Suppose we know that \begin{align} \frac{a+b+d}{a'+b'+d'}\le M\qquad\text{ and }\qquad \frac{a+c+d}{a'+c'+d'}\le M \end{align} and \begin{align} \frac{a}{a'}\le \frac{b}{b'}\le\frac{d}{d'}\qquad\text{ and }\qquad \frac{a}{a'}\le \frac{c}{c'}\le\frac{d}{d'} \end{align} and also that $a,b,c,d,a',b',c',d'\in(0,1]$.
Is the following true? \begin{align} \frac{a+b+c+d}{a'+b'+c'+d'}\le M \end{align}
No. Take all numbers equal to $\frac12$, except for $a'$, which is equal to $1$. Besides, take $M=\frac34$. Then$$\frac{a+b+c+d}{a'+b'+c'+d'}=\frac45.$$