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

[anemone]

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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[anemone]

No one answered last week's POTW. (Sadface)

You can find the suggested solution as follows:

Spoiler

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.