Minkowski for two-entry vectors: equality and strict cases

The inequality, from [L1] at $p=2$: since $\mathrm{\mid }z{\mathrm{\mid }}^2=z^2$ by [L3], the three sums are $\mathrm{\mid }{x}_0+y_0{\mathrm{\mid }}^2+\mathrm{\mid }{x}_1+y_1{\mathrm{\mid }}^2=(x_0+y_0{)}^2+(x_1+y_1{)}^2$, $\mathrm{\mid }{x}_0{\mathrm{\mid }}^2+\mathrm{\mid }{x}_1{\mathrm{\mid }}^2=A$ and $\mathrm{\mid }{y}_0{\mathrm{\mid }}^2+\mathrm{\mid }{y}_1{\mathrm{\mid }}^2=B$ by [L2], so $((x_0+y_0{)}^2+(x_1+y_1{)}^2{)}^{1\mathrm{/}2}\le \sqrt{A}+\sqrt{B}$, which is the displayed inequality with the notation of [L4].

The squared quantities: expanding by [L2] and [L4], $(x_0+y_0{)}^2+(x_1+y_1{)}^2=A+2C+B$ and $(\sqrt{A}+\sqrt{B}{)}^2=A+B+2\sqrt{A}\sqrt{B}=A+B+2\sqrt{AB}$, the last step by the product law of [L4]; both quantities are nonnegative.

The equality criterion: by [L5] applied to the two nonnegative sides of step 1.1 — squared through step 1.2 — equality holds exactly when $A+2C+B=A+B+2\sqrt{AB}$, that is exactly when $C=\sqrt{AB}$; and $C=\sqrt{AB}$ holds exactly when $C\ge 0$ and $C^2=AB$, since $\sqrt{AB}$ is the unique nonnegative square root of $AB$ by [L4]; by [L6] one has $AB-C^2=(x_0{y}_1-x_1{y}_0{)}^2$, so $C^2=AB$ holds exactly when $x_0{y}_1=x_1{y}_0$; hence equality in the displayed inequality holds exactly when $x_0{y}_1=x_1{y}_0$ and $C=x_0{y}_0+x_1{y}_1\ge 0$, and this derivation covers zero vectors because it never divides.

First instance, equality: for $x=(3,4)$ and $y=(6,8)$ one has $A=9+16=25$, $B=36+64=100$ and $x+y=(9,12)$ with squared length $9^2+{12}^2=81+144=225={15}^2$, so the left-hand side is $\sqrt{225}=15$ while the right-hand side is $\sqrt{25}+\sqrt{100}=5+10=15$, an equality; the criterion of step 2.1 is met, since $x_0{y}_1-x_1{y}_0=3\cdot 8-4\cdot 6=24-24=0$ and $C=3\cdot 6+4\cdot 8=50\ge 0$.

Second instance, strictness: for $x=(3,4)$ and $y=(4,-3)$ one has $x+y=(7,1)$ with squared length $49+1=50$, so the left-hand side is $\sqrt{50}=\sqrt{25}\sqrt{2}=5\sqrt{2}$ by [L4], while $A=B=25$ gives the right-hand side $5+5=10$; since $2<4=2^2$ gives $\sqrt{2}<2$ by [L5] and $5>0$ preserves the strict inequality, $5\sqrt{2}<10$, so the inequality is strict here; and $x_0{y}_1-x_1{y}_0=3\cdot (-3)-4\cdot 4=-25\mathrm{\ne }0$, consistent with the criterion of step 2.1.

Summary: the displayed inequality is Minkowski at $p=2$ (step 1.1), equality holds exactly under the two stated conditions (step 2.1), the pair $(3,4)$, $(6,8)$ attains equality (step 3.1: the left side is $15$ and so is the right side $5+10$), and the pair $(3,4)$, $(4,-3)$ is strict (step 3.2); both criteria are checked against the determinant in each instance.