A simple algebra equation to prove.

Question

We agree that $a^2 + b^2 > 2\cdot (\frac{a+b}{2})^2$, where $a,b \in N^+$. We move from $\frac{a+b}{2}$ to $a$ and $\frac{a+b}{2}$ to $b$. Let we represent $x$ as the variable which move from $\frac{a+b}{2}$ to $a$ and $y$ as the variable which move from $\frac{a+b}{2}$ to $b$. Can we prove that with increasing $|x-y|$, $x^2+y^2$ also increases?

Answer 1

No, as a counterexample, take $a = 1$, $b = 2$, so that $x,y \in [1,1.5]$.

\begin{array}{l|l|l|l} x & y & |x - y| & x^2 + y^2 \\ \hline 1.1 & 1.6 & 0.5 & 3.77 \\ 1.05 & 1.61 & 0.56 & 3.6946 \end{array}

Let we represent $x$ as the variable which move from $\dfrac{a+b}{2}$ to $a$
and $y$ as the variable which move from $\dfrac{a+b}{2}$ to $b$.

So I choose $(x,y) = (1.1,1.6)$, then $(x,y) = (1.05,1.61)$.

Can we prove that with increasing $|x-y|$, $x^2+y^2$ also increases?

Observe that we increased $|x-y|$ by $0.06$ as we "move to $a = 1$", but $x^2+y^2$ decreases.

---

(This answers the original version of the question.)

No, as a counterexample, take $a = 1$, $b = 2$, so that $x,y \in [1,1.5]$.

\begin{array}{l|l|l|l} x & y & |x - y| & x^2 + y^2 \\ \hline 1.1 & 1.2 & 0.1 & 2.65 \\ 1.05 & 1.17 & 0.12 & 2.4714 \end{array}

Let we represent $x$ as the variable which move from $\dfrac{a+b}{2}$ to $a$
and $y$ as the variable which move from $\dfrac{a+b}{2}$ to $a$.

So I choose $(x,y) = (1.1,1.2)$, then $(x,y) = (1.05,1.17)$.

Can we prove that with increasing $|x-y|$, $x^2+y^2$ also increases?

Observe that we increased $|x-y|$ by $0.02$ as we "move to $a = 1$", but $x^2+y^2$ decreases.