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Let $l(a,b)$ denote the left-hand side of both inequalities. Then $l(b,a)=l(a,b)=l(-a,-b)$. The right-hand sides of both inequalities have the same properties. So, without loss of generality (wlog) $a\ge b$ and $a\ge0$. Also, by homogeneity, wlog $a=1$. So, we have these two cases to consider: $$(i)\ 0\le b\le a=1\quad\text{and}\quad (ii)\ b<0<a=1.$$ In case (i), your first inequality can rewritten as $$r_{11}(b):=\frac{1-b^{p-1}}{(1-b)^{p-1}}\ge2^{2-p} \tag{1}$$ for $p\ge2$ and $b\in[0,1)$. For $p\ge2$ and $b\in(0,1)$, we have $$r_{11}'(b)=(p-1) (1-b)^{-p} \left(1-b^{p-2}\right)>0$$ and $r_{11}(0)=0$, whence $r_{11}(b)\ge1\ge2^{2-p}$, so that (1) holds.

In case (ii), your first inequality can rewritten as $$r_{12}(c):=\frac{1+c^{p-1}}{(1+c)^{p-1}}\ge2^{2-p} \tag{2}$$ for $p\ge2$ and $c:=-b>0$. For such $p$ and $c$, we have $$r_{12}'(c)=(p-1) (1+c)^{-p} \left(c^{p-2}-1\right),$$ so that $r_{12}$ is decreasing on $(0,1]$ and increasing on $[1,\infty)$, with the minimum value $r_{12}(1)=2^{2-p}$, so that (1) holds.

For $0<b<1=a$, your second inequality can rewritten as $$r_2(b):=\frac{(1+b)^{p-2} \left(1-b^{p-1}\right)}{(1-b) (p-1)}\ge1 \tag{3}$$ for $p\in(1,2)$. We have $$r_2(1-)=2^{p-2}<1$$ for $p\in(1,2)$. So, your second inequality is false in general.

Let $l(a,b)$ denote the left-hand side of both inequalities. Then $l(b,a)=l(a,b)=l(-a,-b)$. The right-hand sides of both inequalities have the same properties. So, without loss of generality (wlog) $a\ge b$ and $a\ge0$. Also, by homogeneity, wlog $a=1$. So, we have these two cases to consider: $$(i)\ 0\le b\le a=1\quad\text{and}\quad (ii)\ b<0<a=1.$$

In case (i), your first inequality can rewritten as $$r_{11}(b):=\frac{1-b^{p-1}}{(1-b)^{p-1}}\ge2^{2-p} \tag{1}$$ for $p\ge2$ and $b\in[0,1)$. For $p\ge2$ and $b\in(0,1)$, we have $$r_{11}'(b)=(p-1) (1-b)^{-p} \left(1-b^{p-2}\right)>0$$ and $r_{11}(0)=0$, whence $r_{11}(b)\ge1\ge2^{2-p}$, so that (1) holds.

In case (ii), your first inequality can rewritten as $$r_{12}(c):=\frac{1+c^{p-1}}{(1+c)^{p-1}}\ge2^{2-p} \tag{2}$$ for $p\ge2$ and $c:=-b>0$. For such $p$ and $c$, we have $$r_{12}'(c)=(p-1) (1+c)^{-p} \left(c^{p-2}-1\right),$$ so that $r_{12}$ is decreasing on $(0,1]$ and increasing on $[1,\infty)$, with the minimum value $r_{12}(1)=2^{2-p}$, so that (1) holds.

As for your second inequality, in case (i) it can rewritten as $$r_{21}(b):=\frac{(1+b)^{2-p} \left(1-b^{p-1}\right)}{(1-b) (p-1)}\ge1 \tag{3}$$ for $p\in(1,2)$. Let $$(Dr_{21})(b):=r_{21}'(b)(p-1)(1-b)^2 (1+b)^{p-1} \\ = -(p-1) b^{p-2}+(p-3) b^{p-1}+b (p-1)-p+3.$$ Then $$(Dr_{21})''(b)=-(1-b) (3-p) (2-p) (p-1) b^{p-4}<0$$ for ($p\in(1,2)$ and) $b\in(0,1)$, so that $(Dr_{21})(b)$ is concave in $b$. Also, $(Dr_{21})(1)=0=(Dr_{21})'(1)$. So, $Dr_{21}<0$ and hence $r_{21}$ is decreasing on $[0,1)$. Also, $r_{21}(1-)=2^{2-p}$. So, $r_{21}\ge2^{2-p}\ge1$, so that (3) holds.

