Prove that $x^2+y^2-\sqrt{2}xy < 4$ provided $1\le x,y \le \frac{5}{2}$. If possible I would like an elegant solution to the following problem:

Let $x,y\in\mathbb{R}$ such that $1\le x \le \frac{5}{2}$ and $1\le y \le \frac{5}{2}$. Prove that
$x^2+y^2-\sqrt{2}xy < 4$

I'm aware that you can use Lagrange multipliers but I want an elementary solution using elementary inequalities that are available up to the high school level, or inequalities used in olympiads.
Remarks
I've substituted the obvious specific values for $x,y$ (we can assume without loss of generality that $x\le y$) below
\begin{array}{|c|c|c|}
\hline
x & y &x^2+y^2-\sqrt{2}xy \text{ (rounded to 3 d.p.)}\\ \hline
1 & 1 & \approx 0.587\\ \hline
1 & \frac{5}{2} & \approx 3.714\\ \hline
\frac{5}{2} & \frac{5}{2} & \approx 3.661\\ \hline
\end{array}
Some thoughts:

*

*I've tried solving it in different equivalent forms. Using the following identity


$x^4+y^4 = \left(x^2+y^2+\sqrt{2}xy \right) \left(x^2+y^2-\sqrt{2}xy \right)$

I attempted to prove $x^4+y^4 < 4 \left(x^2+y^2 + \sqrt{2}xy \right)$ (though I'm not sure if it really helps) but to no avail.


*Also, crudely approximating via $x^2+y^2 -\sqrt{2}xy < \left(\frac{5}{2}\right)^2 + \left(\frac{5}{2}\right)^2 - \sqrt{2} \cdot 1 \cdot 1 $ doesn't work since the RHS is clearly larger than 4


The hessian of the function $f(x,y)=x^2+y^2-\sqrt2xy$ is
$$\begin{vmatrix}
2&-\sqrt2\\
-\sqrt2 &2
\end{vmatrix}=2>0,
$$
hence $f(x,y)$ is convex and therefore takes its maximal value on an extreme point of the domain. A simple check (which you have already done) shows that the maximal value is 
$$f\left(1,\frac52\right)=f\left(\frac52,1\right)<4.$$
