If a1- a2- a3- a4- a5 - 0 then a1a2-a3-a2a3-a4-a3a4-a5-a4a5-a1-a5a1-a2-52

The correct option is A Is trueIf we write, the expression as a_1^2a_1a_2+a_1a_3+a_2^2a_2a_3+a_2a_4+a_3^2a_3a_4+a_3a_5+a_4^2a_4a_5+a_4a_1+a_5^2a_5a_1+a_5a_ ...

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Byju's Answer

Standard XII

Mathematics

Variable Separable Method

If a1, a2, ...

Question

If $a_{1}, a_{2}, a_{3}, a_{4}, a_{5} > 0$ then
$\frac{a_{1}}{a_{2} + a_{3}} + \frac{a_{2}}{a_{3} + a_{4}} + \frac{a_{3}}{a_{4} + a_{5}} + \frac{a_{4}}{a_{5} + a_{1}} + \frac{a_{5}}{a_{1} + a_{2}} \geq \frac{5}{2}$

Is true

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Is false

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Is True only for some values of

a_{n}

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Nothing can be said for sure

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Solution

The correct option is A Is true
If we write, the expression as
$\frac{a_{1}^{2}}{a_{1} a_{2} + a_{1} a_{3}} + \frac{a_{2}^{2}}{a_{2} a_{3} + a_{2} a_{4}} + \frac{a_{3}^{2}}{a_{3} a_{4} + a_{3} a_{5}} + \frac{a_{4}^{2}}{a_{4} a_{5} + a_{4} a_{1}} + \frac{a_{5}^{2}}{a_{5} a_{1} + a_{5} a_{2}}$
Which we know is,
$\geq \frac{(a_{1} + a_{2} + a_{3} + a_{4} + a_{5})^{2}}{\sum a_{l} a_{m}}$
Now, we have
$(a_{1} + a_{2} + a_{3} + a_{4} + a_{5})^{2} = \sum (a_{n}^{2}) + 2 (a_{l} a_{m})$
If the condition in option (A) was True, then
$2 (a_{1}^{2} + a_{2}^{2} + a_{3}^{2} + a_{4}^{2} + a_{5}^{2}) + 4 \sum a_{l} a_{m} \geq 5 \sum a_{l} a_{m}$
which reduces to $\sum (a_{l} - a_{m})^{2} \geq 0$
Thus option (A) itself is correct.

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If a1,a2,a3,a4,a5>0 thena1a2+a3+a2a3+a4+a3a4+a5+a4a5+a1+a5a1+a2≥52

If $a_{1}, a_{2}, a_{3}, a_{4}, a_{5} > 0$ then
$\frac{a_{1}}{a_{2} + a_{3}} + \frac{a_{2}}{a_{3} + a_{4}} + \frac{a_{3}}{a_{4} + a_{5}} + \frac{a_{4}}{a_{5} + a_{1}} + \frac{a_{5}}{a_{1} + a_{2}} \geq \frac{5}{2}$