In case (ii), your second inequality can rewritten as $$r_{22}(c):=\frac{1+c^{p-1}}{(1+c)^{p-1}}\ge1 \tag{4}$$ for $p\in(1,2)$ and $c:=-b>0$. For such $p$ and $c$, we have $$r_{22}'(c)=(1 + c)^{-p} (c^{p-2}-1),$$ so that $r_{22}$ is increasing on $(0,1]$ and decreasing on $[1,\infty)$. Also, $r_{22}(0)=r_{22}(\infty-)=\frac1{p-1}$. So, $r_{22}\ge\frac1{p-1}>1$, so that (4) holds.

Let $l(a,b)$ denote the left-hand side of both inequalities. Then $l(b,a)=l(a,b)=l(-a,-b)$. So, without loss of generality (wlog) $a\ge b$ and $a\ge0$. Also, by homogeneity, wlog $a=1$. So, we have these two cases to consider: $$(i)\ 0\le b\le a=1\quad\text{and}\quad (ii)\ b<0<a=1.$$ In case (i), your first inequality can rewritten as $$r_{11}(b):=\frac{1-b^{p-1}}{(1-b)^{p-1}}\ge2^{2-p} \tag{1}$$ for $p\ge2$ and $b\in[0,1)$. For $p\ge2$ and $b\in(0,1)$, we have $$r_{11}'(b)=(p-1) (1-b)^{-p} \left(1-b^{p-2}\right)>0$$ and $r_{11}(0)=0$, whence $r_{11}(b)\ge1\ge2^{2-p}$, so that (1) holds.

In case (ii), your first inequality can rewritten as $$r_{12}(c):=\frac{1+c^{p-1}}{(1+c)^{p-1}}\ge2^{2-p} \tag{2}$$ for $p\ge2$ and $c:=-b>0$. For such $p$ and $c$, we have $$r_{12}'(c)=(p-1) (1+c)^{-p} \left(c^{p-2}-1\right),$$ so that $r_{12}$ is decreasing on $(0,1]$ and increasing on $[1,\infty)$, with the minimum value $r_{12}(1)=2^{2-p}$, so that (1) holds.

For $0<b<1=a$, your second inequality can rewritten as $$r_2(b):=\frac{(1+b)^{p-2} \left(1-b^{p-1}\right)}{(1-b) (p-1)}\ge1 \tag{3}$$ for $p\in(1,2)$. We have $$r_2(1-)=2^{p-2}<1$$ for $p\in(1,2)$. So, your second inequality is false in general.

Let $l(a,b)$ denote the left-hand side of both inequalities. Then $l(b,a)=l(a,b)=l(-a,-b)$. The right-hand sides of both inequalities have the same properties. So, without loss of generality (wlog) $a\ge b$ and $a\ge0$. Also, by homogeneity, wlog $a=1$. So, we have these two cases to consider: $$(i)\ 0\le b\le a=1\quad\text{and}\quad (ii)\ b<0<a=1.$$ In case (i), your first inequality can rewritten as $$r_{11}(b):=\frac{1-b^{p-1}}{(1-b)^{p-1}}\ge2^{2-p} \tag{1}$$ for $p\ge2$ and $b\in[0,1)$. For $p\ge2$ and $b\in(0,1)$, we have $$r_{11}'(b)=(p-1) (1-b)^{-p} \left(1-b^{p-2}\right)>0$$ and $r_{11}(0)=0$, whence $r_{11}(b)\ge1\ge2^{2-p}$, so that (1) holds.

