# Prove that $a^6-a^5+a^4-a^3+1>0$

Prove that $$a^6-a^5+a^4-a^3+1>0$$.

What I tried and why this question is very tricky: I tired to decompose all the $$a$$ and find a common "denominator", I tried multiple methods, some with radicals another with multiplying with numbers to have a common base, I also tried this: $$a^6-a^5+a^4-a^3>0-1=-1$$ and multiply both sides by $$-1$$ to have a positive number ($$1$$) and to change the symbol ($$>$$ becomes $$<$$) My ideas didn't work, maybe they were incorrect or I didn't solve them correct. Tell me what do you think.

• I also tried to factor the entire expression! Commented Oct 10, 2023 at 14:30
• For some basic information about writing mathematics at this site see, e.g., basic help on mathjax notation, mathjax tutorial and quick reference, main meta site math tutorial and equation editing how-to.m/q/1773). Commented Oct 10, 2023 at 14:34
• FWIW, the minimum value of the expression occurs when $a = \frac{1}{18} \left( 5 - \frac{47}{\sqrt[3]{1043 + 18 \sqrt{3678}}} + \sqrt[3]{1043 + 18 \sqrt{3678}} \right) \approx 0.7903088603839408$, giving $f(a) \approx 0.8318434071389804$. I'm not posting that as an answer because it's very difficult to find without computer assistance.
– Dan
Commented Oct 10, 2023 at 15:46
• @Dan,thanks for your response! Commented Oct 10, 2023 at 16:36

Let $$\displaystyle f(a)=a^6-a^5+a^4-a^3+1$$

$$\bullet$$ For $$a\leq 0,$$ Then $$f(a)>0$$

$$\bullet$$ For $$0 Then $$f(a)=a^6+a^4(1-a)+(1-a^3)>0$$

$$\bullet$$ For $$a\geq 1,$$ Then $$f(a)=a^5(a-1)+a^3(a-1)+1>0$$

So we have $$f(a)>0$$ for $$a\in\mathbb{R}$$

• How do you know the second claim is true?
– dmh
Commented Oct 10, 2023 at 15:02
• In $a\in(0,1),$ We have $(1-a),(1-a^3)>0$ Commented Oct 10, 2023 at 15:03

A general method that only works with Discriminants :

Let's transform the given polynomial into the quadratic-like polynomial with respect to $$\color{#c00}{a^2}$$ :

\begin{align}P(a):&=a^6-a^5+a^4-a^3+1\\ &=\small{\underbrace{(a^2-a+1)}_{>\thinspace 0,\thinspace \forall a\thinspace\in\thinspace\mathbb R}\color{#c00}{\left(a^2\right)^2}-a\cdot \color{#c00}{a^2}+1} \end{align}

Then, we determine the Discriminant $$\Delta_{\color{#c00}{a^2}}$$ :

\begin{align}\Delta_{\color{#c00}{a^2}}&=a^2-4(a^2-a+1)\\ &=-\underbrace{\left(3\color{#0a0}{a^2}-4\color{#0a0}{a}+4\right)}_{\Delta_{\color{#0a0}{a}}=\thinspace-8\thinspace <\thinspace 0}\\ &<0,\thinspace \forall a\in\mathbb R\thinspace . \end{align}

Since $$\Delta_{\color{#c00}{a^2}}<0$$ for all $$a\in\mathbb R$$, this means that $$P(a)>0,\thinspace \forall a\in\mathbb R$$ which completes the proof .

• Nice way Lone student. Commented Oct 18, 2023 at 13:39
• Can you please tell me How can I solve it above way....math.stackexchange.com/questions/2274331/… Commented Oct 18, 2023 at 13:39
• @jacky If I see a way with the same method, I will leave you a comment/solution . Commented Oct 18, 2023 at 15:13

Another way :

$$f(a)=a^6-a^5+a^4-a^3+1$$

$$\displaystyle 2f(a)=2a^6-2a^5+2a^4-2a^3+2$$

$$\displaystyle 2f(a)=a^6+(a^6-2a^3\cdot a^2+a^4)+(a^4-2a^2\cdot a+a^2)+(a^4-a^2+\frac{1}{4})+\frac{3}{4}$$

$$\displaystyle 2f(a)=a^6+(a^3-a^2)^2+(a^2-a)^2+(a^2-\frac{1}{2})^2+\frac{3}{4}>0$$

So we have $$2f(a)>0\Longrightarrow f(a)>0 \;\forall a\in\mathbb{R}$$

• Completing the square seems incorrect : $$\small {2f(a)=2 a^6 - 2 a^5 + 3 a^4 - 2 a^3 + 1}$$ Commented Oct 10, 2023 at 16:22
• @jacky Thanks for the help!!!! Commented Oct 10, 2023 at 16:35