Question

$\textsf{\scriptsize{a, b, c, d, e are real numbers such that :-}}$
$\quad \quad \bullet \sf{a + b + c + d + e = 8, \: and}$
$\quad \quad \bullet \sf{{a}^{2} + {b}^{2} + {c}^{2} + {d}^{2} + {e}^{2} = 16}$
$\textsf{What is the largest possible value of e?}$​

Answer 1

$\bf \red{\underline{\underline{Given:-}}}$

$\tt \bullet {a, b, c, d, e \ are \ real \ numbers}$

$\bullet \tt{a + b + c + d + e = 8, \: and}$

$\bullet \tt{{a}^{2} + {b}^{2} + {c}^{2} + {d}^{2} + {e}^{2} = 16}$

$\bf \orange{\underline{To \ Find:-}}$

• The largest possible value of e.

$\bf \purple{\underline{Solution:-}}$

We know,

$\tt From \ tchebycheff's \ inequality \ :$

$\\ \\ \implies \tt ( \frac{a + b + c + d}{4} ) {}^{2} \leqslant \frac{ {a}^{2} + {b}^{2} + {c}^{2} + {d}^{2} }{4}$

And given that ,

$\quad \quad \bullet \tt{a + b + c + d + e = 8, \: and}$

$\quad \quad \bullet \tt{{a}^{2} + {b}^{2} + {c}^{2} + {d}^{2} + {e}^{2} = 16}$

$\tt Now,\ substituting \ these\ values \ in\ the\ above \ equation,\ we \ get \ :$

$\\ \implies \tt \: ( \frac{8 - {e}^{2} }{4} ) {}^{2} \leqslant \frac{16 - {e}^{2} }{4}$

$\\ \\ \implies \tt \: 64 + {e}^{2} - 16e \: \leqslant 4(16 - {e}^{2} ) \\ \\ \\ \implies \tt \: 5 {e}^{2} - 16e \leqslant 0 \\ \\ \\ \implies \tt \: e(5e - 16) \leqslant 0 \\ \\ \\ \implies \tt 0 \leqslant e \leqslant \frac{16}{5}$

Therefore,

$\implies \tt The \ Range \ of \ e \: \in \: [0,16/5].$

$\tt Hence, \: the\: largest \: possible\:value\: of \: e\: is\: :$

$❥\boxed{\pink{16/5}}$

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