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

$a,b,c,d,e$ 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)

• cauchy inequality Mar 15, 2016 at 2:32
• @ChenJiang how? that looks different from the one given here: mathworld.wolfram.com/CauchysInequality.html Mar 15, 2016 at 2:35
• take all $b_i$ to be $\frac{1}{4}$ Mar 15, 2016 at 2:37
• 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 en.m.wikipedia.org/wiki/Chebyshev%27s_sum_inequality Mar 15, 2016 at 2:45
• @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. Mar 15, 2016 at 2:51

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

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

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

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).^{} \end{aligned}

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

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.