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The problem, which I encountered in a highschool book, goes as following:

Prove that if $a^5-a^3+a=3,$ then $a^6\geq 5$ must hold.

Now, obviously, I have tried a lot of things such as: $a^6=a^4-a^2+3a\Rightarrow a^4-a^2+3a\geq 5$ or multiplying that again by $a^2$ to create a new expression for $a^6$ and simplify some terms, but higher powers keep appearing and it just doesnt seem to end

Peter Allen
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1 Answers1

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The equation $a^5-a^3+1-3=0$ has at least one real solution. As $$3=a(a^4-a^2+1)=a\bigl((a^2-1/2)^2+3/4\bigr)$$ we have $a>0$ assuming $a$ is real.

Now $$a^4-a^2=\frac3a-1,$$ hence $$a^6=a^4-a^2+3a=\frac3a+3a-1 =3(\sqrt{a}-1/\sqrt{a})^2+6-1.$$

Michael Hoppe
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