Noting that no positive $x$ satisfy the equation, We substitute $x\to -x.$ Note that $x>0.$ Now $x^4+ax^3+bx^2+cx+1=0\iff x^2-ax+b-\frac{c}{x}+\frac{1}{x^2}=0$ $\iff (x-a/2)^2+(1/x-c/2)^2=-b+\frac{a^2}{4}+\frac{c^2}{4}.$ Lemma: $-b+\frac{a^2}{4}+\frac{c^2}{4} \leq 0.$ Proof: By Extended LoS, this is equivalent to $4b\geq a^2+c^2\iff \sin B \geq \sin A^2+\sin C^2\iff \sin A\cos C+\cos A\sin C\leq \sin A ^2+\sin C^2$ Now note that $\angle B\geq 90^{\circ}\implies \angle A \leq 90^{\circ}-\angle C,$ so since $\cos$ is decreasing for acute angles, $\cos A\geq \sin C.$ Similarly $\cos C\geq \sin A$ so the inequality follows. Now obviously equality has to hold everywhere so $\angle B=90^{\circ}.$ Thus $a^2+c^2=16.$ Also, we need $x=a/2=2/c$ so $ac=4.$ Solving this gives $a,c=\sqrt 6\pm \sqrt 2$ in some order, so it follows that $x=\frac{\sqrt6\pm \sqrt 2}{2}.$ We substitute back to get $x=\frac{-\sqrt6\pm\sqrt2}{2}.$

The answer is $x = -\frac{1}{2} (\sqrt6 \pm \sqrt 2)$ I think? We prove this the only possible answer. Evidently $x < 0$. Now, note that \[ a^2+c^2 \le b^2 \le 4b \]since $b \le 4$ (the diameter of its circumcircle). Then, \begin{align*} 0 &= x^4 + ax^3 + bx^2 + cx + 1 \\ &= x^2 \left[ \left( x + \frac{1}{2} a \right)^2 + \left( \frac1x + \frac{1}{2} c \right)^2 + \left( b - \frac{a^2+c^2}{4} \right) \right] \\ &\ge 0+0+0 = 0. \end{align*}In order for equality to hold, we must have $x = -\frac{1}{2} a$, $1/x = -\frac{1}{2} c$, and $a^2+c^2 = b^2 = 4b$. This gives us $b = 4$, $ac = 4$, $a^2+c^2=16$. Solving for $a,c > 0$ implies \[ \{a,c\} = \left\{ \sqrt6 \pm \sqrt 2 \right\}. \]This gives the $x$ values claimed above; by taking $a$, $b$, $c$ as deduced here, we find they work too.

Fake geo > Spent three hours on this but got nowhere.

if i screwed up finding equality case after getting $a^2+c^2=16$ how many points? anyway here is a (correct) solution

djmathman wrote: $50 and an icecream cake that Titu wrote this Harder than #5.

I used 3 hours to get ~1 point on this problem. OOF

Note that for positive $x$ we have \begin{align*} x^4+bx^2+1 &\geqslant 2\sqrt{b}x\sqrt{x^4+1} \\ &\geqslant xb\sqrt{x^4+1} \\ &\geqslant x\sqrt{(a^2+c^2)(x^4+1)} \\ &\geqslant x(ax^2+c) \\ &= ax^3+cx. \end{align*}Equality for $b^2=a^2+c^2=16$ and $bx^2=x^4+1$ which gives the desired solutions.

leequack wrote: Fake geo > Spent three hours on this but got nowhere. Spent 4.5 hours and got nowhere

@above spent 6 hours on this and got nowhere (after the test too)

Wait, how do you even guess the problem author... for dj's AIME problem, "David" was the key but here there's no clear suggestion.

The answer is $\tfrac12(-\sqrt6\pm\sqrt2)$. We complete the square: \begin{align*} 0&=x^4+ax^3+bx^2+cx+1\\ &=x^2\left(x+\frac a2\right)^2+\left(\frac c2x+1\right)^2+\left(b-\frac{a^2+c^2}4\right). \end{align*}However since $\angle B\ge90^\circ$ and $b\le 4$, $a^2+c^2\le b^2\le 4b$, so $b-\tfrac14(a^2+c^2)\ge0$. Then by the Trivial Inequality all terms of the above expression are zero. In particular, $\angle B=90^\circ$, and since $x\ne 0$ (as $x=0$ contradicts $\tfrac c2x+1=0$), \[0=x+\frac a2=\frac c2x+1\implies x=-\frac a2=-\frac2c.\]Thus we have $ac=4$ and $a^2+c^2=16$. Solving, $\{a,c\}=\{\sqrt6\pm\sqrt2\}$, and the desired result follows.

aftermaths wrote: Noting that no negative $x$ satisfy the equation, Negative numbers do satisfy the equation though.

redbomb1 wrote: aftermaths wrote: Noting that no negative $x$ satisfy the equation, Negative numbers do satisfy the equation though. I'm almost 100% sure the solution writer meant positive.

