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# urgent pls help

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Let x, y  and z be positive real numbers. Find the minimum value of

May 19, 2019

#1
+7649
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$$P=(x+2y+4z)\left(\dfrac{4}{x}+\dfrac{2}y+\dfrac{1}z\right)\\ \;\;\;\!= 12 + \dfrac{2x}{y}+\dfrac{x}{z}+\dfrac{8y}{x}+\dfrac{2y}{z}+\dfrac{16z}{x}+\dfrac{8z}{y}\\ \text{WLOG, assume } x \le y \le z\\ \text{Let }u=\dfrac{x}{y}, v = \dfrac{y}{z}, w = \dfrac{x}{z} \implies u,v,w \in (0,1]\\ \quad P\\ =12 + 2u+w+\dfrac{8}{u}+2v+\dfrac{8}{v}+\dfrac{16}{w}\\ \quad \min P\\ = 12 + 10 + 10 + 17\\ = 49$$

Remarks: The minimum is attained when x = y = z = 1.

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May 20, 2019
edited by MaxWong  May 20, 2019
#2
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Hi Max

That is an interesting answer but...

Where did you get 10+10+17 from?

Melody  May 21, 2019