$\begin{array}{r}x=cy+bz\\ y=az+cx\\ z=bx+ay\end{array}$

where $x,y,z$ are not all zero, prove that ${a}^{2}+{b}^{2}+{c}^{2}+2abc=1$

Further if at least one of $a,b,c$ is a proper fraction, prove that:

(i) ${a}^{2}+{b}^{2}+{c}^{2}<3$

(ii) $abc>-1$

Mayra Berry
2022-06-27
Answered

If

$\begin{array}{r}x=cy+bz\\ y=az+cx\\ z=bx+ay\end{array}$

where $x,y,z$ are not all zero, prove that ${a}^{2}+{b}^{2}+{c}^{2}+2abc=1$

Further if at least one of $a,b,c$ is a proper fraction, prove that:

(i) ${a}^{2}+{b}^{2}+{c}^{2}<3$

(ii) $abc>-1$

$\begin{array}{r}x=cy+bz\\ y=az+cx\\ z=bx+ay\end{array}$

where $x,y,z$ are not all zero, prove that ${a}^{2}+{b}^{2}+{c}^{2}+2abc=1$

Further if at least one of $a,b,c$ is a proper fraction, prove that:

(i) ${a}^{2}+{b}^{2}+{c}^{2}<3$

(ii) $abc>-1$

You can still ask an expert for help

Kaydence Washington

Answered 2022-06-28
Author has **32** answers

Assuming $a$ is a proper fraction, then

${a}^{2}<1$

Considering quadratic in $c$,

${c}^{2}+2abc+{a}^{2}+{b}^{2}-1=0$

Admitting real value of $c$,$\begin{array}{rl}\mathrm{\Delta}& \ge 0\\ 4({a}^{2}-1)({b}^{2}-1)& \ge 0\\ {b}^{2}-1& \le 0\end{array}$

Hence,

${b}^{2}\le 1$

Similarly,

${c}^{2}\le 1$

Hence,

${a}^{2}+{b}^{2}+{c}^{2}<3$

Also

$abc=\frac{1-{a}^{2}-{b}^{2}-{c}^{2}}{2}>-1$

${a}^{2}<1$

Considering quadratic in $c$,

${c}^{2}+2abc+{a}^{2}+{b}^{2}-1=0$

Admitting real value of $c$,$\begin{array}{rl}\mathrm{\Delta}& \ge 0\\ 4({a}^{2}-1)({b}^{2}-1)& \ge 0\\ {b}^{2}-1& \le 0\end{array}$

Hence,

${b}^{2}\le 1$

Similarly,

${c}^{2}\le 1$

Hence,

${a}^{2}+{b}^{2}+{c}^{2}<3$

Also

$abc=\frac{1-{a}^{2}-{b}^{2}-{c}^{2}}{2}>-1$

asked 2021-02-25

Graph the solution set for the system of linear inequalities

asked 2022-06-20

Let $u,v:[0,\mathrm{\infty})\to [0,\mathrm{\infty})$ satsfying the following system of differential inequalities:

${u}^{\prime}(t)\le {a}_{1}\phantom{\rule{thinmathspace}{0ex}}u(t)+{a}_{2}\phantom{\rule{thinmathspace}{0ex}}v(t)+{a}_{0}\phantom{\rule[-4pt]{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}{v}^{\prime}(t)\le {b}_{1}\phantom{\rule{thinmathspace}{0ex}}u(t)+{b}_{2}\phantom{\rule{thinmathspace}{0ex}}v(t)+{b}_{0}$

for suitable coefficients ${a}_{0},{a}_{1},{a}_{2},{b}_{0},{b}_{1},{b}_{2}\in \mathbb{R}\phantom{\rule{thinmathspace}{0ex}}.$ In particular I have ${a}_{1},{b}_{2}<0\phantom{\rule{thinmathspace}{0ex}}$ and $0<{b}_{1}<|{b}_{2}|\phantom{\rule{thinmathspace}{0ex}}$

Is there a Gronwall lemma for this system of linear differential inqualities? Namely an (optimal) inequality of type

$u(t)\le F(t)\phantom{\rule{0ex}{0ex}}v(t)\le G(t)$

where the functions $F,G:[0,\mathrm{\infty})\to [0,\mathrm{\infty})$ depend on $u,v$ only through their initial values $u(0),v(0)$?

