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Find the smallest real number $Z$ such that for all triangle angles $A$,$B$ , and $C$, the inequality $\sin^2 (A) + \sin^2 (B) - \cos(C) \leq Z$ holds.

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    $\begingroup$ The inequality you wrote doesn't mention Z at all. $\endgroup$
    – mniip
    Feb 25, 2014 at 9:36
  • $\begingroup$ Check your question. Maybe instead of $A$ at the end you would have meant $Z$. Or tell the source. $\endgroup$
    – Sawarnik
    Feb 25, 2014 at 10:41
  • $\begingroup$ Very related: math.stackexchange.com/questions/687974/… $\endgroup$
    – Sawarnik
    Feb 25, 2014 at 12:36

1 Answer 1

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Supposing that the inequality is: $$ \sin^2(A)+\sin^2(B)-\cos(\pi-A-B)\leq Z, $$ Here you have a 2-variable function, namely: $$ f(x,y)=\sin^2(x)+\sin^2(y)-\cos(\pi-x-y), $$ Then you have to solve: $$ \nabla f=0, $$ which translates to: $$ \left\{\begin{array}_2\sin(x)\cos(x)+\sin(\pi-x-y)\cdot(-1)&=&0\\ 2\sin(y)\cos(y)+\sin(\pi-x-y)\cdot(-1)&=&0\\ \end{array}\right. $$ Therefore: $$ \left\{\begin{array}_\sin(2x)-\sin(\pi-x-y)&=&0\\ \sin(2y)-\sin(\pi-x-y)&=&0\\ \end{array}\right. $$ Thus: $$ \sin(2x)=\sin(2y), $$ so: $$ 2x=2y+k\pi,\ k\in\mathbb{Z}, $$ i.e.: $$ x=y+k\frac{\pi}{2},\ k\in\mathbb{Z}. $$ Therefore: $$ \sin(2y)=\sin(\pi-y-k\pi/2-y), $$ Finally: $$ \sin(2y)=\sin(\pi(1-k/2)-2y), $$ Find the correct $y^*$, find the minimum of $f(x(y^*),y^*)$ and find $z^*=f(x(y^*),y^*)$.

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    $\begingroup$ That is exactly what we have to solve! Could you elaborate on how to solve your last equation? $\endgroup$
    – Sawarnik
    Feb 25, 2014 at 12:11
  • $\begingroup$ Edited, hope it helps. $\endgroup$
    – 7raiden7
    Feb 25, 2014 at 12:32

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