# Prove that the sum of cosines of the two smallest angles in a triangle is $\geq 1$

Claim: If $$\alpha, \beta$$ are the two smallest angles in a given triangle, then $$\cos(\alpha)+\cos(\beta)\ge1$$.

Is there a short/illuminating proof of this, perhaps purely geometrical?

I was able to prove this, but in way I feel to be a little tortuous. I'll outline my proof. If $$\gamma$$ is the largest angle (not necessarily unique) and $$M=180° -\gamma$$, then I want to minimize the function $$f(x)=\cos(x)+\cos(M-x)$$ and prove that its minimal value $$\ge1$$. The lower bounds on $$x$$ are $$x>0$$ and $$x \ge 180°-2\gamma$$, the latter to ensure $$\gamma \ge M-x = 180°-\gamma-x$$. By solving $$f'(x)=0$$, we see that the only extremum value is at $$x=M/2$$, but this is a maximum (by looking at the second derivative or a simple example).

Therefore $$f(x)$$ attains its minimum on the boundary, and by assuming $$x$$ to be the smaller angle of the two, we can take it to be either $$0$$ in the limit or $$180°-2\gamma$$.
In the former case, $$\gamma>90°, M<90°$$, and $$f(x)$$ tends to $$f(0)=1+\cos M >1$$, so $$f(x)\ge 1$$. In the latter case, we have the angles $$x,\gamma,\gamma$$, so $$\gamma=90°-\frac{x}{2}$$ and the minimal value of $$f(x)$$ is $$\cos(x)+\cos(90°-\frac{x}{2})=\cos(x)+\sin(\frac{x}{2})=1-2\sin^2(\frac{x}{2})+\sin(\frac{x}{2}))$$ $$=1+\sin{\frac{x}{2}}(1-2\sin\frac{x}{2})$$ and since $$x$$ as the smallest angle $$\le 60°$$, we have $$\sin(\frac{x}{2})\le\frac{1}{2}$$, and the minimal value of $$f(x)$$ greater or equal to $$1$$ as required.

• vague thought: if you drop a perpendicular from the largest angle, you have a more geometric feel for the cosines (not necessarily helpful) Jul 12, 2022 at 13:04
• Since these angles are acute, averaging to at most $\pi/3$, Jensen's inequality provides an easy alternative to your non-geometric proof strategy, since $\cos$ is concave on $(0,\,\pi/2)$.
– J.G.
Jul 12, 2022 at 13:54
• Let $0<\alpha\leq \beta\leq \gamma$, and $\alpha+\beta+\gamma=\pi$. $3\alpha\leq \alpha+\beta+\gamma=\pi\Rightarrow \alpha\leq \frac{\pi}{3}$. $\beta \leq \gamma =\pi-\alpha-\beta \Rightarrow \beta \leq \frac{\pi}{2} -\frac{\alpha}{2}$. Then $\cos\beta \geq \cos\left(\frac{\pi}{2} -\frac{\alpha}{2}\right)$. After that one can use simplification given in the end of question. Jul 12, 2022 at 14:46

## 1 Answer

Using @barrycarter ‘s comments,

The central premise of this proof will be that the side of a triangle opposite to the largest angle, is the longest in length.

Let, without loss of generality, $$\angle A$$ be the largest angle of $$\triangle ABC$$. Also, again WLOG, let $$\alpha \geq\beta$$. Let $$AD$$ be the perpendicular from A to BC. Now, since $$\alpha \geq \beta$$, we will have $$\displaystyle AC\geq AB\implies \frac{1}{AB}\geq \frac{1}{AC}$$.

Now, $$\cos \alpha +\cos\beta = \displaystyle \frac{BD}{AB}+\frac{CD}{AC}$$$$\geq \frac{BD}{AC}+\frac{CD}{AC} =\frac{BD+DC}{AC}=\frac{BC}{AC}$$ However, by assumption, $$\angle A$$ is the largest angle, hence $$BC$$ must be the longest side, which implies $$BC\geq AC$$ so that $$\displaystyle \cos\alpha +\cos \beta\geq\frac{BC}{AC}\geq 1.$$