If $a^{2} + b^{2} + c^{2} + d^{2} = 36$ , then

a b + b c + c d + d a \leq 36

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a b + b c + c d + d a \geq 36

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a b + b c + c d + d a \geq 18

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a b + b c + c d + d a \leq 18

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Solution

The correct option is A $a b + b c + c d + d a \leq 36$
$G i v e n a^{2} + b^{2} + c^{2} + d^{2} = 36$

Now,

$(a - b)^{2} \geq 0$

$(b - c)^{2} \geq 0$

$(c - d)^{2} \geq 0$

$(d - b)^{2} \geq 0$

$O r (a - b)^{2} + (b - c)^{2} + (c - d)^{2} + (d - b)^{2} \geq 0$

$a^{2} + b^{2} - 2 a b + b^{2} + c^{2} - 2 b c + c^{2} + d^{2} - 2 c d + d^{2} + a^{2} - 2 d a \geq 0$

$2 (a^{2} + b^{2} + c^{2} + d^{2}) - 2 (a b + b c + c d + d a) \geq 0$

$(a b + b c + c d + d a) \leq (a^{2} + b^{2} + c^{2} + d^{2})$

$(a b + b c + c d + d a) \leq 36$

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Similar questions

In the figure, prove that:

$(i) C D + D A + A B + B C > 2 A C$

$(i i) C D + D A + A B > B C$

Let ABCD be a quadrilateral with diagonals AC and BD. Prove the following statements (Compare these with the previous problem);

(a) AB + BC + CD > AD;

(b) AB + BC + CD + DA > 2AC;

(d) AB + BC + CD + DA > AC + BD.

In the adjoining figure, ABCD is a quadrilateral and AC is one of its diagonals. Prove that

(i) AB + BC + CD + DA > 2AC

(ii) AB + BC + CD > DA

(iii) AB + BC + CD + DA > AC + BD.

Why is ab+bc+cd+da a monomial?

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