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how we show the following inequalities

$(a|a|^{p-2}-b|b|^{p-2})(a-b)\geq 2^{2-p}|a-b|^p$,if $p\geq 2$

$(a|a|^{p-2}-b|b|^{p-2})(a-b)\geq (p-1)\frac{|a-b|^2}{(|a|+|b|)^{2-p}}$,if $ 1<p< 2 $

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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.


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.

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  • $\begingroup$ @ losif Pinelis thank you so much but For $0<b<a=1$,second inequality can rewritten as $r_2(b)=\frac{(1+b)^{2-p}(1-b^{p-1})}{(p-1)(1-b)}$ $r_3(b)=\frac{1-b^{p-1}}{(p-1)(1-b)}$ $r_3'(b)=\frac{-(p-1)b^{p-2}+(p-2)b^{p-1}+1}{(p-1)(1-b)^2}\geq 0$ $r_3(0)=\frac{1}{p-1} >1$ then second inequality is true $\endgroup$ – sidi mohamd deval Jun 22 at 17:50
  • $\begingroup$ @sidimohamddeval : I did copy the exponent $2-p$ incorrectly. Now this is fixed. $\endgroup$ – Iosif Pinelis Jun 22 at 21:56
  • $\begingroup$ For $0<b<a=1$,second inequality can rewritten as $r_2(b)=\frac{(1+b)^{2-p}(1-b^{p-1})}{(p-1)(1-b)}>1$\\ $r_3(b)=\frac{1-b^{p-1}}{(p-1)(1-b)}$\\ $r_3'(b)=\frac{-(p-1)b^{p-2}+(p-2)b^{p-1}+1}{(p-1)(1-b)^2}\geq 0$ $r_3(0)=\frac{1}{p-1} >1$ then second inequality is true $\endgroup$ – sidi mohamd deval Jun 23 at 2:07
  • $\begingroup$ @ losif Pinelis thank you so much $\endgroup$ – sidi mohamd deval Jun 23 at 2:21

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