Prove that $x^2+y^2-\sqrt{2}xy < 4$ provided $1\le x,y \le \frac{5}{2}$.

Question

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:

1. 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.

2. 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

Answer 1

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 $$f\left(1,\frac52\right)=f\left(\frac52,1\right)<4.$$

Answer 2

Let $f(x,y)=x^2+y^2-\sqrt2xy.$

Thus, $f$ is a convex function of $x$ and $f$ is a convex function of $y$.

Id est, $$\max{f}=\max\{f(1,1),f(1,2,5),f(2.5,2.5)\}=f(1,2.5)=7.25-2.5\sqrt2<4.$$

Answer 3

Hint. Make the substitution $$x=t+u,\\y=\sqrt 2 u.$$