A12-a22- an2b12-b22-bn2 -a1 b1-a2 b2-an bn2

We know that (a_1x + b_1)^2 is + ive i.e. ≥ 0 (being a perfect square) Similarly (a_2 x + b_2)^2 ≥ 0 …… (a_n x + b_n)^2 ≥ 0 Adding we get (a_1^2 + a_2^2 + …… +a ...

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Byju's Answer

Standard IX

Mathematics

Inequality

a12+a22+…… an...

Question

$(a_{1}^{2} + a_{2}^{2} + \dots \dots a_{n}^{2}) (b_{1}^{2} + b_{2}^{2} + \dots \dots + b_{n}^{2}) \geq (a_{1} b_{1} + a_{2} b_{2} + \dots \dots + a_{n} b_{n})^{2}$

True

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False

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Solution

We know that $(a_{1} x + b_{1})^{2}$ is + ive i.e. $\geq 0$
(being a perfect square)
Similarly $(a_{2} x + b_{2})^{2} \geq 0 \dots \dots (a_{n} x + b_{n})^{2} \geq 0$
Adding we get
$(a_{1}^{2} + a_{2}^{2} + \dots \dots + a_{n}^{2}) x^{2} + 2 x (a_{1} b_{1} + a_{2} b_{2} + \dots \dots a_{n} b_{n}) + (b_{1}^{2} + b_{2}^{2} + \dots \dots b_{n}^{2}) \geq 0 \dots \dots (1)$
L.H.S. is of the form $A x^{2} + B x + C$
The sign of this expression is the same as of 1st term which is +ive provided
$B^{2} - 4 A C < 0$
or $4 A C \geq B^{2}$
$4 (a_{1}^{2} + a_{2}^{2} + \dots \dots a_{n}^{2}) (b_{1}^{2} + b_{2}^{2} + \dots \dots b_{n}^{2}) \geq 4 (a_{1} b_{1} + a_{2} b_{2} + \dots \dots a_{n} b_{n})^{2} e t c$ .

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(a21+a22+……a2n)(b21+b22+……+b2n)≥(a1b1+a2b2+……+anbn)2

$(a_{1}^{2} + a_{2}^{2} + \dots \dots a_{n}^{2}) (b_{1}^{2} + b_{2}^{2} + \dots \dots + b_{n}^{2}) \geq (a_{1} b_{1} + a_{2} b_{2} + \dots \dots + a_{n} b_{n})^{2}$