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

Question

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.

Answer 1

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

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

$\bullet $ For $0<a<1,$ 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 $a\in\mathbb{R}$

Answer 2

A general method that only works with Discriminants :

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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 .

Answer 3

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