Show that , if $a \ge 9/8$, then $f '(x) \ge 0$

Question

My questions are to do with the solution of the following question:

Let $f(x)=ax-\frac{x^3}{1+x^2}$ , where $a$ is a positive constant. Show that if $a\ge\frac{9}{8}$ then $f'(x)\ge0$ for all $x$.

Differentiating we get : $$f'(x)=\frac{a+(2a-3)x^2+(a-1)x^4}{(1+x^2)^2}$$

Next the following is said/done: "We can use the inequality $a\ge\frac{9}{8}$ immediately after finding f'(x) since $a$ appears in f'(x) with a plus sign always: $$f'(x)=\frac{a+(2a-3)x^2+(a-1)x^4}{(1+x^2)^2}\ge\frac{\frac{9}{8}+(\frac{9}{4}-3)x^2+(\frac{9}{8}-1)x^4}{(1+x^2)^2}=\frac{(x^2-3)^2}{8(1+x^2)^2}\ge0$$

My questions pertain to everything in bold and after the bold statement:

1. How is the author using the inequality $a\ge\frac{9}{8}$ to get to the final result?
2. What is the significance of $a$ appearing with a positive sign always?

Answer 1

First the sign of $f'(x)$ is the sign of its numerator, which is a biquadratic polynomial, so setting $u=x^2$, we obtain a quadratic polynomial: $$p(u)=(a-1)u^2+(2a-3)u+a.$$ Supposing $a\ne 1$, so we have a quadratic polynomial, it has a constant sign (or is $0$) if and only if is discriminant is non-positive: $$\Delta=(2a-3)^2-4a(a-1)=9-8a\le 0.$$ So the condition is indeed $a\ge \frac98$. Last, since the sign of $p(u)$ is constant, it is also the sign of $p(0)=a$.

Answer 2

First, he's saying that $$a\ge\frac98\implies ay\ge\frac{9y}{8}$$ for various values of $y$. This is only true when $y\ge0.$ I'm sure you know that multiplying by a negative number reverses the sense of an inequality.

As for how he's using it, he's just pulling a factor of $\frac18$ out of the numerator of the next-to-last fraction, then factoring it.

Answer 3

1) If $a \ge k$ then $a$ times anything positive plus or minus anything else will be greater than $k$ times the same thing plus or minus the thing.

So:

$a\ge \frac 98$ means $(2a - 3)x^2 \ge (2*\frac 98 - 3)x^2$ and $(a-1)x^4 \ge (\frac 98 - 1)x^4$ etc.

2) If $a \ge \frac 98$ and if $k > 0$ then $ak > \frac 98k$. But if $k < 0$ then $ak < \frac 98 k$.

Basically, if $a \ge \frac 98$ we can remplace $a$ with $\frac 98$ and get something less than or equal to the expression with $a$ presuming we never multiply $a$ by anything negative. Which as $x^2, x^4, 2, 1+x^2$ are all non-negative, is a safe assumption.

....

Basically there are two basic axioms of inequalities.

I) If $a \ge k$ then $a + c \ge k + c$ for all possible (including negative) $c$. and

II) If $a \ge k$ and $b > 0$ then $ab \ge ak$.

Given those two axioms than

$a \ge \frac 98 \implies 2a-3 \ge 2\frac 98 - 3; a-1 \ge \frac 98 - 1\implies$

$a\ge 98;(2a-3)x^2> (2\frac 98 - 3)x^2; (a-1)x^4 \ge (\frac 98 - 1)x^4\implies$

$a + (2a -3)x^2 + (2a-1)x^4\ge \frac{9}{8}+(\frac{9}{8}-3)x^2+(\frac{9}{8}-1)x^4\implies$

$\frac{a+(2a-3)x^2+(a-1)x^4}{(1+x^2)^2}\ge\frac{\frac{9}{8}+(\frac{9}{4}-3)x^2+(\frac{9}{8}-1)x^4}{(1+x^2)^2}$

Okay, we also need the definition

0) $a \le b; b\le c \implies a \le c$

To get $a \le \frac 98; (2a-3)x^2 \le (2\frac 98 -3)x^2; (a-1)x^4 \le (\frac 98 -1)x^4\implies$

$a + (2a-3)x^2 + (a-1)x^4 \ge a + (2a-3)x^2 + (\frac 98 - 1)x^4;$

$a + (2a-3)x^2 + (\frac 98 - 1)x^4\ge a + (2\frac 98 - 3)x^2 + (\frac 98 - 1)x^4;$

$a + (2\frac 98 - 3)x^2 + (\frac 98 - 1)x^4\ge \frac 98 + (2\frac 98 - 3)x^2 + (\frac 9/8 - 1)x^4;\implies$

$a + (2a-3)x^2 + (a-1)x^4\ge\frac 98 + (2\frac 98 - 3)x^2 + (\frac 98 - 1)x^4$

Answer 4

This is because since the denominator is always non-negative so must be the numerator. Also the numerator has a quadratic form whose coefficient of the term with highest power must be positive i.e. $$a-1>0\\a>1$$ also the $\Delta$ must be negative therefore $$(2a-3)^2<4a(a-1)\\4a^2-12a+9<4a^2-4a\\$$$$\Large a>\dfrac{9}{8}$$

Answer 5

We must show $(a-1)x^4+(2a-3)x^2+a\geq 0$ , and can easily assume $a\geq 1$.

1: exception case If $a=1$, $f'(x)=-x^2+1$

This eventually become negative.

2: If axis$\leq0⇔ -\frac{2a-3}{a-1}\leq0⇔a\geq\frac{3}{2}$ or $a<1 $, this need only $f(0)=a\geq0$

3: If axis $>0⇔ 1<a<\frac32$, this need $D\leq0$

$D=(2a-3)^2-4a(a-1)=-8a+9\leq0$

We get $a\geq\dfrac98$ by these 3 cases.