Let a, b, c, d in R^+ such that 256 abcd geq | vedclass

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(B) By the Arithmetic Mean-Geometric Mean Inequality ($AM$-$GM$),we have $\frac{a+b+c+d}{4} \geq (abcd)^{1/4}$.Raising both sides to the power of $4$,

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$8$

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(B) By the Arithmetic Mean-Geometric Mean Inequality ($AM$-$GM$),we have $\frac{a+b+c+d}{4} \geq (abcd)^{1/4}$.Raising both sides to the power of $4$,we get $\frac{(a+b+c+d)^4}{256} \geq abcd$,which implies $(a+b+c+d)^4 \leq 256 abcd$.The given condition is $256 abcd \geq (a+b+c+d)^4$,which is the reverse of the standard $AM$-$GM$ inequality.Equality holds if and only if $a = b = c = d$.Given the equation $3a + b + 2c + 5d = 11$,and substituting $a = b = c = d = k$,we get $3k + k + 2k + 5k = 11$,which simplifies to $11k = 11$,so $k = 1$.Thus,$a = b = c = d = 1$.Substituting these values into the expression $a^3 + b + c^2 + 5d$,we get $1^3 + 1 + 1^2 + 5(1) = 1 + 1 + 1 + 5 = 8$.

Let $a, b, c, d \in R^+$ such that $256 abcd \geq (a+b+c+d)^4$ and $3a + b + 2c + 5d = 11$. Then,the value of $a^3 + b + c^2 + 5d$ is equal to:

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