Implications of elementary inequalities

Question

I would like your help to prove that the following statement involving some elementary inequalities is correct (or, if wrong, to construct a counterexample)

Assume $$ \begin{cases} Z+X\geq R+W\\ Z+Y\geq R+Q\\ Z\geq R\\ -------\\ Z+X\in [0,1]\\ R+W\in [0,1]\\ Z+Y\in [0,1]\\ R+Q\in [0,1]\\ -------\\ X\in [0,1]\\ Z\in [0,1]\\ W\in [0,1]\\ R\in [0,1]\\ Y\in [0,1]\\ Q\in [0,1]\\ \end{cases} $$

I want to show that $$ Z+X+Y\geq R+W+Q $$

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What I have done so far: notice that $$ Z+X+Y\equiv Z+X+Z+Y-Z $$ and $$ R+W+Q\equiv R+W+R+W-R $$ I guess that this may help but I can't proceed.

Answer 1

Let, according to $Z\ge R$, $Z=kR$ for some $k\ge 1$, then from the first two inequalities we have

$$kR+X\geq R+W \implies (k-1)R\ge W-X\tag{1}$$ $$kR+Y\geq R+Q\implies (k-1)R\ge Q-Y\tag{2}$$

and from the required inequality

$$Z+X+Y\ge R+Q+W \implies (k-1)R\ge (Q+W)-(X+Y)\tag{3}$$

From the given conditions we can construct infinitely many solutions indeed for $\alpha\in(0,1)$, for $R$ and $k$ fixed, it suffices satisfy the following

$$\alpha(k-1)R\ge W-X\tag{1'}$$ $$(1-\alpha)(k-1)R\ge Q-Y\tag{2'}$$

inorder to satify also $(3)$.

Let consider for example

• $R=3$ and $k=5$ then $Z=15$ and $(k-1)R=12$
• assume $\alpha =\frac 23$

then we need to satisfy

• $8\ge W-X$
• $4\ge Q-Y$

then select for example

• $W=20,Q=16, X=Y=12$

and we obtain

• $Z+X=27\ge R+W=23$
• $Z+Y=27\ge R+Q=19$
• $Z+X+Y=39\ge R+W+Q=39$

finally to satisfy the condition for tha range it suffices to divide all quantities for

$$d=\max\{Z+X,R+W,Z+Y,R+Q\}=27$$