Minimum of $x^2+\frac{a}{x}$ without Calculus How can I find the minimum of $x^2+\frac{a}{x}$ on $\mathbb{R}_+$ without calculus?
 A: 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.)
A: 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 $3({a^2 \over 4})^{2 \over 3}$.
A: 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 $x^2=\frac{a}{2x}$.
A: 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.
A: 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.
