Proving a polynomial is positive

Question

How do I prove that the polynomial $$ f(x) = x^5-x^3/2-x+4/5 $$ is positive for all real numbers $x > 0$? One obvious way is to show that $$ f'(x) = 5x^4-3x^2/2-1 $$ has a unique positive root $x_0$ for which $f(x_0) > 0$ and $f''(x_0) > 0$, so the minimum of $f$ must be $f(x_0) > 0$, but the computation of the root and then inserting it into $f, f''$ seems so inelegant to me. I tried completing the square but to no avail. Any elegant ideas?

Answer 1

By AM-GM: $$10x^5-5x^3-10x+8=(3x^5-5x^3+2)+(7x^5-10x+6)\geq$$ $$\geq5\sqrt[5]{(x^5)^3\cdot1^2}-5x^3+\left(5\sqrt[5]{7\left(\frac{3}{2}\right)^4}-10\right)x>0.$$

Answer 2

We have $$x^5-\frac{x^3}{2}-x+\frac{4}{5} = x \left({x}^{2}-\frac 34 \right)^{2}+x \left(x-\frac 3 4 \right) ^{2}+\frac 32 \left( x-{\frac{17}{24}} \right)^{2}+{\frac{91}{1920}}>0.$$ See here.

Answer 3

For $x\ge1$ the result is trivial since

$$f'(x) = 5x^4-\frac 32x^2-1 \ge 5x^2-\frac 32x^2-1=\frac 7 2x^2-1>0$$

then for $0<x=\frac1y<1$ with $y>1$ we have

$$x^5-x^3/2-x+4/5>0 \iff 8y^5-10y^4-5y^2+10>0$$

and by subsuquent elementary completion of the squares

$$8y^5-10y^4-5y^2+10=(8y^5-16y^4+8y^3)+6y^4-8y^3-5y^2+10=$$

$$=8y^3(y-1)^2+(6y^4-12y^3+6y^2)+4y^3-11y^2+10=$$

$$=8y^3(y-1)^2+6y^2(y-1)^2+(4y^3-8y^2+4y)-3y^2-4y+10=$$

$$=8y^3(y-1)^2+6y^2(y-1)^2+4y(y-1)^2-3(y-1)^2-10(y-1)+3>$$

$$>18y(y-1)^2-3(y-1)^2-10(y-1)+3=(18y-3)(y-1)^2-10(y-1)+3>$$

$$>15(y-1)^2-10(y-1)+3>0$$