The inequalities where the author uses extended law of sines are reversed too.

My answer was wrong, but only because I messed up when writing the answers (I wrote the value of a^2 as a and wrote the value of c^2 as c). All of my other steps are correct, only I messed up those answers. Will I lose points?

Note that $x < 0$. Completing the square twice, note \begin{align*} 0 & = x^4 + ax^3 + bx^2 + cx + 1\\ & = \left(x^4 + ax^3 + \frac{a^2}{4}x^2\right) + \left(1 + cx + \frac{c^2}{4}x^2\right) + \left(b - \frac{a^2}{4} - \frac{c^2}{4}\right)x^2\\ & = x^2\left(x + \frac{a}{2}\right)^2 + \left(1 + \frac{c}{2}x\right)^2 + \frac{1}{4}(4b - a^2 - c^2)x^2. \end{align*} Since $\angle B \ge 90^{\circ}$ and $\overline{AC}$ lies inside a circle with diameter 4, $4b \ge b^2 \ge a^2 + c^2$. Hence all three terms are nonnegative and \[x + \frac{a}{2} = 1 + \frac{c}{2}x = 4b - a^2 - c^2 = 0.\]In particular $b = 4$ and $a^2 + c^2 = 16$; also $x = -\frac{a}{2} = -\frac{2}{c}$. Hence $ac = 4$ and \[\{a, c\} = \{\sqrt{6} + \sqrt{2}, \sqrt{6} - \sqrt{2}\} \implies \boxed{x = -\frac{\sqrt{6} \pm \sqrt{2}}{2}}\]

What's the motivation to complete the square? Sorry if someone already mentioned it, but I didn't read through everything in this thread.

Ok, just got home, so I'll post my solution now. I actually only completed the square for c, even though everyone else seems to have completed the square for both a and c. First of all, from the problem condition we can see that $b^2 \geq a^2+c^2$ and $b \leq 4$. Now, $$x^4+ax^3+bx^2+cx+1=0$$$$x^4+ax^3+\left(b-\frac{c^2}{4} \right)x^2 + \frac{c^2}{4} + cx+1=0$$$$x^2 \left(x^2+ax+b-\frac{c^2}{4} \right) + \left(\frac{c}{2}x+1 \right)^2 = 0$$From this, we can conclude that $x^2+ax+b-\frac{c^2}{4} \leq 0$. We can manipulate this to get: \begin{align*} 4x^2+4ax+4b &\leq c^2 \\ &\leq b^2-a^2 \end{align*}So \begin{align*} \left(2x+a \right)^2 &\leq \left(b-2 \right)^2-4 \\ &\leq \left(4-2 \right)^2-4 \\ &=0 \end{align*}From this, we can conclude that $a=-2x$ and $b=4$. Also, note that this is the equality, so $4x^2+4ax+4b-c^2=0$. Plugging this into $x^2 \left(x^2+ax+b-\frac{c^2}{4} \right) + \left(\frac{c}{2}x+1 \right)^2 = 0$ and applying the Trivial Inequality yields $c=-\frac{2}{x}$. Finally, plugging in all the values into the original equation gives $-x^4+4x^2-1=0$, or $x^4-4x^2+1=0$. Applying the quadratic formula, we obtain $$x^2 = \frac{4 \pm \sqrt{16-4}}{2}=2 \pm \sqrt{3}.$$This simplifies to $$x=\frac{-\sqrt{6} \pm \sqrt{2}}{2}.$$It is easy to verify that these both work.

