MarkBcc168 wrote: Assume $c$ is a real number. If there exists $x\in[1,2]$ such that $\max\left\{\left |x+\frac cx\right |, \left |x+\frac cx + 2\right |\right\}\geq 5$, please find the value range of $c$. $$t=x+\frac{c}{x}$$$$\bullet Max\left \{ \left | t+2 \right |,\left | t \right | \right \} \geq 5\leftrightarrow \sqrt{\left ( \left | t+2 \right |+\left | t \right | \right )^2}+ \sqrt{\left ( \left | t+2 \right |-\left | t \right | \right )^2}\geq 10$$$$\leftrightarrow \left ( \left | t+1 \right |+1 \right )^2\geq 25\leftrightarrow \left | t+1 \right |\geq 4\leftrightarrow t\leq -5\vee t\geq 3$$$$\bullet x+\frac{c}{x}\leq -5\vee x+\frac{c}{x}\geq 3\leftrightarrow c\leq -x^2-5x\vee c\geq -x^2+3x,\forall x\in \left [ 1,2 \right ]$$$$\leftrightarrow c\leq \underset{x\in \left [ 1,2 \right ]}{Min}\left \{ -x^2-5x \right \}\vee c\ge \underset{x\in \left [ 1,2 \right ]}{Max}\left \{ -x^2+3x \right \}$$$$\leftrightarrow \boxed {c\leq -14\vee c\geq \frac{9}{4} }$$

MarkBcc168 wrote: Assume $c$ is a real number. If there exists $x\in[1,2]$ such that $\max\left\{\left |x+\frac cx\right |, \left |x+\frac cx + 2\right |\right\}\geq 5$, please find the value range of $c$. $\max\left\{\left |x+\frac cx\right |, \left |x+\frac cx + 2\right |\right\}\geq 5$ is equivalent to $x+\frac{c}{x} \leq -5$ or $x+\frac{c}{x} \geq 3$ Which is $c \leq -x^2-5x$ or $c \geq -x^2+3x$ When $x\in[1,2]$ $c \leq -x^2-5x \leq -6$ and $c \geq -x^2+3x \geq 2$ So, range of $c$ is $c \leq -6$ and $c \geq 2$

平凡恒等式用起来比较方便

The answer is any $c\in(-\infty, -6]\cup[2,\infty)$, which works because If $c\leqslant -6$, then $x=1$ gives $\left|x+\tfrac{c}{x}\right|\geqslant 5.$ If $c\geqslant 2$, then $x=1$ gives $\left|x+\tfrac{c}{x}+2\right|\geqslant 5$. Now we prove that any $c\in (-6,2)$ does not work, in other words $$\max\left\{\left |x+\frac cx\right |, \left |x+\frac cx + 2\right |\right\} < 5\iff -5 < x+\frac{c}{x} < 3$$for all $x\in[1,2]$. To prove that $-5 < x+\tfrac{c}{x}$, rewrite it as $x^2+5x+c>0\iff (x+2.5)^2 + c - 6.25 > 0$, which is clearly true. To prove that $x+\tfrac{c}{x} < 3$, rewrite it as $x^2-3x+c<0\iff (x-1.5)^2 + c - 2.25 < 0$, which is clearly true. Hence we are done.