91影视

Let ${\stackrel{藱}{\mathrm{v}}}_1,{\stackrel{藱}{\mathrm{v}}}_2$be unit orthogonal eigenvectors with eigenvalues 1 and 2 respectively.

So, write localid="1659679849030" $\stackrel{藱}{\mathrm{x}}={\mathrm{c}}_1{\stackrel{藱}{\mathrm{v}}}_1+{\mathrm{c}}_2{\stackrel{藱}{\mathrm{v}}}_2\mathrm{for}\mathrm{some}{\mathrm{c}}_1,{\mathrm{c}}_2.\mathrm{Also}\left|\left|\stackrel{藱}{\mathrm{x}}\right|\right|=\sqrt{{\mathrm{c}}_1^2+{\mathrm{c}}_2^2}$

$$\begin{gathered} 鈬�\mathrm{A}\stackrel{藱}{\mathrm{x}}={\mathrm{c}}_1\mathrm{A}{\stackrel{藱}{\mathrm{v}}}_1+{\mathrm{c}}_2\mathrm{A}{\stackrel{藱}{\mathrm{v}}}_2 \\ 鈬�\mathrm{A}\stackrel{藱}{\mathrm{x}}={\mathrm{c}}_1{\stackrel{藱}{\mathrm{v}}}_1+2{\mathrm{c}}_1{\stackrel{藱}{\mathrm{v}}}_2 \\ 鈬�\left|\left|\stackrel{藱}{\mathrm{x}}\right|\right|=\sqrt{{\mathrm{c}}_1^2+4{\mathrm{c}}_2^2} \\ 鈬�\stackrel{藱}{\mathrm{x}}.\mathrm{A}\stackrel{藱}{\mathrm{v}}={\mathrm{c}}_1^2+2{\mathrm{c}}_2^2 \end{gathered}$$

Let is the angle between $\stackrel{藱}{\mathrm{x}}$and $\mathrm{A}\stackrel{藱}{\mathrm{x}}$. Then,

$$\begin{gathered} \cos 胃=\frac{\stackrel{藱}{x}.A\stackrel{藱}{a}}{||\stackrel{藱}{x}||||A\stackrel{藱}{x}||} \\ =\frac{c_1^2+2c_2^2}{\sqrt{c_1^2+c_2^2}\sqrt{c_1^2+4_2^2}} \\ =\frac{\sqrt{\left(c_1^2+2c_2^2\right)}}{\sqrt{c_1^4+5c_1^2{c}_2^2+4c_2^4}} \\ ={\left(\frac{c_1^4+4c_1^2{c}_2^2+4c_2^4}{c_1^4+4c_1^2{c}_2^2+4c_2^4+c_1^2{c}_2^2}\right)}^{1/2} \end{gathered}$$

$$\begin{gathered} =\left(\frac{1}{1+{\displaystyle \frac{{\mathrm{c}}_1^2{\mathrm{c}}_2^2}{{\mathrm{c}}_1^4+4{\mathrm{c}}_1^2{\mathrm{c}}_2^2+4{\mathrm{c}}_2^4}}}\right) \\ =\left(\frac{1}{1+{\displaystyle \frac{{\mathrm{c}}_1^2{\mathrm{c}}_2^2}{{\mathrm{c}}_1^4+4{\mathrm{c}}_1^2{\mathrm{c}}_2^2+4{\mathrm{c}}_2^4}}}\right) \\ ={\left(\frac{1}{1+{\displaystyle \frac{{\mathrm{c}}_1^2{\mathrm{c}}_2^2}{{({\mathrm{c}}_1^2+2{\mathrm{c}}_2^2)}^2}}}\right)}^{1/2} \\ ={\left(\frac{1}{1+{\displaystyle \frac{{\mathrm{c}}_1{\mathrm{c}}_2}{{({\mathrm{c}}_1^2+2{\mathrm{c}}_2^2)}^2}}}\right)}^{1/2} \end{gathered}$$

Here, $\cos \mathrm{胃}$takes minimum value when localid="1659678852211" $\frac{{\mathrm{c}}_1{\mathrm{c}}_2}{{\mathrm{c}}_1^2+{\mathrm{c}}_2^2}$takes maximum value.

To find the maximum value of $\frac{{\mathrm{c}}_1{\mathrm{c}}_2}{{\mathrm{c}}_1^2+{\mathrm{c}}_2^2}$, differentiate it with respect to ${\mathrm{c}}_1$and ${\mathrm{c}}_2$.

$$\begin{gathered} \frac{鈭�}{鈭�c_1}\left(\frac{{\mathrm{c}}_1{\mathrm{c}}_2}{{\mathrm{c}}_1^2+2{\mathrm{c}}_2^2}\right)=\frac{\left({\mathrm{c}}_1^2+2{\mathrm{c}}_2^2\right)c_2-c_1{c}_2\left(2c_2\right)}{{\left({\mathrm{c}}_1^2+2{\mathrm{c}}_2^2\right)}^2} \\ =\frac{-c_1^2{c}_2+2c_2^3}{{\left({\mathrm{c}}_1^2+2{\mathrm{c}}_2^2\right)}^2} \\ =\frac{c_2(-c_1^2+2c_2^2)}{{\left({\mathrm{c}}_1^2+2{\mathrm{c}}_2^2\right)}^2} \\ \frac{鈭�}{鈭�c_1}\left(\frac{{\mathrm{c}}_1{\mathrm{c}}_2}{{\mathrm{c}}_1^2+2{\mathrm{c}}_2^2}\right)=\frac{\left({\mathrm{c}}_1^2+2{\mathrm{c}}_2^2\right)c_2-c_1{c}_2\left(4c_2\right)}{{\left({\mathrm{c}}_1^2+2{\mathrm{c}}_2^2\right)}^2} \\ =\frac{c_1^3-2c_1{c}_2^2}{{\left({\mathrm{c}}_1^2+2{\mathrm{c}}_2^2\right)}^2} \\ =\frac{c_2(c_1^2-2c_2^2)}{{\left({\mathrm{c}}_1^2+2{\mathrm{c}}_2^2\right)}^2} \end{gathered}$$

In order to find the extreme value, equate both the partial derivatives with 0. Ignore the solution ${\mathrm{c}}_1,{\mathrm{c}}_2=0$since it corresponds to minimum. The other condition we obtain is ${\mathrm{c}}_1^2=2{\mathrm{c}}_2^2$. Thus, the maximum is achieved when this condition is satisfied.

When ${\mathrm{c}}_1^2=2{\mathrm{c}}_2^2$, we have

$$\begin{gathered} \frac{{\mathrm{c}}_1^2{\mathrm{c}}_2^2}{\left({\mathrm{c}}_1^2+2{\mathrm{c}}_2^2\right)}=\frac{2{\mathrm{c}}_2^4}{{\left(4{\mathrm{c}}_2^2\right)}^2} \\ =\frac{2{\mathrm{c}}_2^4}{16{\mathrm{c}}_2^4} \\ =\frac{1}{8} \end{gathered}$$

Hence, $\mathrm{肠辞蝉胃}={\left(\frac{1}{1+{(1/8)}^2}\right)}^{1/2}=\sqrt{8/9}$

This means that the minimum possible value of is $\cos \mathrm{胃}\mathrm{is}\sqrt{8/9}>\sqrt{3}/2$. Thus, in any case $\cos \mathrm{胃}>\sqrt{3}/2$which implies that $胃<蟺/6$.

Thus, the given statement is TRUE.