GameMaster402 wrote: What's the motivation to complete the square? Sorry if someone already mentioned it, but I didn't read through everything in this thread. idk most people's motivation, but for me, I played around with special cases and noticed that $45-45-90$ turned into $x^2\left(x+\sqrt{2}\right)^2+\left(\sqrt{2}x+1\right)^2$. Then, I played around with other cases and noticed I never got a negative result, so I tried to show that is was always $\ge 0$ with completing the square and showing that remaining term (which nicely has a degree of $2$) was always positive edit: 666th post!

Porky623 wrote: The inequalities where the author uses extended law of sines are reversed too. Sorry, I fixed the typoes. If I forgot to substitute back on the exam, would that warrant a 6?

GameMaster402 wrote: What's the motivation to complete the square? Sorry if someone already mentioned it, but I didn't read through everything in this thread. To be honest, one way to motivate is to notice that the problem would be really hard if something like this couldn't be done

It is obvious that $x$ doesn't have any positive solution. So, we will replace $x$ with $-x$ where $x$ is positive. The equation becomes $x^4+bx^2+1=ax^3+cx$ Now, $x^4+bx^2+1 \ge 2 \sqrt{bx^2(x^4+1)} = 2 \sqrt{b}x \sqrt{(x^4+1)} \ge bx \sqrt{(x^4+1)}$ [because $\frac{b}{\sin B}=4 \implies b=4 \sin B \le 4 \implies 2 \ge \sqrt{b}$] Now, $b = \sqrt{a^2+c^2-2ac \cos B} \ge \sqrt{a^2+c^2}$ as $\cos B \le \cos 90 =0$ $\therefore x^4+bx^2+1 \ge bx \sqrt{(x^4+1)} \ge x \sqrt{(a^2+c^2)(x^4+1)} \ge x \sqrt{(ax^2+c)^2} = ax^3+cx$ So, we must have equality cases in all the ineqs. $(1) x^4+1=bx^2$ $(2) 2 = \sqrt{b} \implies b=4$ and $\angle{B} = 90^{\circ}$ $(3) b^2=a^2+c^2 \implies a^2+c^2=16$ $(4) \frac{a}{x^2}=\frac{c}{1} \implies a=cx^2$ Now, $16=a^2+c^2=c^2x^4+c^2=bc^2x^2=bac \implies ac=4$ So, $a+c=\pm 2 \sqrt{6}$ and $a-c= \pm 2 \sqrt{2}$ This gives us $a=\sqrt{6}+\sqrt{2}, c=\sqrt{6}-\sqrt{2}$ or $a=\sqrt{6}-\sqrt{2}, c=\sqrt{6}+\sqrt{2}$ From (4) $x = \sqrt{\frac{a}{c}}$. $\therefore x = \sqrt{\frac{\sqrt{6}+\sqrt{2}}{\sqrt{6}-\sqrt{2}}}$ or $x=\sqrt{\frac {\sqrt{6}-\sqrt{2}}{\sqrt{6}+\sqrt{2}}}$. Thus $x=\frac{\sqrt{6} \pm \sqrt{2}}{2}$. But this is the value of $-x$. So, our answer is $x=-\frac{1}{2}(\sqrt{6}+\sqrt{2})$ or $x=-\frac{1}{2}(\sqrt{6}-\sqrt{2})$

Remark that $x\ne 0$. Consider a solution $x$. Observe that \[0=x^4+ax^3+bx^2+cx+1 = (x+a/2)^2x^2+(cx/2+1)^2 + (b-c^2/4-a^2/4)x^2\ge\]\[(b-c^2/4-a^2/4)x^2.\]As $x\ne 0$, this implies \[b \le c^2/4+a^2/4 \le b^2/4.\]Rearranging yields $b\ge 4$. It is clear that $b\le 4$, so we get $b=4$. Remark that equality holds, so $\angle ABC = 90^\circ$. Note $a^2/4+c^2/4=b$, so by the previous rearrangement we have \[0=(x+a/2)^2x^2+(cx/2+1)^2.\]By the Trivial Inequality, this forces $x=-a/2,x=-2/c$, so $2/c=-x=a/2$. Rearranging yields \[ac=4.\]Thus, we can write \[(a+c)^2=a^2+c^2+2ac = 16+8=24.\]Then, we get that $a,c$ are roots to \[0=x^2-(a+c)x+ac = x^2-2\sqrt{6}x+4.\]By the quadratic formula, \[a = \frac{2\sqrt{6}\pm \sqrt{24-16}}{2} = \sqrt{6}\pm \sqrt{2}.\]Finally, we conclude \[x = \frac{-\sqrt{6}\pm \sqrt{2}}{2}.\]Equality can be attained by taking \[(a,b,c) = \left(\sqrt{6}+\sqrt{2},4,\sqrt{6}-\sqrt{2}\right),\left(\sqrt{6}-\sqrt{2},4,\sqrt{6}+\sqrt{2}\right).\] Since everyone on this thread asked about motivation, here was mine: Whenever I solve a quartic, it typically is because the quartic is symmetric; that is, $a=c$ in this case. However, I didn't see any convenient way to force that (in fact, that isn't even the equality case). Then, I noticed that I wanted to use $b^2\ge c^2+a^2$, so I tried to get $c^2+a^2$ to materialize. After a bit work, I managed to get the $c^2+a^2$ to show up by completing the square, then the rest of the proof was mostly abusing the equality case and manipulating to solve for $a,b,c$.

