Ishaan Booker

2022-07-21

Fraction inequality

Suppose we know that

$\begin{array}{r}\frac{a+b+d}{{a}^{\prime}+{b}^{\prime}+{d}^{\prime}}\le M\phantom{\rule{2em}{0ex}}\text{and}\phantom{\rule{2em}{0ex}}\frac{a+c+d}{{a}^{\prime}+{c}^{\prime}+{d}^{\prime}}\le M\end{array}$

and

$\begin{array}{r}\frac{a}{{a}^{\prime}}\le \frac{b}{{b}^{\prime}}\le \frac{d}{{d}^{\prime}}\phantom{\rule{2em}{0ex}}\text{and}\phantom{\rule{2em}{0ex}}\frac{a}{{a}^{\prime}}\le \frac{c}{{c}^{\prime}}\le \frac{d}{{d}^{\prime}}\end{array}$

and also that $a,b,c,d,{a}^{\prime},{b}^{\prime},{c}^{\prime},{d}^{\prime}\in (0,1]$

Is the following true?

$\begin{array}{r}\frac{a+b+c+d}{{a}^{\prime}+{b}^{\prime}+{c}^{\prime}+{d}^{\prime}}\le M\end{array}$

Suppose we know that

$\begin{array}{r}\frac{a+b+d}{{a}^{\prime}+{b}^{\prime}+{d}^{\prime}}\le M\phantom{\rule{2em}{0ex}}\text{and}\phantom{\rule{2em}{0ex}}\frac{a+c+d}{{a}^{\prime}+{c}^{\prime}+{d}^{\prime}}\le M\end{array}$

and

$\begin{array}{r}\frac{a}{{a}^{\prime}}\le \frac{b}{{b}^{\prime}}\le \frac{d}{{d}^{\prime}}\phantom{\rule{2em}{0ex}}\text{and}\phantom{\rule{2em}{0ex}}\frac{a}{{a}^{\prime}}\le \frac{c}{{c}^{\prime}}\le \frac{d}{{d}^{\prime}}\end{array}$

and also that $a,b,c,d,{a}^{\prime},{b}^{\prime},{c}^{\prime},{d}^{\prime}\in (0,1]$

Is the following true?

$\begin{array}{r}\frac{a+b+c+d}{{a}^{\prime}+{b}^{\prime}+{c}^{\prime}+{d}^{\prime}}\le M\end{array}$

frisiao

Beginner2022-07-22Added 13 answers

No. Take all numbers equal to $\frac{1}{2}$, except for ${a}^{\prime}$, which is equal to $1$. Besides, take $M=\frac{3}{4}$. Then

$\frac{a+b+c+d}{{a}^{\prime}+{b}^{\prime}+{c}^{\prime}+{d}^{\prime}}=\frac{4}{5}.$

$\frac{a+b+c+d}{{a}^{\prime}+{b}^{\prime}+{c}^{\prime}+{d}^{\prime}}=\frac{4}{5}.$

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