Equality occurring at two different values for a three way inequality

Question

The question (which was posted by another user) says to prove $$\dfrac{a^2+b^2}{6}\leq \dfrac{a^2+b^2+ab}{3}\leq \dfrac{a^2+b^2}{2}$$ for any $a,b\in\mathbb R$.
This part I was able to solve but while solving I noticed a certain oddity. In my solution, for the first part, I get,
$$\dfrac{a^2+b^2}{6}\leq \dfrac{a^2+b^2+ab}{3}$$ $$\Rightarrow \dfrac{a^2+b^2}{2}\leq {a^2+b^2+ab}$$ $$\Rightarrow a^2 + b^2 \leq a^2 + b^2 + \Bigl(a+b\Bigl)^2$$ $$\Rightarrow \Bigl(a+b\Bigl)^2 \geq 0$$ With equality occuring at $a=-b.$
Now while solving the second part, long story short I get $$\Bigl(a-b\Bigl)^2 \geq 0$$ With equality occuring at $a= b$. I just wanted to know if this something that can be allowed, that is can a double inequality (not strict) have two different values for which equality occurs. Put succinctly in one pf the comments, if
$f(a,b) \leq g(a,b) \leq h(a,b)$ is it allowed to have a scenario where $f(a_1,b_1) = g(a_1,b_1)$ and $g(a_2,b_2) = h(a_2,b_2)$?

Answer 1

As far as I understand your question, the answer is yes.

The constraint $$ p \leq q \leq r $$ means the same thing as $$ p \leq q \qquad \text{and} \qquad q \leq r $$ and does indeed permit you to have both $p=q$ and $q < r$. Similarly, it also permits both $p < q$ and $q = r$.

You can find examples from the very work you've done; e.g. plug in $a=1$ and $b=-1$ to get one such arrangement, and plug in $a=1$ and $b=1$ to get the other arrangement. (these are chosen simply to be the simplest nontrivial numbers one can plug in that satisfy your solutions)

So, as far as I can tell, your stated question really amounts to nothing more than:

Is it possible to have six numbers $u,v,w,x,y,z$ such that the following both hold?

• $u=v$ and $v < w$
• $x < y$ and $y = z$

Answer 2

I will give an explicit example that shows that the answer is yes. Let us consider

\begin{cases} f(a,b)=2ab\\ g(a,b)=a^2+b^2\\ h(a,b)=a^2+b^2+(b-a)^2. \end{cases}

It is $$f(a,b) \leq g(a,b) \leq h(a,b).$$ Both inequalities are equalities if and only if $a=b.$ And we have the equalities

$$f(1,1)=g(1,1)$$ and $$g(3,3)=h(3,3).$$