Prove $a ≤ x ≤ b ⇒ x ≤ a +b ∀a,b ∊ R^+$ It seems too obvious to be proved algebraically.
What i thought was:
We know that the sum of two positive numbers is a positive number so:
$a + b &gt; b$ and $a + b &gt; a$
Using the hypotesis of $a &gt; x$ and $b &gt; x$ then
$a + b &gt; b &gt; x$ and $a + b &gt; a &gt; x$
Is it correct? If you know, please tell me another way to prove it.
It seems you are on the right track, but you made one error in the final line with the $a + b &gt; a &gt; x$. I would solve this as:
$\forall a,b \in \mathbb{R}^+$, $b&lt;a+b$ and $a&lt;a+b$ (definition of addition between two positive numbers)
Given $x, a, b$ such that $a \leq x \leq b$ and $b&lt;a+b$ then we can conclude through the transitive property $a \leq x &lt; b$ or
$$x&lt;a+b \ \ \  \text{and} \ \ \ x \geq a$$
The only item I can't seem to prove is that $x \leq a+b$. I would say by the above proof, $x$ is strictly less than $a+b$.
It seems you are on the right track, but you made one error in the final line with the $a + b &gt; a &gt; x$. I would solve this as:
$\forall a,b \in \mathbb{R}^+$, $b&lt;a+b$ and $a&lt;a+b$ (definition of addition between two positive numbers)
Given $x, a, b$ such that $a \leq x \leq b$ and $b&lt;a+b$ then we can conclude through the transitive property $a \leq x &lt; b$ or
$$x&lt;a+b \ \ \  \text{and} \ \ \ x \geq a$$
The only item I can't seem to prove is that $x \leq a+b$. I would say by the above proof, $x$ is strictly less than $a+b$.