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Exercise:

Prove that $$x^{2n}+x^{2n-1}y+x^{2n-2}y^2+\cdots +y^{2n}\geq 0$$

Attempt:

I'm not sure where to begin with this problem, so I don't really have anything to show.

Request:

There's no need to give me the whole solution (though it is welcome so that those who need it will have it); I just need a hint to kickstart my attempt.

3 Answers 3

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If $x=0$ then $f(x,y)=y^{2n}\ge 0$, else pose $c=y/x$

$f(x,y)=x^{2n}(1+c+c^2+...+c^{2n})=x^{2n}(\frac{1-c^{2n+1}}{1-c})$ numerator and denominator of the fraction are the same sign, the whole thing multiplied by a square, so it is positive.

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    yep, corrected, thx. – 2017-01-04
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Hint: Consider the factorisation of $x^{2n+1}-y^{2n+1}$. What if $x \le y$? What if $x>y$?

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Hint: the expression is symmetric in $x,y$ so we can assume WLOG that $x \ge y$. If $x=y$ then the expression reduces to $(2n+1)x^{2n} \ge 0$ otherwise multiplying by $x-y \gt 0$ gives $x^{2n+1}-y^{2n+1} \gt 0$ which holds true because $x \gt y$.