Question

If $a, b, c, d$ and $e$ are positive real numbers such that $a+b+c+d+e=15$ and $a b^2 c^3 d^4 e^5=(120)^3(50)$, then the value of $a^2+b^2+c^2+d^2+e^2$ is $\ldots \ldots \ldots$

Answer 1

We have given that,
$a+b+c+d+e=15$
$AM=\frac{\left[a + \frac{b}{2} + \frac{b}{2} + \frac{c}{3} + \frac{c}{3} + \frac{c}{3} + \frac{d}{4} + \frac{d}{4} + \frac{d}{4} + \frac{d}{4} + \frac{e}{5} + \frac{e}{5} + \frac{e}{5} + \frac{e}{5} + \frac{e}{5}\right]}{15}$
$AM=1$
$GM=\left(a \cdot \frac{b^{2}}{2^{2}} \cdot \frac{c^{3}}{3^{3}} \cdot \frac{d^{4}}{4^{4}} \cdot \frac{e^{5}}{5^{5}}\right)^{\frac{1}{15}}$
$G M=\frac{(120)^3 \cdot 50}{2^2 \cdot 3^3 \cdot 4^4 \cdot 5^5}=1$
$GM=1$
$\therefore AM=GM$
Hence, $a=\frac{b}{2}=\frac{c}{3}=\frac{d}{4}=\frac{e}{5}$
$\therefore a^{2}+b^{2}+c^{2}+d^{2}+e^{2}$
$=1^{2}+2^{2}+3^{2}+4^{2}+5^{2}$
$=\frac{5 \times 6 \times 11}{6}=55$