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$a,b,c,d,e$ 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 $e$.

My book tells me to use tchebycheff's inequality $$\left(\frac{a+b+c+d}{4}\right)^2\le \frac{a^2+b^2+c^2+d^2}{4}$$

But this not the Chebyshev's inequality given in wikipedia. Can someone state the actual name of the inequality so I can read more about it?

(I got $e\in\left[0,\frac{16}{5}\right]$ using the inequality)

Aditya Dev
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  • cauchy inequality – Chen Jiang Mar 15 '16 at 02:32
  • @ChenJiang how? that looks different from the one given here: http://mathworld.wolfram.com/CauchysInequality.html – Aditya Dev Mar 15 '16 at 02:35
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    take all $b_i$ to be $\frac{1}{4}$ – Chen Jiang Mar 15 '16 at 02:37
  • This is the Arithmetic Mean-Root Mean Square inequality (squared). You have equality in the arithmetic mean-root mean square inequality for the sum 8 and the product 16, because $(8/4)^2=4=16/4$, which happens if and only if all of $a,b,c,d,e$ are equal. so $e=16/5$ is the only value. – Sarvesh Ravichandran Iyer Mar 15 '16 at 02:38
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    It can also be considered applied Chebyshev's inequality. Perhaps you will recognise it in the form: $$\frac{a+b+c+d}4\cdot\frac{a+b+c+d}4\le \frac{a^2+b^2+c^2+d^2}4$$ Check https://en.m.wikipedia.org/wiki/Chebyshev%27s_sum_inequality – Macavity Mar 15 '16 at 02:45
  • @Macavity i couldnt understand much about chebyshev's inequality. (I just understood its something related to probability/statistics). What form did you specify? – Aditya Dev Mar 15 '16 at 02:50
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    @AdityaDev I gave above the link which has the Sum form. It relies on ordering and is at times a good shortcut to mean inequalities. – Macavity Mar 15 '16 at 02:51

3 Answers3

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As @ChenJiang stated, its a case of cauchy's inequality

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

$$(a+b+c+d)^2\le 4(a^2+b^2+c^2+d^2)$$ $$(8-e)^2\le 4(16-e^2)$$ $$5e^2-16e\le 0$$ $$e(5e-16)\le 0$$ $$\implies 0\le e\le \frac{16}{5}$$

Community
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Aditya Dev
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I think you may misunderstand the "Chebyshev inequality" to the inequality which is focused on the probability theory.

Apparently, here, the "Chebyshev inequality" refers to the "Chebyshev sum inequality," which should be stated as follows:

In mathematics, Chebyshev's sum inequality, named after Pafnuty Chebyshev, states that if \begin{aligned} &a_1\geq a_2\geq\cdots\geq a_n\quad\mathrm{~and~}\quad b_1\geq b_2\geq\cdots\geq b_n, \\ &\text{then} \\ &\frac1n\sum_{k=1}^na_kb_k\geq\left(\frac1n\sum_{k=1}^na_k\right)\biggl(\frac1n\sum_{k=1}^nb_k\biggr). \\ &\text{Similarly, if} \\ &a_1\leq a_2\leq\cdots\leq a_n\quad\text{ and }\quad b_1\geq b_2\geq\cdots\geq b_n, \\ &\text{then} \\ &\frac1n\sum_{k=1}^na_kb_k\leq\left(\frac1n\sum_{k=1}^na_k\right)\biggl(\frac1n\sum_{k=1}^nb_k\biggr).^{[1]} \end{aligned}

See https://en.wikipedia.org/wiki/Chebyshev%27s_sum_inequality;

Take the $a$ and $b$ as $\frac{a+b+c+d}{4}$, By using the Chebyshev sum inequality, one yields the hint in your book $$\left(\frac{a+b+c+d}{4}\right)^2\le \frac{a^2+b^2+c^2+d^2}{4}$$

D.y.s
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The book means the special case $x=y$, $n=4$ of the Chebyshev inequality $\overline{xy} \geq \bar{x} \bar{y}$ where $x,y$ are sequences of length $n$, both arranged in the same order (such as both increasing or both decreasing), and the bar means averaging.

zyx
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