Appendix

• The interpolation inequality for $e^x$
  
  $$e^{ta+(1-t)b}\le te^a+(1-t)e^b,\;\;\;\;\;\; t\in[0,1],\;\;\;\;\;\;a,b\ge 0$$
  
  

  Proof: Evaluating the exponential function $y=f(x)=e^x$ at two points $x=a$ and $x=b$ we get two points in the x-y plane $(a,\;e^a)$ and $(b,\;e^b)$. The parametric expression of the secant line through these two points is
  

  $$( ta+(1-t)b,\;te^a+(1-t)e^b ),\;\;\;\;\;(0\le t\le 1)$$
  
  
  The parametric expression of the function $y=e^x$ between these two points is:
  
  $$( ta+(1-t)b,\;e^{ta+(1-t)b} ),\;\;\;\;\;(0\le t\le 1)$$
  
  
  As the $y=e^x$ is convex downward (or concave upward), it is never higher than the secant line function in the interval $a\le x\le b$, and we have
  
  $$e^{ta+(1-t)b} \le te^a+(1-t)e^b$$
  
  
  The equality holds only when $t=0$ or $t=1$.

• Young's inequality

  For any two real positive values $p,\,q>0$ satisfying
  

  $$\frac{1}{p}+\frac{1}{q}=1,\;\;\;\;\;\;q=\frac{p}{p-1},\;\;\;p=\frac{q}{q-1}$$
  
  
  and any two real numbers $\alpha$ and $\beta$, we have
  
  $$\alpha\beta\le\frac{\alpha^p}{p}+\frac{\beta^q}{q}$$
  
  
  Proof: Substituting $a=p\log \alpha$ and $b=q\log\beta$ into the interpolation inequality for $f(x)=e^x$ above, we get:
  

  $$e^{tp\log\alpha+(1-t)q\log\beta} =e^{tp\log\alpha}e^{(1-t)q\log\beta} =\alpha^{tp}\beta^{(1-t)q}$$

  $$\le te^{p\log\alpha}+(1-t)e^{q\log\beta} =t\alpha^p+(1-t)\beta^q$$

  
  
  We further let $t=1/p$, and get $1-t=1-1/p=(p-1)/p=1/q$, the above becomes Young's inequality:
  
  $$\alpha\beta\le\frac{\alpha^p}{p}+\frac{\beta^q}{q}$$
  
  

• Holder's inequality
  
  $$\sum_i \vert x_i y_i\vert\le\vert\vert{\bf x}\vert\vert _p \vert\vert{\bf y}\vert\vert _q$$
  
  
  Proof: Applying Young's inequality to the following
  
  $$\alpha=\frac{\vert x_i\vert}{\vert\vert{\bf x}\vert\vert _p... ...ac{\vert y_i\vert}{\left(\sum_i\vert y_i\vert^q\right)^{1/q}}$$
  
  
  we get
  
  $$\alpha\beta=\frac{\vert x_i\vert\,\vert y_i\vert}{\left[\le... ...ac{1}{q}\frac{\vert y_i\vert^q}{\sum_i\vert y_i\vert^q\right}$$
  
  
  Taking summation on both sides we get
  
  $$\frac{\sum_i \vert x_i\vert\,\vert y_i\vert}{\left[\left(\s... ..._i\vert^q\right)^{1/q}\right]} \le \frac{1}{p}+\frac{1}{q}=1$$
  
  
  i.e.,
  
  $$\sum_i \vert x_i y_i\vert\le\sum_i \vert x_i\vert\,\vert y_... ...vert\vert{\bf x}\vert\vert _p\;\vert\vert{\bf y}\vert\vert _q$$
  
  

• Minkowski's inequality

  
  

  $$\vert\vert{\bf x}+{\bf y}\vert\vert _p\le\vert\vert{\bf x}\vert\vert _p+\vert\vert{\bf y}\vert\vert _p,\;\;\;\;\;\;p\ge 1$$
  
  
  Proof:
  

  $$\vert\vert{\bf x}+{\bf y}\vert\vert _p^p = \sum_i\vert x_i+y_i\vert^p=\sum_i\vert x_i+y_i\vert\;\vert x_i+y_i\vert^{p-1}$$

  $$\le \sum_i\vert x_i\vert\;\vert x_i+y_i\vert^{p-1}+\sum_i\vert y_i\vert\;\vert x_i+y_i\vert^{p-1}$$

  
  
  Applying Holder's inequality to each of the two terms on the right-hand side, we get
  
  $$\vert\vert{\bf x}+{\bf y}\vert\vert _p^p\le \vert\vert{\bf ... ...t\vert _p\left(\sum_i\vert x_i+y_i\vert^{(p-1)q}\right)^{1/q}$$
  
  
  But since $1/p+1/q=1$, and
  
  $$q=p/(p-1),\;\;\;\;\;1/q=(p-1)/p,\;\;\;\;\;(p-1)q=p$$
  
  
  the above becomes:
  
  $$\vert\vert{\bf x}+{\bf y}\vert\vert _p^p\le \vert\vert{\bf... ...}\vert\vert _p) \vert\vert{\bf x}+{\bf y}\vert\vert _p^{p-1}$$
  
  
  Multiplying both sides by $\vert\vert{\bf x}+{\bf y}\vert\vert _p^{1-p}$, we get Minkowski's inequality.