Let p, q, r in R^+. If 27pqr geq (p + q + r)^ | vedclass

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(A) Given the inequality $27pqr \geq (p + q + r)^3$,we can rewrite it as $\sqrt[3]{pqr} \geq \frac{p + q + r}{3}$.By the Arithmetic Mean-Geometric Mea

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$3$

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(A) Given the inequality $27pqr \geq (p + q + r)^3$,we can rewrite it as $\sqrt[3]{pqr} \geq \frac{p + q + r}{3}$.By the Arithmetic Mean-Geometric Mean ($AM$-$GM$) inequality,we know that $\frac{p + q + r}{3} \geq \sqrt[3]{pqr}$.For both inequalities to hold simultaneously,we must have $p = q = r$.Given the equation $3p + 4q + 5r = 12$,substitute $p = q = r = x$:$3x + 4x + 5x = 12$ $\Rightarrow 12x = 12$ $\Rightarrow x = 1$.Thus,$p = 1, q = 1, r = 1$.Now,calculate $p^3 + q^4 + r^5 = 1^3 + 1^4 + 1^5 = 1 + 1 + 1 = 3$.

Let $p, q, r \in R^+$. If $27pqr \geq (p + q + r)^3$ and $3p + 4q + 5r = 12$,then find the value of $p^3 + q^4 + r^5$.

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