If $r_1, r_2, t_1,$ and $t_2$ are real numbers and if $\left|r_{1}\right|<\left|r_{2}\right|$, $\left|t_{1}\right|<\left|t_{2}\right|$ and $$ \left|\left(r_{1}-r_{2}\right)\left(t_{1}-t_{2}\right)\right|>\left|\left(r_{1}+r_{2}\right)\left(t_{1}+t_{2}\right)-2\left(r_{1}r_{2}+t_{1}t_{2}\right)\right|, $$ does it hold that $ \left|\left(r_{1}+r_{2}\right)-\left(t_{1}+t_{2}\right)\right|\ge\left|\left|r_{1}\right|-\left|t_{1}\right|\right|+\left|\left|r_{2}\right|-\left|t_{2}\right|\right|? $ Thanks in advance.
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The second inequality $$|r_{1}+r_{2}-(t_{1}+t_{2})|>||r_{1}|-|t_{1}||+||r_{2}|-|t_{2}||$$holds if $r_{1}> t_{1}>0$, $r_{2}> 0> t_{2}$, then we have $$|r_{1}+r_{2}-(t_{1}+t_{2})|=|r_{1}-t_{1}|+|r_{2}-t_{2}|>||r_{1}|-|t_{1}||+||r_{2}|-|t_{2}||$$
By assumption $r_{1}<r_{2}$, $0<t_{1}<-t_{2}$, then the left hand of the first inequality become $$(r_{2}-r_{1})(t_{1}-t_{2})=r_{2}t_{1}-r_{1}t_{1}-r_{2}t_{2}+r_{1}t_{2}$$
The right hand side now is depended on whether $$-2t_{1}t_{2}<2r_{1}r_{2}-(r_{1}+r_{2})(t_{2}+t_{1})$$
If this is true when we want to have $$r_{2}t_{1}-r_{1}t_{1}-r_{2}t_{2}+r_{1}t_{2}>2r_{1}r_{2}+2t_{1}t_{2}-t_{1}r_{1}-t_{2}r_{2}-t_{1}r_{2}-t_{2}r_{1}$$
Re-arranging the terms we need to have $r_{2}t_{1}+r_{1}t_{2}>r_{1}r_{2}+t_{1}t_{2}$.
Let $r_{2}=50,r_{1}=1,t_{2}=-4,t_{1}=3$, then left hand become $150-4=146$, right hand become $50-12=38$. While $-2t_{1}t_{2}=24$,$2r_{1}r_{2}-(r_{1}+r_{2})(t_{2}+t_{1})=100+50=150$.
Now go back to the original inequality we have:
$$49*7=343>|51*-1-2*(50-12)|=127$$ and $$|51-(-1)|=52>48=2+46$$
Bombyx mori
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user32240, what specific a and b would be a counter-example? Again, a=1.5 and b=1 still do not make the possibility. – lisana Jan 13 '13 at 18:58
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@user32240, it is not valid to change |(a−b)−1| into |(b−a)−1| so this example does not work. – Shard Jan 13 '13 at 19:16
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Dear user32240, thank you for your answer, but did you prove ( http://en.wikipedia.org/wiki/Mathematical_proof) it or give a counter example? – lisana Jan 22 '13 at 07:30
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If you were giving a counter example, then what specific $r_1, r_2, t_1,$ and $t_2$ that would satisfy the assumption but do not satisfy the conclusion were you giving? Thanks. – lisana Jan 22 '13 at 16:31
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A counter example in the sense satisfy the requirements you give in your problem statement. – Bombyx mori Jan 22 '13 at 17:40
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Dear user32240, if it satisfies the requirements in both the assumption and the conclusion, it won't be a counterexample to the statement. A counterexample (http://en.wikipedia.org/wiki/Counterexample) should satisfy the requirement in the assumption but should not satisfy the requirement in the conclusion. – lisana Jan 22 '13 at 18:26
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I think you are confused; this is a counter example in the sense you wanted to show this does not exist, but it exists. – Bombyx mori Jan 22 '13 at 18:40
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@user32240 It is not a counter-example as it satisfies all four inequalities, to be a counter-example it would have to satisfy the first 3 inequalities but not the last. Personally I believe the statement to be true, and that no counter-example exists, but I cannot prove it. – Shard Jan 26 '13 at 13:09
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@Shard: He wants this not to hold, so an example satisfying all four inequalities is a counter example in this sense. – Bombyx mori Jan 26 '13 at 15:03
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@user32240 My reading of the question is that we want to know if the fourth inequality always holds if the other three do. This requires either a proof that it will always hold given the previous conditions, or a counter-example whereby numbers are found which satisfy the first three, but do not satisfy the fourth. – Shard Jan 26 '13 at 15:25
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@Shard: No - he/she means if this condition ever holds; in other words he is looking for a counter example. And here it is. – Bombyx mori Jan 27 '13 at 00:40