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If $a,b,c$ are angles of a triangle then show that $$\sin(a)+\sin(b)-\cos(c)\leq \frac{3}{2}$$

We'll use $\sin(x)\leq x, x$ positive

  1. $\sin(a)\leq a$
  2. $\sin(b)\leq b$
  3. $-\cos(c)=\cos(a+b)$

Case I: $0\leq a+b \leq\frac{\pi}{2}$ $\cos(a+b)=\sin(\frac{\pi}{2}-(a+b))\leq \frac{\pi}{2}-(a+b)$ Add with 1, 2 and as LHS $ \leq \pi/2$ we still need a tighter bound for this aswell

Case II: $\frac{\pi}{2}\leq a+b\leq \pi$ This is the part where I am stuck.

Previously I though to solve this question using Jensen's inequality and also had a look at some other trigonometric inequalities but those didn't help a bit.

All I need is hints towards the solution.

Aditya Teraiya
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2 Answers2

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your suggestion of using jensens works too. by jensens inequality, we have: $$\frac{\sin a + \sin b}{2} \leq \sin \left(\frac{a+b}{2}\right) = \cos \left(\frac{c}{2}\right)$$ $$\implies \sin a + \sin b -\cos c \leq 2\cos \left(\frac{c}{2}\right) -\cos c$$ $$\implies \sin a + \sin b -\cos c \leq 2\cos \left(\frac{c}{2}\right) - 2\cos ^2 \left(\frac{c}{2}\right) + 1$$ since $x(1-x)\leq \frac{1}{4}\quad\forall x\in [0,1]$, we have $$\sin a + \sin b -\cos c \leq 2\left(\frac{1}{4}\right) + 1 = \frac{3}{2}$$ our desired result

Aniruda Suswaram
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The following solution is interesting enough. I think.

I'll rewrite the problem so:

Let $\alpha$, $\beta$ and $\gamma$ be measures-angles of a triangle. Prove that: $$\sin\alpha+\sin\beta-\cos\gamma\leq\frac{3}{2}.$$

Now, let $a$, $b$ and $c$ be lengths sides of the triangle.

Thus, the inequality is equivalent to $$(a-b)^2(a+b-c)^2(a+b+c)^2+c^2(a^2+b^2+ab-c^2)^2\geq0.$$

Michael Rozenberg
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