MHB Prove $a+b>c+d$ for Real Numbers with Sine Inequalities

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The problem presented involves proving that for real numbers a, b, c, and d, the inequalities a + sin b > c + sin d and b + sin a > d + sin c imply that a + b > c + d. The discussion invites participants to engage with this mathematical challenge, emphasizing the need for a rigorous proof. Additionally, there is a note about the lack of responses to the previous week's problem, highlighting a potential decline in participation. The thread encourages readers to refer to guidelines for submitting solutions. The focus remains on the mathematical proof and community engagement in solving the problem.
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Here is this week's POTW:

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Let $a,\,b,\,c$ and $d$ be real numbers such that

$a+\sin b > c+ \sin d$ and

$b+\sin a > d + \sin c$.

Prove that $a+b>c+d$.

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Remember to read the https://mathhelpboards.com/showthread.php?772-Problem-of-the-Week-%28POTW%29-Procedure-and-Guidelines to find out how to https://mathhelpboards.com/forms.php?do=form&fid=2!
 
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No one answered last week's POTW. (Sadface)

You can find the suggested solution as follows:
For $x\ge0$, $|\sin x|\le x$. Let $s=a-c$ and $t=d-b$. We have

$\begin{align*}s &=a-c \\&>\sin d-\sin b\\&=2\cos\left(\dfrac{d+b}{2}\right)\sin\left(\dfrac{d-b}{2}\right)\\& \ge -2|\sin \dfrac{t}{2}| \end{align*}$

and

$\begin{align*}t &=d-b \\&<\sin a-\sin c\\&=2\cos\left(\dfrac{a+c}{2}\right)\sin\left(\dfrac{a-c}{2}\right)\\& \le -2|\sin \dfrac{s}{2}| \end{align*}$

If $s\ge 0$, then $t<2|\sin\dfrac{s}{2}|\le s$.

Similarly,

if $t\le 0$, then $s>-2|\sin\left(-\dfrac{t}{2}\right)|\ge -2\left(-\dfrac{t}{2}\right)=t$

Finally, if $s<0<t$, then $-s<2|\sin\dfrac{t}{2}|\le t$ and $t<2|\sin\dfrac{s}{2}|=|\sin\left(-\dfrac{s}{2}\right)|\le -s$, which leads to a contradiction.
 
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