Find all values of the parameter a $\in\Bbb R$ for which the following inequality is valid for all x $\in\Bbb R$.

$$ 1+\log_5(x^2+1)\ge \log_5(ax^2+4x+a) $$

I'm lost when I got to this stage: $ 5x^2-4x+5\ge ax^2+a$

I did this but still don't know how to proceed: $ (5-a)x^2-4x+(5-a)\ge0 $

My reasoning is that the discriminant for $ (5-a)x^2-4x+(5-a)\ge0 $ must be $\ge0$ because x $\in\Bbb R$. And from that I get a <= 7 or a <= 3.

Then because $log_5(ax^2+4x+a)$, then $(ax^2+4x+a)$ > 0. Then..? Here I use same reasoning as for finding the above a <= 3 or 7 too, that is x $\in\Bbb R$. Then I get a <= -2 and a <= 2 by using discriminant.

Where does my reasoning go wrong? Can anyone explain to me how to solve this?

Answer given is $(2,3]$.


Hint: To find where a quadratic function $f$ satisfies $f(x)\geq 0$, first find where $f(x)=0$. On each of the remaining intervals, the sign of $f$ must be constant. The same method is useful for any such inequality ($<,>,\leq,\geq$).

This method works with any polynomial $f$, not just quadratics.

It also extends easily to rational functions. Look for zeroes of the numerator and of the denominator. Be careful where the denominator vanishes -- such points are typically excluded from the domain, but they still partition the line into intervals along with the zeroes of the numerator.

Addendum: There are two things to worry about: (1) both sides have to be defined for all real $x$; and (2) the inequality must be true for all real $x$.

First part: The LHS is defined for all real $x$. For the RHS, we require that the quadratic expression $ax^2 + 4x + a$ be strictly positive so that the logarithm is defined. In terms of the corresponding graph, which is a parabola, this means the parabola must not meet the $x$-axis and it must open upwards. Denoting the discriminant by $\Delta$, we must have $\Delta<0$ (no real zeroes) and $a>0$ (opens upward): $$16-4a^2<0$$ $$(4-2a)(4+2a)<0$$$$(a-2)(a+2)>0$$This is satisfied when $a<-2$ or $a>2$. Since we also require $a>0$, we eliminate the case $a<-2$ and conclude that we must have $$\boxed{a>2}\tag{req. for RHS to be defined}$$

Second part: Exponentiating both sides of the original equation with base $5$ and rearranging, we arrive at $$(a-5)x^2 + 4x + (a-5)\leq 0$$ We now ask: for which values of $a$ is this true for all real $x$? Denote the LHS by $f(x)$. If $a-5=0$, the inequality is linear, and so is not satisfied by all real $x$ (some $x$ would produce positive values of $f(x)$). If $a-5>0$, the function is a quadratic with a positive leading coefficient; its graph is a parabola that opens upward and so $f(x)>0$ for sufficiently large $x$ and the original inequality fails.

We are therefore left with the requirement that $a-5<0$, i.e., $a<5$, in order to have the parabola open downward. This is not enough to guarantee that $f(x)<0$ for all $x$; for this, we require that the vertex of the parabola lie below the $x$-axis as well. The $x$-coordinate of the vertex is the midpoint between the zeroes, so it is at $$x_v=\frac{-4}{2(a-5)}$$ (To see this, recall the roots of the quadratic $Ax^2+Bx+C=0$ are $x=\frac{-B\pm\sqrt{\Delta}}{2A}$, so the midpoint is $x=\frac{-B}{2A}$.)

It is easy enough to compute the $y$-coordinate of the vertex (I leave this to you -- tedious but not hard) as $$y_v=f(x_v)=f\left(\tfrac{-4}{2(a-5)}\right)=\frac{(a-9)(a-3)}{a-5}$$ Since $a<5$, for this to be nonpositive requires that $$\boxed{a\leq3}\tag{req. for inequ. always true}$$

Together: So, in the end, we require $$\boxed{2<a\leq 3}$$ as desired.

  • $\begingroup$ I have tried once more and my reasoning have added into the edited post. I still can't solve it $\endgroup$
    – JJChai
    Mar 21 '16 at 21:16
  • $\begingroup$ Thanks thanks thanks thanks !!!!!! I understand a lot more now! Thanks!!! :) $\endgroup$
    – JJChai
    Mar 23 '16 at 11:18
  • $\begingroup$ Glad it helped! One thing to take away from this: break a big problem into smaller manageable steps. Be methodical, and you can tackle anything! $\endgroup$
    – MPW
    Mar 23 '16 at 11:43

$\log(x)$ is monotonically increasing. Thus, assuming $ax^2+4x+a\gt0$ we can deduce $$ \begin{array}{c} 1+\log_5(x^2+1)\ge\log_5(ax^2+4x+a)\\ \Updownarrow\\ 5x^2+5\ge ax^2+4x+a\\ \Updownarrow\\ (5-a)x^2-4x+(5-a)\ge0 \end{array}\tag{1} $$ The last inequality in $(1)$ is satisfied for $x$ on a (possibly empty) finite closed interval when $a\gt5$ and on the complement of a (possibly empty) finite open interval when $a\lt5$. We want the last inequality in $(1)$ to be satisfied for all $x\in\mathbb{R}$, so we want the solution to be the complement of an empty interval. That means we want $a\lt5$.

The endpoints of the (possibly empty) finite interval are given by $$ x=\frac{2\pm\sqrt{-21+10a-a^2}}{5-a}\tag{2} $$ When $3\lt a\lt7$, the discriminant is positive and so the interval is not empty. Combined with the constraint that $a\lt5$, we want $a\le3$.

We also want that $ax^2+4x+a\gt0$ for all $x$. Obviously, we need $a\gt0$. Next, we need $$ ax^2+4x+a=a\left(x+\frac2a\right)^2+a-\frac4a\gt0\tag{3} $$ So that $(3)$ is positive for all $x$, we need $a\gt2$.

Thus, we want $2\lt a\le3$.


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