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Question Number 195331 by Shlock last updated on 30/Jul/23

	Answered by mr W last updated on 06/Aug/23
	$say a=pc, b=qc (p+q+1)((1/p)+(1/q)+1)=10 ((a+b)/c)=p+q (p+q)_(max) =? due to symmetry: at maximum, p=q (2p+1)((2/p)+1)=10 (2p+1)(p+2)=10p 2p^2 −5p+2=0 ⇒p={2,(1/2)} ⇒(p+q)_(max) =2×2=4 ✓ ⇒(p+q)_(min) =2×(1/2)=1$
	$s a y a = p c, b = q c$ $(p + q + 1) (\frac{1}{p} + \frac{1}{q} + 1) = 10$ $\frac{a + b}{c} = p + q$ ${(p + q)}_{m a x} = ?$ $d u e t o s y m m e t r y : a t m a x i m u m, p = q$ $(2 p + 1) (\frac{2}{p} + 1) = 10$ $(2 p + 1) (p + 2) = 10 p$ $2 p^{2} - 5 p + 2 = 0$ $\Rightarrow p = {2, \frac{1}{2}}$ $\Rightarrow {(p + q)}_{m a x} = 2 \times 2 = 4 ✓$ $\Rightarrow {(p + q)}_{m i n} = 2 \times \frac{1}{2} = 1$
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