Let p , q , r are positive real numbers, such that 27 p q r ( p + q + r ) 3 and 3 p + 4 q + 5 r = 12 , then p 3 + q 4 + r 5 =

Question

Let $p, q, r$ are positive real numbers, such that $27 p q r \geq {(p + q + r)}^{3}$ and $3 p + 4 q + 5 r = 12$ , then $p^{3} + q^{4} + r^{5} =$

Smart Achievers · Accepted Answer

(1)

By AM-GM,

$\frac{p + q + r}{3} \geq \sqrt[3]{p q r}$

So,

${(p + q + r)}^{3} \geq 27 p q r$

But given,

$27 p q r \geq {(p + q + r)}^{3}$

Hence equality must hold:

${(p + q + r)}^{3} = 27 p q r$

Equality in AM-GM gives

$p = q = r$

Also,

$3 p + 4 q + 5 r = 12$

Therefore,

$3 p + 4 p + 5 p = 12$

$12 p = 12 \Rightarrow p = q = r = 1$

Thus,

$p^{3} + q^{4} + r^{5} = 1 + 1 + 1 = 3$

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