Consider six numbers in $(0,1)$ $\{a,b,c,d,e,f\}$. Suppose $$ a<c+e\\ b<d\\ b<f $$
Can we conclude that $$ ab<cd+ef $$
It seems to me yes since $$ a<c+e \Leftrightarrow ab<cb+eb<cd+ef $$
Do you agree?
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1yes that is correct – Math Lover May 26 '21 at 19:07
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1Yes it is true since $b>0$, $c>0$ and $e>0$. – Axel May 26 '21 at 19:07
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1Your arument is correct if you replace $\Leftrightarrow$ with $\Rightarrow$ (which is all you need) – Hagen von Eitzen May 26 '21 at 19:13
1 Answers
1
If $z>0$ then for any real numbers $x,y$ we have,
$$x < y \iff zx<zy$$
Here,
$$a<c+e$$
Thus as $b>0$,
$$ab<(c+e)b = cb+eb \quad (0)$$
Moreover because $b<d$ and $c>0$ we have,
$$cb<cd \quad (1)$$
Likewise, as $b<f$ and $e>0$,
$$eb<ef \quad (2)$$
Adding $(1)$ and $(2)$ we get,
$$cb+eb<cd+ef$$
Therefore combining it with $(0)$ we can deduce,
$$\boxed{ ab<cb+eb<cd+ef}$$
Axel
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