If $a, b, c, d, e$ are positive real numbers, such that $a + b + c + d + e = 8$ and $a^{2} + b^{2} + c^{2} + d^{2} + e^{2} = 16$ , find the range of $e$ .

e ϵ [0, \frac{16}{5}]

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e ϵ [- 5, \frac{16}{5}]

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e ϵ [0, \frac{4}{5}]

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e ϵ [0, \frac{16}{25}]

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Solution

The correct option is A $e ϵ [0, \frac{16}{5}]$
As we know
${(\frac{a + b + c + d}{4})}^{2} \leq \frac{a^{2} + b^{2} + c^{2} + d^{2}}{4}$ (i)(Using Tchebycheff's Inequality)
Where $a + b + c + d + e = 8$ and $a^{2} + b^{2} + c^{2} + d^{2} + e^{2} = 16$
$∴$ Eq. (i), reduces to
${(\frac{8 - e}{4})}^{2} \leq \frac{16 - e^{2}}{4}$
$\Rightarrow$ $64 + e^{2} - 16 e \leq 4 (16 - e^{2})$
$\Rightarrow$ $5 e^{2} - 16 e \leq 0$
$\Rightarrow$ $e (5 e - 16) \leq 0$ (Using number line rule)
$\Rightarrow$ $0 \leq e \leq \frac{16}{5}$
Thus, range of $e ϵ [0, \frac{16}{5}]$

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