Question
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(A) 3(B) 6(C) 2(D) None of these
Step 1
Taking the cube root of both sides, we get $\sqrt[3]{27pqr} \geq p+q+r$. Simplifying the left side, we get $3\sqrt[3]{pqr} \geq p+q+r$. Show more…
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