In case (ii), your first inequality can rewritten as $$r_{12}(c):=\frac{1+c^{p-1}}{(1+c)^{p-1}}\ge2^{2-p} \tag{2}$$ for $p\ge2$ and $c:=-b>0$. For such $p$ and $c$, we have $$r_{12}'(c)=(p-1) (1+c)^{-p} \left(c^{p-2}-1\right),$$ so that $r_{12}$ is decreasing on $(0,1]$ and increasing on $[1,\infty)$, with the minimum value $r_{12}(1)=2^{2-p}$, so that (1) holds.

For $0<b<1=a$, your second inequality can rewritten as $$r_2(b):=\frac{(1+b)^{p-2} \left(1-b^{p-1}\right)}{(1-b) (p-1)}\ge1 \tag{3}$$ for $p\in(1,2)$. We have $$r_2(1-)=2^{p-2}<1$$ for $p\in(1,2)$. So, your second inequality is false in general.

Let $l(a,b)$ denote the left-hand side of both inequalities. Then $l(b,a)=l(a,b)=l(-a,-b)$. So, without loss of generality (wlog) $a\ge b$ and $a\ge0$, and then $l(a,b)\ge l(a,|b|)$. So, wlog $b\ge0$, so that $a\ge b\ge0$. Also, by homogeneity, wlog $a=1$. So, the first inequality can rewritten as $$r_1(b):=\frac{1-b^{p-1}}{(1-b)^{p-1}}\ge2^{2-p} \tag{1}$$ for $p\ge2$ and $b\in[0,1)$. For $p\ge2$ and $b\in(0,1)$, we have $$r_1'(b)=2^{p-2} (p-1) (1-b)^{-p} \left(1-b^{p-2}\right)>0$$ and $r_1(0)=0$, whence $r_1(b)\ge1\ge2^{2-p}$, so that (1) holds.

Similarly, your second inequality can rewritten as $$r_2(b):=\frac{(1+b)^{p-2} \left(1-b^{p-1}\right)}{(1-b) (p-1)}\ge1 \tag{2}$$ for $p\in(1,2)$ and $b\in[0,1)$. We have $$r_2(1-)=2^{p-2}<1$$ for $p\in(1,2)$. So, your second inequality is false i n general.

Let $l(a,b)$ denote the left-hand side of both inequalities. Then $l(b,a)=l(a,b)=l(-a,-b)$. So, without loss of generality (wlog) $a\ge b$ and $a\ge0$. Also, by homogeneity, wlog $a=1$. So, we have these two cases to consider: $$(i)\ 0\le b\le a=1\quad\text{and}\quad (ii)\ b<0<a=1.$$ In case (i), your first inequality can rewritten as $$r_{11}(b):=\frac{1-b^{p-1}}{(1-b)^{p-1}}\ge2^{2-p} \tag{1}$$ for $p\ge2$ and $b\in[0,1)$. For $p\ge2$ and $b\in(0,1)$, we have $$r_{11}'(b)=(p-1) (1-b)^{-p} \left(1-b^{p-2}\right)>0$$ and $r_{11}(0)=0$, whence $r_{11}(b)\ge1\ge2^{2-p}$, so that (1) holds.

In case (ii), your first inequality can rewritten as $$r_{12}(c):=\frac{1+c^{p-1}}{(1+c)^{p-1}}\ge2^{2-p} \tag{2}$$ for $p\ge2$ and $c:=-b>0$. For such $p$ and $c$, we have $$r_{12}'(c)=(p-1) (1+c)^{-p} \left(c^{p-2}-1\right),$$ so that $r_{12}$ is decreasing on $(0,1]$ and increasing on $[1,\infty)$, with the minimum value $r_{12}(1)=2^{2-p}$, so that (1) holds.

For $0<b<1=a$, your second inequality can rewritten as $$r_2(b):=\frac{(1+b)^{p-2} \left(1-b^{p-1}\right)}{(1-b) (p-1)}\ge1 \tag{3}$$ for $p\in(1,2)$. We have $$r_2(1-)=2^{p-2}<1$$ for $p\in(1,2)$. So, your second inequality is false in general.