How can I find the minimum of $x^2+\frac{a}{x}$ on $\mathbb{R}_+$ without calculus?
Minimum of $x^2+\frac{a}{x}$ without Calculus
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2Why do you want to avoid calculus and make it yourself hard? – 2011-02-06
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0I think you want $a\geq0$. – 2011-02-06
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5@Jonas: the solution without calculus is actually much easier, as bobobinks shows. The use of AM-GM is a justification of a typical intuition in these situations: if you want to optimize f(x) + g(x) where f is increasing and g is decreasing, you roughly want a value of x where f and g are approximately equal. I find this much more enriching than a computation of derivatives. In general it is an interesting exercise to split up a sum in such a way that 1) AM-GM applies and 2) the equality case actually occurs. – 2011-02-06
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0If a<0 you have a problem: 1/x will get near -infinity when x tends towards 0. – 2011-02-06
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0@Qiaochu: You're right. Thanks for the advice. – 2011-02-06
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0@Jonas: Your comment is interesting since I go the other way around. Usually when I want to find a maximum or minimum the first thing I go after is some sort of bound obtained from simple inequalities. If I don't get it, my final resort is to use the big hammer, Calculus. – 2011-02-06
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0@Sivaram: I will try that too from now on, the reply is some kind of eye-opener. I would use calculus because it one of the first things I have learnt I think and it would be a no-brainer. – 2011-02-06
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0@Jonas: Thanks for taking my previous comment in the right spirit. The reason for my previous comment is that you do not need a hammer to kill a mosquito. And in general there is an elegance to solve problems without advanced tools. – 2011-02-06
5 Answers
We have: $x^2+\frac{a}{x}=x^2+\frac{a}{2x}+\frac{a}{2x}\geq3\sqrt[3]{(x^2)(\frac{a}{2x})(\frac{a}{2x})}=3\sqrt[3]{\frac{a^2}{4}}$, where AM-GM was used. Equality occurs if $x^2=\frac{a}{2x}$.
Revised (neater) answer: Fix $a>0$. For any $x>0$, $$ \Big(x^2 + \frac{a}{x}\Big) - \Big(m^2 + \frac{a}{m}\Big) = \frac{{(x - m)(x^2 m + m^2 x - a)}}{{mx}} \geq 0, $$ if $2m^3 = a$. We are done. (The above also shows that the minimum is strict.)
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0It could be argued that the existence of local minima (which is what you're using here) counts as calculus. – 2011-02-06
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0The comment above is no longer relevant. – 2011-02-07
In the overkill-but-doesn't-use-calculus department: We assume $a > 0$, and the goal is to determine the minimal $K$ such that the equation $x^2 + {a \over x} = K$ has a positive root. This is equivalent to finding how many positive roots $p_K(x) = x^3 - Kx +a$ has. The total number of real roots is determined by the discriminant of $p_K(x)$, which is given by $D_K = 4k^3 - 27a^2$ here. If $D_K > 0$, it has three distinct real roots. If $D_K < 0$ is has one real and two complex conjugate roots. If $D_K = 0$ it has three real roots including at least one repeated root.
Next, note from the graphs of $x^2$ and ${a \over x}$ that the function $x^2 + {a \over x}$ decreases from $\infty$ to $-\infty$ as $x$ goes from $-\infty$ to zero, so for any $K$ the equation $x^2 + {a \over x} = K$ has exactly one negative root. $p_K(0) \neq 0$ as long as $a > 0$. Hence by the above considerations $p_K(x)$ has a real positive root if and only if $D_K \geq 0$, which translates into $K^3 \geq 27{a^2 \over 4}$, or $K \geq 3({a^2 \over 4})^{2 \over 3}$. Hence the minimal value of $x^2 + {a \over x}$ for $x > 0$ is given by $3({a^2 \over 4})^{2 \over 3}$.
By the change of variable $x=az/2$, the answer is $(a/2)^{2/3}$ times the minimum $m$ of $u(z)=z^2+2/z$ over $z\ge0$. Since $u(1)=3$, $m\le3$. Hence $m=3$ iff $u(z)\ge3$ for every $z\ge0$ iff $p(z)\ge0$ for every $z\ge0$, where $p$ is the polynomial defined as $p(z)=z(u(z)-3)=z^3-3z+2$.
Since $p(1)=0$, $p(z)=(z-1)q(z)$ where $q$ is a polynomial of degree $2$. One finds $q(z)=z^2+z-2$. Since $q(1)=0$, $q(z)=(z-1)r(z)$ where $r$ is a poynomial of degree $1$. One finds $r(z)=z+2$. Hence $p(z)=(z-1)^2(z+2)$ and $p(z)\ge0$ for every $z\ge0$.
I do not know whether you consider this proof as being without calculus.
You can plot it for a few values of $a$ and guess the answer. Then plug it in and prove that small changes of $x$ increase the value. You may argue that this approach just hides the calculus and there is some truth to that. But you can do this without knowing how to take a derivative.