2+(m-2p)^2+(m-2q)^2+(m-2)^2 \ge 0 (đúng)
Dấu " = " xảy ra khi { m=2nm=2pm=2qm=2\begin{matrix} m = 2n \\ m = 2p \\ m = 2q \\ m= 2 \end{matrix}
<=> m=2n=2p=2q=2m = 2n = 2p = 2q = 2
<=> n=p=q=1n = p = q = 1
 
Last edited by a moderator:
V

vipboycodon

m2+n2+p2+q2+1m(n+p+q+1)m^2+n^2+p^2+q^2+1 \ge m(n+p+q+1)
<=> m2+n2+p2+q2+1mnmpmpm0m^2+n^2+p^2+q^2+1-mn-mp-mp-m \ge 0

<=> (m24mn+n2)+(m24mp+p2)+(m24mq+q2)+(m24m+1)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

<=> (m2n)2+(m2p)2+(m2q)2+(m21)20(\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.
 
Last edited by a moderator:
Top Bottom