 # Suppose we know that (a+b+d)/(a'+b'+d')<= M and (a+c+d)/(a'+c'+d')<= M and (a)/(a')<=(b)/(b')<=(d)/(d') and (a)/(a')<=(c)/(c')<=(d)/(d') and also that a,b,c,d,a′,b′,c′,d′ in (0,1]. Is the following true? (a+b+c+d)/(a'+b'+c'+d')<= M Ishaan Booker 2022-07-21 Answered
Fraction inequality
Suppose we know that

and

and also that $a,b,c,d,{a}^{\prime },{b}^{\prime },{c}^{\prime },{d}^{\prime }\in \left(0,1\right]$
Is the following true?
$\begin{array}{r}\frac{a+b+c+d}{{a}^{\prime }+{b}^{\prime }+{c}^{\prime }+{d}^{\prime }}\le M\end{array}$
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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
$\frac{a+b+c+d}{{a}^{\prime }+{b}^{\prime }+{c}^{\prime }+{d}^{\prime }}=\frac{4}{5}.$

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