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Question

If $$a_1, a_2, a_3, a_4, a_5 > 0$$ then
$$\dfrac{a_1}{a_2+a_3}+\dfrac{a_2}{a_3+a_4}+\dfrac{a_3}{a_4+a_5}+\dfrac{a_4}{a_5+a_1}+\dfrac{a_5}{a_1+a_2} \ge \dfrac{5}{2}$$


A
Is true
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B
Is false
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C
Is True only for some values of an
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D
Nothing can be said for sure
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Solution

The correct option is A Is true
If we write, the expression as
$$\dfrac{a_1^2}{a_1a_2+a_1a_3}+\dfrac{a_2^2}{a_2a_3+a_2a_4}+\dfrac{a_3^2}{a_3a_4+a_3a_5}+\dfrac{a_4^2}{a_4a_5+a_4a_1}+\dfrac{a_5^2}{a_5a_1+a_5a_2}$$
Which we know is, 
$$\ge \dfrac{(a_1+a_2+a_3+a_4+a_5)^2}{\sum a_la_m}$$
Now, we have
$$(a_1+a_2+a_3+a_4+a_5)^2=\sum (a_n^2) + 2(a_la_m)$$
If the condition in option (A) was True, then
$$2(a_1^2+a_2^2+a_3^2+a_4^2+a_5^2) + 4 \sum a_la_m \ge 5 \sum a_la_m$$
which reduces to $$\sum (a_l - a_m)^2 \ge 0$$
Thus option (A) itself is correct.

Mathematics

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