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 \(\ge\) C²

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 :

\(\left(a_1x-b_1\right)^2\ge0\Rightarrow a^2_1.x^2+2.a_1b_1x+b_1^2\ge0\)

\(\left(a_2x-b_2\right)^2\ge0\Rightarrow a^2_2.x^2+2.a_2b_2x+b_2^2\ge0\)

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\(\left(a_nx-b_n\right)^2\ge0\Rightarrow a^2_n.x^2+2.a_nb_nx+b_n^2\ge0\)

=> \(\left(a_1^2+a_2^2+........+a_n^2\right).x^2-2.\left(a_1b_1+a_2b_2+.....+a_nb_n\right)x+\left(b_1^2+b_2^2+.......+b_n^2\right)\ge0\)(1)

That mean \(A.x^2-2.C.x+B\ge0\)

Because (1) right with all x so that , change x = \(\dfrac{C}{A}\) into (1) , we have :

\(A.\dfrac{C^2}{A^2}-2.\dfrac{C^2}{A^2}+B\ge0\)

\(\Rightarrow AB-C^2\ge0\)

\(\Rightarrow AB\ge C^2\)

=> \(\left(a_1^2+a_2^2+....+a_n^2\right)\). \(\left(b_1^2+b_2^2+..........+b_n^2\right)\) \(\ge\left(a_1b_1+a_2b_2+......+a_nb_n\right)^2\)

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

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(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

Is there anyone have another way shorter than that way ???