Let a- b- c- d - R- and 256 a b c d - a - b - c - d4 and 3a - b - 2c - 5d - 11 then a3 - b - c2 - 5d is

Given: nbsp;256 a b c d ≥ (a + b + c + d)^4 ⇒ (abcd)^1/4≥(a + b + c + d)/4 We know that A.M ≥ G.M (a + b + c + d)/4≥ (abcd)^1/4 So both expression satisfies o ...

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Standard XI

Mathematics

AM,GM,HM Inequality

Let a, b, c, ...

Question

Let $a, b, c, d \in R^{+}$ and $256 a b c d \geq (a + b + c + d)^{4}$ and $3 a + b + 2 c + 5 d = 11$ then $a^{3} + b + c^{2} + 5 d$ is

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Solution

The correct option is B $8$
Given: $256 a b c d \geq (a + b + c + d)^{4} \Rightarrow (a b c d)^{\frac{1}{4}} \geq \frac{(a + b + c + d)}{4}$
We know that $A . M \geq G . M$
$\frac{(a + b + c + d)}{4} \geq (a b c d)^{\frac{1}{4}}$
So both expression satisfies only when $A . M = G . M$
$∴ a = b = c = d$
We know that
$3 a + b + 2 c + 5 d = 11 \Rightarrow a = 1 = b = c = d$
Hence the value of the expression
$a^{3} + b + c^{2} + 5 d = 8$

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Similar questions

Q. Let

a, b, c, d \in R^{+}

and

256 a b c d \geq (a + b + c + d)^{4}

and

3 a + b + 2 c + 5 d = 11

then

a^{3} + b + c^{2} + 5 d

Q. Let

a, b, c, d \in R^{+}

and

256 a b c d \geq (a + b + c + d)^{4}

and

3 a + b + 2 c + 5 d = 11

then

a^{3} + b + c^{2} + 5 d

Q. Let a, b, c, d,

ϵ R^{+}

and 256 abcd

\geq (a + b + c + d)^{4}

and 3a+b+2c+5d=11 then

a^{3} + b + C^{2} + 5 d

is equal to :-

Q. If

2200 = 2^{a} \times 5^{b} \times 11^{c} \times 3^{d},

then

a + b - c + d =

Q. Given

\frac{a}{b} = \frac{c}{d}

, prove that

\frac{3 a - 5 b}{3 a + 5 b} = \frac{3 c - 5 d}{3 c + 5 d}

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Let a,b,c,d∈R+ and 256abcd≥(a+b+c+d)4 and 3a+b+2c+5d=11 then a3+b+c2+5d is

The correct option is B 8Given: 256abcd≥(a+b+c+d)4⇒(abcd)14≥(a+b+c+d)4 We know that A.M≥G.M (a+b+c+d)4≥(abcd)14 So both expression satisfies only when A.M=G.M ∴a=b=c=d We know that 3a+b+2c+5d=11⇒a=1=b=c=d Hence the value of the expression a3+b+c2+5d=8

Let $a, b, c, d \in R^{+}$ and $256 a b c d \geq (a + b + c + d)^{4}$ and $3 a + b + 2 c + 5 d = 11$ then $a^{3} + b + c^{2} + 5 d$ is