This problem is vomit-inducing. Consider the inequality chain \begin{align*} 0 &= x^4 + ax^3 + bx^2 + cx + 1 \\ &\geq x^4 + ax^4 + \frac{b^2}{4} x^2 + cx + 1\\ &\geq x^4 + ax^4 + \frac{a^2 + c^2}{4} x^2 + cx + 1 \\ &\geq x^2 \left(x + \frac{a}{2} \right)^2 + \left(\frac{cx}{2} + 1 \right)^2 \\ &\geq 0, \end{align*}and so everything is an equality. In particular, $b = 4$, $a^2 + c^2 = 16$, and $x = -\frac{a}{2} = -\frac{2}{c}$. Now solving yields the equality cases of $x = -\frac{\sqrt{6} \pm \sqrt{2}}{2}$.

First note since $a, b, c$ are positive, $x$ is negative. Let $x=-y.$ So we have $$y^4-ay^3+by^2-cy+1=0 \iff y^2-ay+b-\frac{c}{y}+\frac{1}{y^2}=0,$$now seeing the $y^2$ and $\frac{1}{y^2},$ that gives us motivation to factor it as $$(y-a/2)^2+(\frac{1}{y}-c/2)^2=\frac{a^2}{4}+\frac{c^2}{4}-b \ge 0.$$Claim: $\frac{a^2}{4}+\frac{c^2}{b}-b \le 0.$ Proof: This is equivalent to $4b \ge a^2+c^2,$ but we have $b^2 \ge a^2+c^2$ since $\angle{B} \ge 90^{\circ},$ and $b \le 4,$ we get that $4b \ge a^2+c^2.$ $\Box$ Now obviously equality has to hold from the above arguments so $\angle{B}=90^{\circ},$ also $a^2+c^2=16.$ From $(x-a/2)^2+(1/x-c/2)^2=\frac{a^2}{4}+\frac{c^2}{4}-b \ge 0$ we get that $y=\frac{a}{2}=\frac{2}{c} \implies ac=4.$ Solving the equation gives us $y=\frac{\sqrt{6} \pm \sqrt{2}}{2} \implies x=\boxed{\frac{-\sqrt{6} \pm \sqrt{2}}{2}}.$ $\blacksquare$

After failing to solve this problem, I'm wondering how most people find claims for these types of problems? It seems very unmotivated to me to claim that $-b+\frac{a^2}{4}+\frac{c^2}{4} \leq 0,$ unless messing around somehow revealed this.

