Finding maxima of a function $f(x) = \sqrt{x} - 2x^2$ without calculus 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
 A: 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 $$x\geq 0$$
A: $f\left(\frac{1}{4}\right)=\frac{3}{8}.$
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.
A: 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
$$2e+4b=0$$ 
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 $x=1/4.$
A: $$\dfrac14=\dfrac{2t^3+3\cdot\dfrac{1-2t^3}3}{1+3}\ge\sqrt[4]{2t^3\left(\dfrac{1-2t^3}3\right)^3}$$
