Does $\left\lVert (x,y) \right\rVert := \sqrt{x^2+y^2+xy} \phantom{1}$ satisfy the triangular inequality?

Question

I´m trying to prove that the following expression defines a norm in $\mathbb{R^2}$: $$ \left\lVert (x,y) \right\rVert := \sqrt{x^2+y^2+xy} $$

We can see that:

$0 \le (x - y)^2 \Rightarrow 0 \le x^2 - 2xy + y^2$

$\phantom{20000000000}\Rightarrow 2xy \le x^2 + y^2$

$\phantom{20000000000}\Rightarrow xy \le \frac{x^2 + y^2}{2}$

$\phantom{20000000000}\Rightarrow xy \le x^2 + y^2$

$\phantom{20000000000}\Rightarrow (x^2+y^2+xy)\ge 0 , \phantom{5}\forall x,y \in \mathbb{R^2}$

Hence $\phantom{15}\left\lVert (x,y) \right\rVert : \mathbb{R^2} \times \mathbb{R^2} \rightarrow \mathbb{R} \phantom{15}$ so it is well defined

On the other hand it is trivial that:

$\left\lVert (x,y) \right\rVert \ge 0, \phantom{1} \forall x, y \in \mathbb{R^2}$

$\left\lVert (x,y) \right\rVert = 0 \iff (x,y) = 0$

And we can observe that

$\left\lVert \lambda (x,y) \right\rVert = \sqrt{\lambda^2x^2 + \lambda^2y^2 + \lambda x\lambda y} = \lvert \lambda \rvert\sqrt{x^2+y^2+xy} = \lvert \lambda \rvert \left\lVert (x,y) \right\rVert, \phantom{1} \forall(x,y) \in \mathbb{R^2}$

But I´m having problems to demonstrate that the expression verifies triangular inequality:

$\left\lVert (x,y) + (x',y') \right\rVert \le \left\lVert (x,y) \right\rVert + \left\lVert (x',y') \right\rVert$

How could I prove that $\sqrt{(x+x')^2 +(y+y')^2+(x+x')(y+y')} \le \sqrt{x^2+y^2+xy} + \sqrt{(x')^2+(y')^2+x'y'} \phantom{5}$?

Answer 1

Other have suggested variants, but it is possible to conduct the direct calculation.

$\bigg(\sqrt{x^2+y^2+xy}+\sqrt{u^2+v^2+uv}\bigg)^2-\bigg(\sqrt{x+u)^2+(y+v)^2+(x+u)(y+v)}\bigg)^2$

$ = 2\sqrt{\cdot}\sqrt{\cdot}-A$

Squaring separately both terms:

$\bigg(2\sqrt{x^2+y^2+xy}\sqrt{u^2+v^2+uv}\bigg)^2-\bigg(\overbrace{-2ux-2yv-xv-uy}^A\bigg)^2$

$=3x^2v^2+3y^2u^2-6xyuv=3(xv-uy)^2\ge 0$

You can now go the chain backward and set the inequalities between square roots.

Answer 2

Observe that for the quadratic $\;q((x_1,x_2),\,(y_1,y_2))=x_1y_1+\frac12\left(x_1y_2+x_2y_1\right)+x_2y_2\;$

we have that

$$q((x_1,x_2),\,(y_1,y_2))=(x_1,x_2)\begin{pmatrix}1&\cfrac12\\\cfrac12&1\end{pmatrix}\binom{y_1}{y_2}$$

and the above matrix is positive definite, thus it is an inner product. Of course,

$$q((x,y),\,(x,y))=x^2+y^2+xy\ldots$$

Answer 3

If we change basis by $$ (x,y) = \left( \frac{u}{\sqrt{3}}+v, \frac{u}{\sqrt{3}}-v \right) $$ we get $$\begin{align} x^2 + y^2 + xy &= \left(\frac{u}{\sqrt{3}}+v\right)^2 + \left(\frac{u}{\sqrt{3}}-v\right)^2 + \left(\frac{u}{\sqrt{3}}+v\right)\left(\frac{u}{\sqrt{3}}-v\right) \\ &= \left( \frac{u^2}{3} + \frac{2uv}{\sqrt{3}} + v^2 \right) + \left( \frac{u^2}{3} - \frac{2uv}{\sqrt{3}} + v^2 \right) + \left( \frac{u^2}{3} - v^2 \right) \\ &= u^2 + v^2 \end{align}$$ i.e. we have the Euclidean norm in $(u,v)$.