Show that the function $f(x) = x^5 - 10x^3 + 50x - 21$ is an increasing function for all x values. I have worked out that $dy/dx = 5x^4 - 30x^2 + 50$ and this must be greater than $0$ if it is an increasing function.
I simplified this to $x^4 - 6x^2 + 10 > 0$. I know that I must prove this inequality true but I cannot work out how.
I am able to input various values of x and show them to be larger than 0 but am unsure how to do it algebraically.
I know for a quadratic, I can put it into a 'complete the square' form and then you can prove it to be larger than 0 as squaring something = positive and then adding a positive will remain positive. But am unsure how to do it where power of x > 2.

We can proof it also by definition.
Indeed, let $a>b$.
Thus, by AM-GM we obtain:
$$f(a)-f(b)=a^5-10a^3+50a-21-(b^5-10b^3+50b-21)=$$
$$=(a-b)(a^4+a^3b+a^2b^2+ab^3+b^4+50-10(a^2+ab+b^2))\geq$$
$$=(a-b)\left(2\sqrt{50(a^4+a^3b+a^2b^2+ab^3+b^4)}-10(a^2+ab+b^2)\right)=$$
$$=\frac{10(a-b)(2(a^4+a^3b+a^2b^2+ab^3+b^4)-(a^4+2a^3b+3a^2b^2+2ab^3+b^4))}{\sqrt{2(a^4+a^3b+a^2b^2+ab^3+b^4)}+a^2+ab+b^2}=$$
$$=\frac{10(a-b)(a^4-a^2b^2+b^4)}{\sqrt{2(a^4+a^3b+a^2b^2+ab^3+b^4)}+a^2+ab+b^2}>0$$ and we are done!
