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This is to extend Christian Remling's comment to all real $t>0$, with an explicit lower bound on $K/M$, where $K:=K_n(a,t)$ and $M:=M_n(a,t)$.

$\newcommand\norm[1]{\lVert#1\rVert}$The key here, as in Christian Remling's comment, is the observation that $\norm x_1\ge\norm x_2$ for all $x\in\mathbb R^n$. Indeed, this observation implies $$K \ge\inf_{x\in\mathbb R^n}\bigl(\lvert\norm x_2-\norm a_2\rvert+t\norm x_2\bigr) =\inf_{u\ge0}\bigl(\lvert u-\norm a_2|+tu\bigr) =\min(1,t)\norm a_2$$ and $$M\le\max(1,t)\min(\norm a_2,\norm a_1)=\max(1,t)\norm a_2,$$ whence $$\frac KM\ge\frac{\min(1,t)}{\max(1,t)}.$$ (In particular, $K\ge M$ if $t=1$.)

This is to extend Christian Remling's comment to all real $t>0$, with an explicit lower bound on $K/M$, where $K:=K_n(a,t)$ and $M:=M_n(a,t)$.

$\newcommand\norm[1]{\lVert#1\rVert}$The key here, as in Christian Remling's comment, is the observation that $\norm x_1\ge\norm x_2$ for all $x\in\mathbb R^n$. Indeed, this observation implies $$K \ge\inf_{x\in\mathbb R^n}\bigl(\lvert\norm x_2-\norm a_2\rvert+t\norm x_2\bigr) =\inf_{u\ge0}\bigl(\lvert u-\norm a_2|+tu\bigr) =\min(1,t)\norm a_2$$ and $$M\le\norm a_2,$$ whence $$\frac KM\ge\min(1,t).$$ (In particular, $K\ge M$ if $t\ge1$.)

This is to extend Christian Remling's comment to all real $t>0$, with an explicit lower bound on $K/M$, where $K:=K_n(a,t)$ and $M:=M_n(a,t)$.

The key here, as in Christian Remling's comment, is the observation that $\|x\|_1\ge\|x\|_2$ for all $x\in\mathbb R^n$. Indeed, this observation implies $$K \ge\inf_{x\in\mathbb R^n}\big(|\|x\|_2-\|a\|_2|+t\|x\|_2\big) =\inf_{u\ge0}\big(|u-\|a\|_2|+tu\big) =\min(1,t)\|a\|_2$$ and $$M\le\max(1,t)\min(\|a\|_2,\|a\|_1)=\max(1,t)\|a\|_2,$$ whence $$\frac KM\ge\frac{\min(1,t)}{\max(1,t)}.$$ (In particular, $K\ge M$ if $t=1$.)

This is to extend Christian Remling's comment to all real $t>0$, with an explicit lower bound on $K/M$, where $K:=K_n(a,t)$ and $M:=M_n(a,t)$.

$\newcommand\norm[1]{\lVert#1\rVert}$The key here, as in Christian Remling's comment, is the observation that $\norm x_1\ge\norm x_2$ for all $x\in\mathbb R^n$. Indeed, this observation implies $$K \ge\inf_{x\in\mathbb R^n}\bigl(\lvert\norm x_2-\norm a_2\rvert+t\norm x_2\bigr) =\inf_{u\ge0}\bigl(\lvert u-\norm a_2|+tu\bigr) =\min(1,t)\norm a_2$$ and $$M\le\max(1,t)\min(\norm a_2,\norm a_1)=\max(1,t)\norm a_2,$$ whence $$\frac KM\ge\frac{\min(1,t)}{\max(1,t)}.$$ (In particular, $K\ge M$ if $t=1$.)

This is to extend Christian Remling's comment to all real $t>0$, with an explicit lower bound on $K/M$, where $K:=K_n(a,t)$ and $M:=M_n(a,t)$.

The key here, as in Christian Remling's comment, is the observation that $\|x\|_1\ge\|x\|_2$ for all $x\in\mathbb R^n$. Indeed, this observation implies $$K\ge\inf_{x\in\mathbb R^n} \big(|\|x\|_2-\|a\|_2|+t\|x\|_2\big)=\min(1,t)\|a\|_2$$ and $$M\le\max(1,t)\min(\|a\|_2,\|a\|_1)=\max(1,t)\|a\|_2,$$ whence $$\frac KM\ge\frac{\min(1,t)}{\max(1,t)}.$$ (In particular, $K\ge M$ if $t=1$.)

This is to extend Christian Remling's comment to all real $t>0$, with an explicit lower bound on $K/M$, where $K:=K_n(a,t)$ and $M:=M_n(a,t)$.

The key here, as in Christian Remling's comment, is the observation that $\|x\|_1\ge\|x\|_2$ for all $x\in\mathbb R^n$. Indeed, this observation implies $$K \ge\inf_{x\in\mathbb R^n}\big(|\|x\|_2-\|a\|_2|+t\|x\|_2\big) =\inf_{u\ge0}\big(|u-\|a\|_2|+tu\big) =\min(1,t)\|a\|_2$$ and $$M\le\max(1,t)\min(\|a\|_2,\|a\|_1)=\max(1,t)\|a\|_2,$$ whence $$\frac KM\ge\frac{\min(1,t)}{\max(1,t)}.$$ (In particular, $K\ge M$ if $t=1$.)