yes, messing around for a while to complete the square on x^4+ax^3 and cx+1

Note that $x<0$, so let $y=-x$. So $y^4-ay^3+by^2-cy+1=0$. Divide everything by $y^2$. So we get \[y^2-ay+b-\frac{c}{y}+\frac{1}{y^2}=0\] Claim: $\frac{a^2}{4}+\frac{c^4}{4}-b\ge 0$. Thus, \[\left(y-\frac{a}{2}\right)^2+\left(\frac{1}{y}-\frac{c}{2}\right)^2=y^2-ay+\frac{a^2}{4}+\frac{c^2}{4}-\frac{c}{y}+\frac{1}{y^2}=\frac{a^2}{4}+\frac{c^2}{4}-b\ge 0,\]as desired $\blacksquare$. Claim: $\frac{a^2}{4}+\frac{c^2}{4}-b\le 0$. We have $b^2\ge a^2+c^2$. Since $b\le 4$, $4b\ge a^2+c^2$, which implies $0\ge a^2+c^2-4b\implies 0\ge \frac{a^2}{4}+\frac{c^2}{4}-b$. $\blacksquare$ So $b=\frac{a^2}{4}+\frac{c^2}{4}$. We have $4b=a^2+c^2$ and $b^2\ge a^2+c^2$, so $b=4$ and $a^2+c^2=16$. Now we have $\left(y-\frac{a}{2}\right)^2+\left(\frac{1}{y}-\frac c2\right)^2= 0$, so $y=\frac{a}{2}$ and $\frac{1}{y}=\frac{c}{2}$. This gives $ac=4$. We solve the equation $a^2+c^2=16$ and $ac=4$. So $a^2+\frac{16}{a^2}=16$. Thus, $a^4-16a^2+16=0$. Solving gives $y=\sqrt{2-\sqrt{3}}=\frac{\sqrt{6}-\sqrt{2}}{2}$ or $y=\sqrt{2+\sqrt{3}}=\frac{\sqrt{6}+\sqrt{2}}{2}$. So $x=\boxed{-\frac{\sqrt{6}\pm \sqrt{2}}{2}}$, which work.

Note the following minor facts: $x\ne0$, $a^2+c^2\le b^2$ (the Pythagorean inequality), and $b\le4$, since the side of the triangle cannot be longer than the diameter of the circumcircle. Then: $$b-\frac{a^2}4-\frac{c^2}4\ge b-\frac{b^2}4\ge0$$so: \begin{align*} 0&=x^4+ax^3+bx^2+cx+1\\ &=\left(x^2+\frac{ax}2\right)^2+\left(\frac{cx}2+1\right)^2+x^2\cdot\left(b-\frac{a^2}4-\frac{c^2}4\right)\\ &\ge0\end{align*}Equality holds everywhere we mentioned an inequailty, so $b=4$ and $a^2+c^2=b^2$ and $x=-\frac a2=-\frac2c$. The latter two equations rearrange to $ac=4$ and $a^2+c^2=16$. Subbing $c=\frac4a$ into the second equation, we have $a^4-16a^2+16=0$, and so $a^2=8\pm4\sqrt3$. Then $a=\sqrt{8\pm4\sqrt3}$. Note that both of these values produce valid triangles subject to the problem conditions and, as a result, $x=-\frac a2=-\sqrt{2\pm\sqrt3}=\frac{-\sqrt2\pm\sqrt6}2$.

Notice that \[ x^4 + ax^3 + bx^2 + cx + 1 = x^2\left(x + \frac{a}{2}\right)^2 + \left(\frac{c}{2}x + 1\right)^2 + \left(b - \frac{a^2}{4} - \frac{c^2}{4}\right)x^2.\]Now, we will show that $b\ge \tfrac{1}{4}(a^2 + c^2)$. Notice that it is equivalent to show \[ b\ge \frac{1}{4}(b^2 + 2ac\cos B)\iff 4\sin B\ge \frac{1}{4}(16\sin^2 B + 2ac\cos B)\]\[ \iff \sin B\ge \sin^2B + \frac{1}{8}ac\cos B.\]Noting that $\sin B\ge \sin^2 B$ and $0\ge \tfrac{1}{8}ac\cos B$, we have that the above inequality is true, so $b^2 \ge \tfrac{1}{4}(a^2 + c^2)$ is as well. Additionally, note that equality holds exactly when $B = 90^\circ$. Thus we have the equation \[ x^2\left(x + \frac{a}{2}\right)^2 + \left(\frac{c}{2}x + 1\right)^2 + \left(b - \frac{a^2}{4} - \frac{c^2}{4}\right)x^2 = 0.\]Note that each individual term on the LHS is at least zero, so for equality to hold, each individual term must be exactly zero. This implies $x = -a/2$, $x = -2/c$, so $ac = 4$. We also get $b = a^2/4 + c^2/4\implies B = 90^\circ$. It follows that $\overline{AC}$ is a diameter of the circumcircle, so $b =4$, and consequently $a^2 + c^2 = 16$. Hence, $a = \sqrt{6} \pm \sqrt{2}$ and $c = \sqrt{6}\mp \sqrt{2}$, both of which are positive. Then, \[ x = -\frac{a}{2} = \boxed{-\frac{1}{2}(\sqrt{6}\pm\sqrt{2})}.\]Constructions for both values of $x$ are $(a,b,c) = (\sqrt{6}\pm\sqrt{2}, 4,\sqrt{6}\mp\sqrt{2})$, and it is easy to see that these satisfy the triangle inequality as well as the Pythagorean inequality $b^2\ge a^2 + c^2$, so we are done.

