Question

# Suppose p, q, r are real numbers such that q=p(4âˆ’p),r=q(4âˆ’q),p=r(4âˆ’r). The maximum possible value of p+q+r is?

A
0
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B
3
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C
9
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D
27
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Solution

## The correct option is C 9q=p(4−p) ...(i)r=q(4−q) ...(ii)p=r(4−r) ...(iii)multiplying by pr in (i) by pq in (ii) and by qr in (iii).we getpqr=p2r(4−p) ...(iv)pqr=q2p(4−q) ...(v)pqr=r2q(4−r) ...(vi)multiplying eq. (iv),(v)and(vi)(pqr)3=(pqr)3(4−p)(4−q)(4−r)1=(4−p)(4−q)(4−r) ...(vii)we know that Arithmatic Mean is always greater or equal to Geometric Mean.(4−p)+(4−q)+(4−r)3≥3√(4−p)(4−q)(4−r)from eq(vii)(4−p)+(4−q)+(4−r)3≥3√1(4−p)+(4−q)+(4−r)3≥1(12−(p+q+r)3≥112≥(p+q+r)+312−3≥(p+q+r)9≥(p+q+r)i.e. p+q+r≤9so maximum value of p+q+r is 9.

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