# on the inequalities in $\mathbb{R}$ [closed]

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

## 1 Answer

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

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.

• @ 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 – sidi mohamd deval Jun 22 at 17:50
• @sidimohamddeval : I did copy the exponent $2-p$ incorrectly. Now this is fixed. – Iosif Pinelis Jun 22 at 21:56
• 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 – sidi mohamd deval Jun 23 at 2:07
• @ losif Pinelis thank you so much – sidi mohamd deval Jun 23 at 2:21