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  4. Let p, q, r ∈ R+and 27 p q r ≥(p+q+r)3 and 3 p+4 q+5 r=12 then p3+q4+r5 is equal to

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Q. Let $p, q, r \in R^{+}$and $27 p q r \geq(p+q+r)^{3}$ and $3 p+4 q+5 r=12$ then $p^{3}+q^{4}+r^{5}$ is equal to

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Solution:

$27 p q r \geq(p+q+r)^{3}$
$\Rightarrow (p q r)^{1 / 3} \geq \frac{p+q+r}{3} $
$\Rightarrow p=q=r$
Also, $3 p+4 q+5 r=12$
$\Rightarrow p=q=r=1$