If the side lengths of a triangle increase and the third side is fixed, the opposite angle decreases

Question

Suppose that I have an Euclidean triangle in the plane with side lengths $a,b,c$. Denote the angle opposite $c$ by $\theta$.

I am trying to prove, that if we keep $c$ fixed, and increase $a,b$, then $\theta$ should get smaller.

That is, if $\tilde a,\tilde b,c$ are side lengths of another triangle, and $\tilde \theta$ is the corresponding angle, then

$$ \tilde a \ge a, \, \tilde b \ge b \Rightarrow \tilde \theta \le \theta.$$

I tried to prove this via the law of cosines:

$$ \cos(\theta)= \frac{a^2+b^2-c^2}{2ab}\le \frac{\tilde a^2+\tilde b^2-c^2}{2\tilde a \tilde b}=\cos(\tilde \theta),$$

but somehow got stucked.

Is there a simple algebraic proof of this inequality? or alternatively, geometric proof?

Answer 1

This is a simple counterexample:

Answer 2

The way I understood the given problem which to me makes sense:

"If we keep positions of $(A,B)$ fixed and at least one of lengths of $(a,b)$ increased, then show that $\theta$ included between them decreases."

An ellipse is a convenient choice since adjacent legs sum up to a constant. Since confocal ellipses do not intersect it suffices to choose on layer as shown: $$h= a \sin \beta=b \sin \alpha\,;\tag1$$ Lawof Sines $$ \frac{\sin \theta}{c}=\frac{\sin \alpha}{a}=\frac{\sin \beta}{b}=\frac{\sin \alpha+\sin \beta}{a+b}=\frac{\sin \alpha+\sin \beta}{p}\tag2$$ where $p$ is major axis $>c$.

Since $\beta=\alpha-\theta$ $$\frac{\sin (\theta-\alpha) +\sin \beta}{\sin \theta}=\frac{p}{c}>1 \tag3$$

$\dfrac{p}{c} $ will be always greater than $1$ and expression at left increases monotonously with $\theta$, proving the proposition... the way I understood it.