${u}^{\prime}(t)\le {a}_{1}\phantom{\rule{thinmathspace}{0ex}}u(t)+{a}_{2}\phantom{\rule{thinmathspace}{0ex}}v(t)+{a}_{0}\phantom{\rule[-4pt]{0ex}{0ex}}\phantom{\rule{0ex}{0ex}}{v}^{\prime}(t)\le {b}_{1}\phantom{\rule{thinmathspace}{0ex}}u(t)+{b}_{2}\phantom{\rule{thinmathspace}{0ex}}v(t)+{b}_{0}$

for suitable coefficients ${a}_{0},{a}_{1},{a}_{2},{b}_{0},{b}_{1},{b}_{2}\in \mathbb{R}\phantom{\rule{thinmathspace}{0ex}}.$ In particular I have ${a}_{1},{b}_{2}<0\phantom{\rule{thinmathspace}{0ex}}$ and $0<{b}_{1}<|{b}_{2}|\phantom{\rule{thinmathspace}{0ex}}$

Is there a Gronwall lemma for this system of linear differential inqualities? Namely an (optimal) inequality of type

$u(t)\le F(t)\phantom{\rule{0ex}{0ex}}v(t)\le G(t)$

where the functions $F,G:[0,\mathrm{\infty})\to [0,\mathrm{\infty})$ depend on $u,v$ only through their initial values $u(0),v(0)$?

asked 2022-06-21

How can we solve this system of linear inequalities?

Let ${c}_{i}$ be a given non-negative integer for all $i\in \{1,\dots ,n\}$. would like to find the non-negative integers ${a}_{i}$ and ${b}_{i}$ for all $i\in \{1,\dots ,n\}$ such that:

$\begin{array}{r}\{\begin{array}{ll}{c}_{i}={a}_{i}+{b}_{i},& \text{for all}i\in \{1,\dots ,n\}\\ {a}_{i}{a}_{i+1},& \text{for all}i\in \{1,\dots ,n-1\}\\ {b}_{i}{b}_{i+1},& \text{for all}i\in \{1,\dots ,n-1\}\\ {b}_{i}\ge 0\text{and integer},& \text{for all}i\in \{1,\dots ,n\}\\ {a}_{i}\ge 0\text{and integer},& \text{for all}i\in \{1,\dots ,n\}\end{array}\end{array}$

Let ${c}_{i}$ be a given non-negative integer for all $i\in \{1,\dots ,n\}$. would like to find the non-negative integers ${a}_{i}$ and ${b}_{i}$ for all $i\in \{1,\dots ,n\}$ such that:

$\begin{array}{r}\{\begin{array}{ll}{c}_{i}={a}_{i}+{b}_{i},& \text{for all}i\in \{1,\dots ,n\}\\ {a}_{i}{a}_{i+1},& \text{for all}i\in \{1,\dots ,n-1\}\\ {b}_{i}{b}_{i+1},& \text{for all}i\in \{1,\dots ,n-1\}\\ {b}_{i}\ge 0\text{and integer},& \text{for all}i\in \{1,\dots ,n\}\\ {a}_{i}\ge 0\text{and integer},& \text{for all}i\in \{1,\dots ,n\}\end{array}\end{array}$

asked 2022-04-10

Inconsistent system of linear inequalities.

$\{x+y-t\ge 1,-x+2y+z\ge 1,-3y-z+t\ge 0\}$

By Fourier's method, I have eliminated the variables x and z and I have arrived to the following inequality: $y-t\le -1$. I do not see why this inequality should imply that the system is inconsistent. I do not know what I am doing wrong.

$\{x+y-t\ge 1,-x+2y+z\ge 1,-3y-z+t\ge 0\}$

By Fourier's method, I have eliminated the variables x and z and I have arrived to the following inequality: $y-t\le -1$. I do not see why this inequality should imply that the system is inconsistent. I do not know what I am doing wrong.

asked 2020-10-21

Graph the solution set of the system of inequalities or indicate that the system has no solution.

asked 2022-07-28

Concentric with thw circle${x}^{2}+{y}^{2}+8x-3y+16=0$ and touching the line 4x+3y=12

asked 2022-06-26

Question: Use the definition of absolute value and systems of inequalities to prove that for any real numbers x and c, and any positive real number δ, the given statement is true:

$|x-c|<\delta \iff x\in (c-\delta ,c+\delta )$

$|x-c|<\delta \iff x\in (c-\delta ,c+\delta )$