By Alexey Kubanov

2019-01-12 12:35:43 8 Comments

My question is how to prove that $f(x) = \sqrt x - 2x^2$ has its maximum at point $x_0 = \frac{1}{4}$

It is easy to do that by finding its derivative and setting it to be zero (this is how I got $x_0 = \frac{1}{4}$). But the task is to do that without using any calculus tools and I`m stuck at it.

My idea was to introduce $t = \sqrt x$ to get $f(t) = t - 2t^4$ and then find its maximum because I know how to do that with parabola which can be transformed to $a(x-x_0)^2 + y_0$. So I was trying to turn $f(t) = t - 2t^4$ into $f(t) = a(x-x_0)^4 + y_0$ but it seems impossible.

My second thought was to use the definition of rising function on an interval so I supposed we have $a, b \in \left(0; \frac{1}{4}\right) \text{and } a < b$. Then I prove that $$\sqrt a - 2a^2 < \sqrt b - 2b^2$$ $$2(b - a)(a + b) < \sqrt b - \sqrt a$$ $$ 2(\sqrt a + \sqrt b)(a + b) < 1$$ which is true since $0 < a < b < \frac{1}{4}$

Exactly the same way I prove that for every $a, b \in \left(\frac{1}{4}; +\infty\right) \text{and } a < b$ $$\sqrt a - 2a^2 > \sqrt b - 2b^2$$

So we have that $f(x)$ is rising on $\left(0; \frac{1}{4}\right)$ and declining on $\left(\frac{1}{4}; +\infty\right)$, consequently at $\frac{1}{4}$ we have a maximum of $f$

But I guess it is not fair to use derivatives to find the maximum and then simply prove that this value is correct. Is there an even better solution? What are your thoughts about my proof?

UPD: there are a lot of solutions which are formally OK but first we have to guess $x_0 = \frac{1}{4}$ or $y_0 = \frac{3}{8}$. My goal is to find a solution which does not rely on guessing and I think the inequality above is the best way to do that


@lab bhattacharjee 2019-01-12 13:43:40


@B. Goddard 2019-01-12 13:34:47

If we rearrange the quartic to instead find the minimum of $2t^4-t$, we can proceed as follows. If there is a minimum point at $(r,s)$, we can subtract $s$ from the polynomial, and the graph of $2t^4-t-s$ will just touch the $t$-axis and then turn around. So we know that $2t^4-t-s$ is divisible by a linear term squared. Write

$$2t^4-t-s = (at^2+bt+c)(dt+e)^2.$$

We seek $a,b,c,d,e$, so multiply out the rhs:

$$2t^4-t-s = ad^2t^4+(2ade+bd^2)t^3+(ae^2+2bde+cd^2)t^2+(be^2+2cde)t+ce^2,$$

and equate like coefficients.

We start with $ad^2=2$ and because the content makes no difference, we have some freedom. I take $a=1/2$ and $d=2$ (but other choices will work fine.) From the cubic coefficient we now have


so that $e=-2b.$ Plug this in the equation for the quadratic term and we get $c=3b^2/2.$ Plug everything we have so far into the equation for the linear term and solve for $b$ to get $b=1/2$. Back substitution gives $e=-1$ and $c=3/8.$

So we have

$$2t^4-t-s = \left(\frac{1}{2}t^2-\frac{1}{2}t+\frac{3}{8}\right)(2t-1)^2.$$

This polynomial has a minimum, then, at $t=1/2$, so the answer for the original function is $x=1/4.$

@Michael Rozenberg 2019-01-12 13:03:25


We'll prove that we got a maximal value.

Indeed, we need to prove that $$\sqrt{x}-2x^2\leq\frac{3}{8},$$ which is true by AM-GM: $$2x^2+\frac{3}{8}=2x^2+3\cdot\frac{1}{8}\geq4\sqrt[4]{2x^2\left(\frac{1}{8}\right)^3}=\sqrt{x}.$$

We can get a value $\frac{3}{8}$ by the following way.

Let $$\max_{x\geq0}f=k.$$ Thus, $$\sqrt{x}-2x^2\leq k$$ or $$2x^2+k\geq\sqrt{x}.$$ Now, since for $x\geq0$ we have $$\sqrt{x}=\sqrt[4]{x^2},$$ we need four addends if we want to use AM-GM.

And indeed, by AM-GM $$2x^2+k=2x^2+3\cdot\frac{k}{3}\geq4\sqrt[4]{2x^2\left(\frac{k}{3}\right)^3}.$$ Id est, we need $$4\sqrt[4]{2x^2\left(\frac{k}{3}\right)^3}=\sqrt{x}$$ or $$512k^3=27$$ or $$k=\frac{3}{8}.$$ The equality occurs for $$2x^2=\frac{k}{3}$$ or $$x=\frac{1}{4},$$ which says that we got a maximal value.

@Alexey Kubanov 2019-01-12 13:08:37

We first have to guess $\frac{1}{4}$ and my goal is to avoid that

@Michael Rozenberg 2019-01-12 13:17:50

@Alexey Kubanov I added something. See now.

@TheSimpliFire 2019-01-12 13:18:51

@MichaelRozenberg Great answer!

@Alexey Kubanov 2019-01-12 13:21:16

Thank you a lot! Now I got it and the proof is really amazing!

@Dr. Mathva 2019-01-12 13:23:49

@MichaelRozenberg I always find it awesome how you play with inequalities. I've seen some other of your posts and they're all brilliant! I know inequalities are practicing and using the same 100 tricks again and again... But I was wondering if there a book you can recommend on inequalities?

@Michael Rozenberg 2019-01-12 13:31:34

@Dr. Mathva I think the best book it's the Inequalities Forum in AoPS. Also, there are books and papers of Vasile Cirtoje.

@Moo 2019-01-12 14:00:45

@MichaelRozenberg: I think you meant

@Michael Rozenberg 2019-01-12 14:03:56

@Moo Yes, of course! There is also the beautiful last book, which he wrote in 2016.

@Dr. Sonnhard Graubner 2019-01-12 12:49:11

We have to pove that $$\sqrt{x}-2x^2\le \frac{3}{8}$$ or $$\sqrt{x}\le 2x^2+\frac{3}{8}$$ after squaring we obtain $$0\le 4x^4+\frac{3}{2}x^2-x+\frac{9}{64}$$ but this is $$0\le \frac{1}{64}(4x-1)^2(16x^2+8x+9)$$ this is true for $$x\geq 0$$

@KM101 2019-01-12 12:51:56

That involves initially assuming the maximum occurs at $x = \frac{1}{4}$.

@Dr. Sonnhard Graubner 2019-01-12 12:54:25

I think no, i have proved that $$f(x)\le \frac{3}{8}$$ for $$x\geq 0$$

@Alexey Kubanov 2019-01-12 12:55:16

The question is how do you get $\frac{3}{8}$. To get that you first have to know that $f$ has its maximum at $\frac{1}{4}$ and then compute $f(\frac{1}{4})$ which is not good I guess...

@Alexey Kubanov 2019-01-12 12:58:23

@Dr.SonnhardGraubner yes, formally it is OK, but choosing $\frac{3}{8}$ among all $r \in \mathbb R$ is tough :)

@Dr. Sonnhard Graubner 2019-01-12 12:59:22

You can use a graphic calculator to get this

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