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1234thhc minhtoannmt 19/12/2017 at 18:48
I don't know
Flash Shit :3
22/08/2017 at 12:24
Denote : \(A=a_1^2+a_2^2+....+a_n^2\)
\(B=b_1^2+b_2^2+.....+b_n^2\)
\(C=\left(a_1b_1+a_2b_2+........+a_nb_n\right)^2\)
Need to prove : AB \(\ge\) C^{2}
If A = 0 so that \(a_1=a_2=....=a_n\), inequality is proven. Similar with B = 0 , so we need to consider if A and B different to 0
With every x , we have got :
\(\left(a_1xb_1\right)^2\ge0\Rightarrow a^2_1.x^2+2.a_1b_1x+b_1^2\ge0\)
\(\left(a_2xb_2\right)^2\ge0\Rightarrow a^2_2.x^2+2.a_2b_2x+b_2^2\ge0\)
.............................................
\(\left(a_nxb_n\right)^2\ge0\Rightarrow a^2_n.x^2+2.a_nb_nx+b_n^2\ge0\)
=> \(\left(a_1^2+a_2^2+........+a_n^2\right).x^22.\left(a_1b_1+a_2b_2+.....+a_nb_n\right)x+\left(b_1^2+b_2^2+.......+b_n^2\right)\ge0\)(1)
That mean \(A.x^22.C.x+B\ge0\)
Because (1) right with all x so that , change x = \(\dfrac{C}{A}\) into (1) , we have :
\(A.\dfrac{C^2}{A^2}2.\dfrac{C^2}{A^2}+B\ge0\)
\(\Rightarrow ABC^2\ge0\)
\(\Rightarrow AB\ge C^2\)
=> \(\left(a_1^2+a_2^2+....+a_n^2\right)\). \(\left(b_1^2+b_2^2+..........+b_n^2\right)\) \(\ge\left(a_1b_1+a_2b_2+......+a_nb_n\right)^2\)
Selected by MathYouLike 
KEITA FC 8C 19/12/2017 at 12:47
Denote : A=a21+a22+....+a2nA=a12+a22+....+an2
B=b21+b22+.....+b2nB=b12+b22+.....+bn2
C=(a1b1+a2b2+........+anbn)2C=(a1b1+a2b2+........+anbn)2
Need to prove : AB ≥≥ C2
If A = 0 so that a1=a2=....=ana1=a2=....=an, inequality is proven. Similar with B = 0 , so we need to consider if A and B different to 0
With every x , we have got :
(a1x−b1)2≥0⇒a21.x2+2.a1b1x+b21≥0(a1x−b1)2≥0⇒a12.x2+2.a1b1x+b12≥0
(a2x−b2)2≥0⇒a22.x2+2.a2b2x+b22≥0(a2x−b2)2≥0⇒a22.x2+2.a2b2x+b22≥0
.............................................
(anx−bn)2≥0⇒a2n.x2+2.anbnx+b2n≥0(anx−bn)2≥0⇒an2.x2+2.anbnx+bn2≥0
=> (a21+a22+........+a2n).x2−2.(a1b1+a2b2+.....+anbn)x+(b21+b22+.......+b2n)≥0(a12+a22+........+an2).x2−2.(a1b1+a2b2+.....+anbn)x+(b12+b22+.......+bn2)≥0(1)
That mean A.x2−2.C.x+B≥0A.x2−2.C.x+B≥0
Because (1) right with all x so that , change x = CACA into (1) , we have :
A.C2A2−2.C2A2+B≥0A.C2A2−2.C2A2+B≥0
⇒AB−C2≥0⇒AB−C2≥0
⇒AB≥C2⇒AB≥C2
=> (a21+a22+....+a2n)(a12+a22+....+an2). (b21+b22+..........+b2n)(b12+b22+..........+bn2) ≥(a1b1+a2b2+......+anbn)2

Flash Shit :3 22/08/2017 at 13:04
Is there anyone have another way shorter than that way ???
Lê Quốc Trần Anh Coodinator
16/11/2017 at 17:50KEITA FC 8C
08/12/2017 at 21:33
The answer is :
\(\widehat{VEZ}=\widehat{AED}\widehat{AEV}\widehat{DEZ}=\dfrac{180^0.3}{5}45^045^0=18^0\)