Find the minimum of $x^3+\frac{1}{x^2}$ for $x>0$

Finding this minimum must be only done using ineaqualities.

$$x^3+\frac{1}{x^2}=\frac{1}{2}x^3+\frac{1}{2}x^3+\frac{1}{3x^2}+\frac{1}{3x^2}+\frac{1}{3x^2}$$

Using inequalities of arithemtic and geometric means:

$$\frac{\frac{1}{2}x^3+\frac{1}{2}x^3+\frac{1}{3x^2}+\frac{1}{3x^2}+\frac{1}{3x^2}+1}{6}\geqslant \sqrt[6]{\frac{1}{2}x^3\frac{1}{2}x^3\frac{1}{3x^2}\frac{1}{3x^2}\frac{1}{3x^2}}=\sqrt[6]{\frac{1}{108}}\Rightarrow x^3+\frac{1}{x^2}\geqslant 6\sqrt[6]{\frac{1}{108}}-1$$

Sadly $$\ 6\sqrt[6]{\frac{1}{108}}-1$$ is not correct answer, it is not the minimum.

• I know that you can't use calculus techniques, but I think the correct answer will be; $$f(x)=3(\frac{2}{5})^\frac{2}{5}-\frac{2}{(\frac{2}{5})^\frac{3}{5}}$$ Jun 20, 2020 at 12:32
• Why would you add 1? Jun 20, 2020 at 12:40

Very similar to what you have done: $$\frac{\frac{1}{2}x^3+\frac{1}{2}x^3+\frac{1}{3x^2}+\frac{1}{3x^2}+\frac{1}{3x^2}}{5}\geq \sqrt[5]{\frac{1}{2}x^3\frac{1}{2}x^3\frac{1}{3x^2}\frac{1}{3x^2}\frac{1}{3x^2}}=\sqrt[5]{\frac{1}{108}}$$ This gives us $$x^3+\frac{1}{x^2}\geq \sqrt[5]{\frac{1}{108}}$$

Desmos screenshot:

P.S. Your method fails because equality holds only when $$\frac{x^3}{2}=\frac{1}{3x^2}=1$$ which is impossible. The extra "one" you added in your AM-GM application screwed your attempt.

• I asked it before too, is it a coincidence that $\frac{x^3}{2}=\frac{1}{3x^2}$ for least value or something more than that Jun 20, 2020 at 12:43
• @AshWhole No. It's not a coincidence. When you apply the AM-GM inequality on a set of numbers, equality holds when all the numbers are equal. Jun 20, 2020 at 12:50
• @AshWhole, that just means the functions are closest together at $x = 1$, not that $x^3 + 1/x^2$ has a minimum there! Try plotting it. You can only conclude that this point would be a minimum if the function that you're using as a lower bound is constant. A less complicated example is that a convex parabola is always bounded below by a tangent, but unless that tangent is horizontal, the point of intersection is not the parabola's minimum value. Jun 20, 2020 at 12:57
• I think this answer needs a bit more work to show that this lower bound is actually the minimum though (ie it is attained by the function). Plotting it does illustrate this, but it's good form to at least point out that $\tfrac 12 x^3 = 1/3x^2$ has a solution, and better yet, find this value of $x$ and illustrate that it attains this minimum. Jun 20, 2020 at 13:02
• @AshWhole well can't do $S=x^3+\frac{1}{x²} \ge 2\sqrt{x}=l$ because the they are greater than keeps changing . For x=1 sure the equality holds, but say x = $\frac{1}{2}$ the S is greater than $\sqrt{2}$ but it could be less than 2 , so we try to break them in such a way that the thing they are greater than is constant . Jun 20, 2020 at 13:03

A bit late answer but I thought it might be worth noting.

I was wondering whether we could squeeze out the minimum directly from Young's inequality:

$$ab\leq \frac{a^p}{p} + \frac{b^q}{q};\: a,b \geq 0;\: p,q>1 \text{ and } \frac 1p + \frac 1q = 1$$

And, yes indeed, this works well, too. Since we need the powers of $$x$$ to cancel out, I start constructing the exponents with $$x^6$$: $$\begin{eqnarray*} x^3 + \frac 1{x^2} & = & \left(x^6\right)^{\frac 12} + \left(\frac 1{x^6}\right)^{\frac 13}\\ & = & \left(\sqrt[5]{x^6}\right)^{\frac 52} + \left(\frac 1{\sqrt[5]{x^6}}\right)^{\frac 53}\\ & = & \frac 25\cdot \left(\sqrt[5]{\frac{5^2}{2^2}}\sqrt[5]{x^6}\right)^{\frac 52} + \frac 35\cdot\left(\sqrt[5]{\frac{5^3}{3^3}}\frac 1{\sqrt[5]{x^6}}\right)^{\frac 53} \\ & \stackrel{Young}{\geq} & \sqrt[5]{\frac{5^2}{2^2}\cdot \frac{5^3}{3^3}} = \frac 5{\sqrt[5]{2^2\cdot 3^3}} \end{eqnarray*}$$

Equality holds for $$a^p = b^q$$ which means $$\frac 52 x^3 = \frac 53 \frac 1{x^2} \Leftrightarrow x=\sqrt[5]{\frac 23}$$

• Thank you for the work! Jun 21, 2020 at 7:15
• Why do we need the powers to cancel out? Jun 22, 2020 at 8:53
• @AshWhole Because we would like to have a constant value (independent of $x$) as minimum. Jun 22, 2020 at 10:29

We need to find a positive number $$a$$ such that

$$x^3+{1\over x^2}\ge a^3+{1\over a^2}$$

for all $$x\gt0$$. The existence of such an $$a$$ is not in doubt, since $$x^3+{1\over x^2}\to\infty$$ as $$x\to0$$ and $$x\to\infty$$. But

\begin{align} x^3+{1\over x^2}\ge a^3+{1\over a^2} &\iff(x^3-a^3)-\left({1\over a^2}-{1\over x^2}\right)\ge0\\ &\iff(x-a)\left((x^2+ax+a^2)-{x+a\over a^2x^2} \right)\ge0 \end{align}

Now $$x-a$$ changes sign at $$x=a$$, so in order for nonnegativity to be maintained, the other factor, $$(x^2+ax+a^2)-{x+a\over a^2x^2}$$, must do so as well. In particular that factor must also equal $$0$$ at $$x=a$$, so we must have

$$3a^2={a+a\over a^2a^2}={2\over a^3}$$

or $$a=\sqrt[5]{2/3}$$.

• Why does it need to be zero ? Jun 21, 2020 at 20:23
• @AshWhole, the only way for a continuous function to change sign is to pass through $0$. Jun 21, 2020 at 21:18