Clearly $x$ is negative, so set $x\rightarrow -x$, now our equation is: $$x^4-ax^3+bx^2-cx+1=0$$we can divide by $x^2$ since $x\ne 0$: $$x^2-ax+b-\frac{c}{x}+\frac{1}{x^2}=0$$$$\Longleftrightarrow \left(x-\frac{a}{2}\right)^2+\left(\frac{1}{x}-\frac{c}{2}\right)^2=\left(\frac{a}{2}\right)^2+\left(\frac{c}{2}\right)^2-b$$Claim: $\boxed{\left(\frac{a}{2}\right)^2+\left(\frac{c}{2}\right)^2-b\le 0}$ Proof: Clearly $b\le 4$ and $b^2\ge a^2+c^2$, thus: $$4b\ge b^2 \ge a^2+c^2$$and the claim follows $\blacksquare$ Thus For our equation to hold we must have: $x=\frac{a}{2}=\frac{2}{c}$, $b=4$, and $a^2+c^2=16$ We can easily solve these to find that the original $x=\dfrac{-\sqrt{6}\pm\sqrt{2}}{2}$

Disaster of a problem. The answer is $x = \frac{-\sqrt{6} \pm \sqrt{2}}{2}$ only. Let $a = 4\sin A$ and cyclic by the extended law of sines. Claim. Given these conditions we have \[ \sin(A)\cos(C) + \cos(A)\sin(C) \ge \sin^2 A + \sin^2 C. \] $\textit{Proof.}$ Divide by $\cos A \cos C$: \[ \tan A + \tan C \ge \tan A \frac{\sin A}{\cos C} + \tan C \frac{\sin C}{\cos A}.\]If $\tan A \ge \tan C$, then clearly since $A + C \le \frac{\pi}{2}$, we must have $A \ge C$, and also since $\frac{\pi}{2} - C \ge A$, it follows that $\cos C \ge \sin A$. Similarly, $\cos A \ge \sin C$. Equality holds if and only if $A + C = \frac{\pi}{2}$. $\square$ We return to the original problem now. Notice that the polynomial can be written as \begin{align*} x^4 + 4\sin(A)x^3 + 4\sin(A+C)x^2 + 4\sin(C)x + 1 &\ge (x^4 + 4\sin(A)x^3 + 4\sin^2(A)x^2) + (4\sin^2(C)x^2 + 4\sin(C)x + 1)\\ &=(x^2 + 2x \sin A)^2 + (2x \sin C + 1)^2 &\ge 0. \end{align*}The first term is $0$ if $x = 0$ or $x = -2\sin A$. The second term is $0$ only if $x = -\frac{1}{2 \sin C}$. Clearly $x=0$ is ruled out, so the only case left is $4 \sin A \sin C = 1$. Since for equality to hold we require $\cos C = \sin A$, it thus follows that $\sin(2A) = \frac{1}{2}$, or $A = 15^\circ, 75^\circ$. Straightforward computations give the answer of $\frac{-\sqrt{6} \pm \sqrt{2}}{2}$.

We have \begin{align*} x^4 + ax^3 + bx^2 + cx + 1 & \geq x^4 + bx^2 + 1 - \sqrt{(ax^3 + cx)^2} \\ & \geq x^4 + bx^2 + 1 - \sqrt{(a^2 + c^2)(x^6 + x^2)} \\ & \geq x^4 + bx^2 + 1 - b \sqrt{x^6 + x^2} \\ & \geq 2 \sqrt{bx^2 (x^4+1)} - b \sqrt{x^6 + x^2} \\ & \geq (2 \sqrt{b} - b) \sqrt{x^6 + x^2} \\ & \geq 0, \end{align*}where we use Cauchy Schwarz on line two, the fact that $a^2 + c^2 \leq b^2$ on line three, AM-GM on line four and $b \leq 4$ on line six. For equality to hold, we must have $b=4$ and $x^4 + 1 = bx^2$ (for the AM-GM equality). Also, $x$ must be negative for equality to hold on line $1$. This implies that $x$ is $\frac{- \sqrt{6} \pm \sqrt{2}}{2}$, which can easily be checked to work.