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.

  • $\begingroup$ vague thought: if you drop a perpendicular from the largest angle, you have a more geometric feel for the cosines (not necessarily helpful) $\endgroup$ Jul 12, 2022 at 13:04
  • 2
    $\begingroup$ 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 $(0,\,\pi/2)$. $\endgroup$
    – J.G.
    Jul 12, 2022 at 13:54
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    $\begingroup$ 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. $\endgroup$ Jul 12, 2022 at 14:46

1 Answer 1


Using @barrycarter ‘s comments,enter image description here

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 $$\displaystyle \cos\alpha +\cos \beta\geq\frac{BC}{AC}\geq 1.$$


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