Question

Let a, b, c, d, $ϵR^{+}$ and 256 abcd $\ge (a+b+c+d{)}^4$ and 3a+b+2c+5d=11 then $a^3+b+C^2+5d$ is equal to :-

Answer 1

Given $256$ abcd $\ge (a+b+c+d{)}^4$

$3a+b+2c+5d=11$ …………$\left(1\right)$

We know that

(Arithmetic mean) $\le$ (Geometric mean)

From equation $\left(1\right)$

$\left(\frac{a+a+a+b+c+c+d+d+d+d+d}{11}\right)\le (a^{3}b{c}^2{d}^5{)}^{1/4}$

$\left(\frac{3a+b+2c+5d}{11}\right)\le (a^{3}b{c}^2{d}^5{)}^{1/4}$

$1\le a^{3}b{c}^2{d}^5$ …………$\left(2\right)$

Now

For terms, $a^3,b,c^2,d^5$

$\frac{a^3+b+c^2+d^5}{4}\le (a^{3}b{c}^2{d}^5{)}^{1/4}$

$\left(\frac{a^3+b+c^2+d^5}{4}\right)\le a^{3}b{c}^2{d}^5$

From equation $\left(2\right)$

${\left(\frac{a^3+b+c^2+d^5}{4}\right)}^4=1$

$a^3+b+c^2+d^5=256$

Hence value of $a^3+b+c^2+d^5$ is $256$.