Find the maximum value of $x+(p/x)$, if $x<0$ and $p>0$

Question

I tried to solve it this way: $$x+(p/x)=t$$ $$x^2+2p+(p/x)^2=t^2$$ $$x^2+(p/x)^2=t^2-2p$$ Because both $x^2,(p/x)^2\ge0$, the whole left side should be $\ge0$. But because it's an equation, this means that

$t^2-2p≥0\\(t-\sqrt{2p})(t+\sqrt{2p})\ge0$

Therefore $t\in(-\infty,-\sqrt{2p}\ ]\cup[\sqrt{2p}, +\infty)$.

My answer is incorrect. What am I doing wrong?

Answer 1

Let $t=-x$. Consider $$y=-(t+\frac{p}{t})$$ Clearly, $$\text{max}(y)=-1*\text{min}(t+\frac{p}{t})=-2\sqrt p\ \ \text{by AM-GM}$$ Note that it occurs when $t=p/t$, or $t=\sqrt p$ ($x=-\sqrt p$)

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The problem with your solution was that you simply put $t^2-2p\geq 0$. But, can $x^2+p^2/x^2$ even attain the value 0? The answer is NO! It's minimum value is $2p$. So, if you $t^2-2p\geq 2p$, and proceed the way you did, you'll get the answer ;)

Answer 2

Denote $$f(x)=x+\frac{p}{x}$$ then $$f'(x)=1-\frac{p}{x^2}$$ Can you solve this? And $$f''(x)=\frac{2p}{x^3}$$

Answer 3

Attempt.

$y:= x +p/x, x<0, p>0.$

$Y:= -y= -x-p/x=$

$ X+p/X$, where $X=-x >0$.

Find the minimum of $Y(X).$

$X+p/X=$

$ (√X-√p/√X)^2 +2√p \gt 0.$

Hence

$X+p/X \ge 2√p, $ (why?)

Minimum value $2√p$ is attained for

$(√X-√p/√X)^2=0$, i.e. at

$X=√p.$