91影视

It is given that \({\rm{u}} = \left( \begin{array}{l}a\\b\end{array} \right)\) and \({\rm{v}} = \left( \begin{array}{l}1\\1\end{array} \right)\). So, according to inner product axioms, \({\left\| {\rm{u}} \right\|^2} = {a^2} + {b^2}\) and \({\left\| {\rm{v}} \right\|^2} = 2\)and\(\left\langle {{\rm{u,v}}} \right\rangle = a + b\).

According to Cauchy Schwarz inequality, an inner product on a vector space \(V\) is a function that, to each pair of vectors \({\bf{u}}\) and \({\rm{v}}\) in \(V\), associates a real number \(\left\langle {{\bf{u}},{\rm{v}}} \right\rangle \)and satisfies the axiom, \(\left| {\left\langle {{\bf{u}},{\rm{v}}} \right\rangle } \right| \le \left\| {\rm{u}} \right\|\left\| {\rm{v}} \right\|\) for all \({\bf{u}}\) and \({\rm{v}}\)in \(V\).

Divide both sides of Cauchy Schwarz inequality by 2 and thence squaring both sides yieldthe following result:

\(\begin{array}{c}{\left\langle {{\rm{u,v}}} \right\rangle ^2} \le {\left\| {\rm{u}} \right\|^2}{\left\| {\rm{v}} \right\|^2}\\\frac{{{{\left\langle {{\rm{u,v}}} \right\rangle }^2}}}{4} \le \frac{{{{\left\| {\rm{u}} \right\|}^2}{{\left\| {\rm{v}} \right\|}^2}}}{4}\\{\left( {\frac{{\left\langle {{\rm{u,v}}} \right\rangle }}{2}} \right)^2} \le \frac{{{{\left\| {\rm{u}} \right\|}^2}{{\left\| {\rm{v}} \right\|}^2}}}{4}\end{array}\)

Plug the above obtained values into theinequality \({\left( {\frac{{\left\langle {{\rm{u,v}}} \right\rangle }}{2}} \right)^2} \le \frac{{{{\left\| {\rm{u}} \right\|}^2}{{\left\| {\rm{v}} \right\|}^2}}}{4}\), as follows:

\(\begin{array}{c}{\left( {\frac{{\left\langle {{\rm{u,v}}} \right\rangle }}{2}} \right)^2} \le \frac{{{{\left\| {\rm{u}} \right\|}^2}{{\left\| {\rm{v}} \right\|}^2}}}{4}\\{\left( {\frac{{a + b}}{2}} \right)^2} \le \frac{{\left( {{a^2} + {b^2}} \right)\left( 2 \right)}}{4}\\{\left( {\frac{{a + b}}{2}} \right)^2} \le \frac{{{a^2} + {b^2}}}{2}\end{array}\)

Thus, the inequality\({\left( {\frac{{a + b}}{2}} \right)^2} \le \frac{{{a^2} + {b^2}}}{2}\)is proved to be true.