看看

\begin{align*}
x^2+y^2+z^3 &\geqslant \frac12(x+y)^2+z^3 \\
& =\frac12(14-z)^2+z^3 \\
& =\frac12(z-2)^2(9+2z)+80 \\
& \geqslant 80
\end{align*}
当$x=y=6$, $z=2$时取等。

\begin{align*}
21(x^2+y^2)+z^2 & = 21(x^2+y^2)+\left( \frac{1-xy}{x+y} \right)^2 \\
& \geqslant \frac{21}2(x+y)^2+\left( \frac{1-\frac{(x+y)^2}4}{x+y} \right)^2 \\
& =\frac{169(x+y)^2}{16}+\frac1{(x+y)^2}-\frac12 \\
& \geqslant 2\sqrt{\frac{169}{16}}-\frac12 \\
& =6
\end{align*}
当$x=y=\frac1{\sqrt{13}}$, $z=\frac6{\sqrt{13}}$时取等；

\begin{align*}
5a^2+4b^2+5c^2+d^2 &= 5(a^2+c^2)+\frac45\left( \frac14+1 \right)(4b^2+d^2) \\
& \geqslant \frac52(a+c)^2+\frac45(b+d)^2 \\
& \geqslant 2\sqrt{2(a+c)(b+d)} \\
& =2\sqrt{2(ab+bc+cd+da)} \\
& =2\sqrt2
\end{align*}
当$a=c=\frac1{\sqrt[4]{50}}$, $b=\frac1{\sqrt[4]{200}}$, $d=\sqrt[4]{\frac{32}{25}}$时取等；

其实我八点多的时候在 http://kkkkuingggg.5d6d.net/thread-525-1-1.html 已经发了一份了……

$f\left( x\right) =\dfrac {e^{x}} {x}+x^{2}+2$

解：不妨设$P \left ( \dfrac{y_1^2}{4},y_1\right),Q\left(\dfrac{y_2^2}{4},y_2\right),M\left(\dfrac{y_0^2}{4},y_0\right)$,依题意可得$$\begin{cases}\frac{y_1}{\dfrac{y_1^2}{4}+1}=\frac{y_0}{\dfrac{y_0^2}{4}+1} \\ \frac{y_2+1}{\dfrac{y_2^2}{4}-1}=\frac{y_0+1}{\dfrac{y_0^2}{4}-1}\end{cases},$$
解得$$\begin{cases}y_1=\frac{4}{y_0}\\ y_2=-\frac{4+y_0}{1+y_0}\end{cases},$$
于是直线$PQ$的方程为：$$y-y_1=\frac{y_2-y_1}{\dfrac{y_2^2}{4}-\dfrac{y_1^2}{4}}\left(x-\frac{y_1^2}{4}\right),$$
整理即得：$$y-\dfrac{4}{y_0}=\dfrac{4y_0(1+y_0)}{4-y_0^2}x-\frac{16(1+y_0)}{(4-y_0^2)y_0},$$
于是$$y=\dfrac{4y_0(1+y_0)}{4-y_0^2}(x-1)-4,$$
即直线$PQ$过定点$(1,-4)$.

\begin{array}{|c|c|c|c|c|}
\hline\\1 & 2 & 3 & 16 & 17 \\
\hline\\
4 & 5 & 6 & 18 & 19 \\
\hline\\
7 & 8 & 6 & 20 & 21 \\
\hline\\
10 & 11 & 12 & 22 & 23 \\
\hline\\
13 & 14 & 15 & 24 & 25 \\
\hline
\end{array}
\begin{cases}1=2 \\ 4-67\\8999\end{cases}