# What Is The Smallest Real Number? [closed]

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.

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

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 $z^*=f(x(y^*),y^*)$.