Inequality involving Lagrange Multipliers #2

Boy oh boy, the next installment in the epic saga of inequalities is finally here! How wonderful.

A while ago, one of my friends gave me this inequality, and, being the analysis enthusiast that I am, I couldn't resist using partial derivatives to solve it. Even though I found a solution, I feel a bit unfulfilled; I used Desmos to find the roots of a polynomial, and that wouldn't be allowed during a contest. Fortunately, this blog isn't a Maths contest, so it shall be allowed here. I suppose there were other methods of finding the roots of the polynomial I'm talking about, but they probably involved a lot of calculations that I wouldn't have had the patience to go through.

Let $x,y,z$ be real, strictly positive numbers satisfying $xyz=1$. Prove that $(x^{10}+y^{10}+z^{10})^2$ is bigger or equal to $3(x^{13}+y^{13}+z^{13})$.

Solution:

Let $f(x,y,z)=(x^{10}+y^{10}+z^{10})^2-3(x^{13}+y^{13}+z^{13})$ and $g(x,y,z)=xyz$. We know that, in order to reach the extremes, $$∇f(x,y,z)=λ∇g(x,y,z) ⇒\frac{𝛛f}{𝛛x}=λ\frac{𝛛g}{𝛛x}$$ The same relation works for $y$ and $z$. By differentiating, we get that $$20x^9(x^{10}+y^{10}+z^{10})-39x^{12}=λyz$$
$$⇒ 20x^{10}(x^{10}+y^{10}+z^{10})-39x^{13}=λ$$ because $xyz=1$. Thus, we get, by multiplying with $y^{10}$, that $$20x^{10}y^{10}(x^{10}+y^{10}+z^{10})-39x^{13}y^{10}=λy^{10}$$
By doing the same for $y$ and subtracting the two relations, we get that $$λ(y^{10}-x^{10})=39x^{10}y^{10}(y^3-x^3)$$ and, by dividing the whole thing by $x^{10}y^{10}$, which we can, since the numbers are strictly positive, we get that $$λ(x^{-10}-y^{-10})=39(y^3-x^3)$$ This implies that $$λx^{-10}+39x^3=λy^{-10}+39y^3=λz^{-10}+39z^3$$ since the relation is symmetrical. With that in mind, let $h(x)=λx^{-10}+39x^3$. By differentiating this function, we get the derivative being equal to $\frac {dh}{dx}=-10λx^{-11}+117x^2$.
Let us now prove that $λ>0$. For this, we shall presume that $x≤y≤z$, so that $x≤1$, and look at $20x^{10}(x^{10}+y^{10}+z^{10})-39x^{13}≥20x^{10}(x^{10}+2x^{-5})-39x^{13}=20x^{20}+40x^{5}-39x^{13}≥0$, with the first part coming from the AM-GM inequality and the second coming from the fact that $x$ is smaller than 1. Thus, $λ$ is positive.
Back to the derivative of $h$, since $λ$ is positive, we get that there is only one value of $x$ for which the derivative is 0, and, thus, there are at most 2 points with the same output of the function. Thus, either $x=y=z=1$, or $x=y,z=x^{-2}$. In the latter case, we shall go back to $\frac{𝛛f}{𝛛x}=λ\frac{𝛛g}{𝛛x}$ and $\frac{𝛛f}{𝛛z}=λ\frac{𝛛g}{𝛛z}$ subtract the derivatives and multiply by $x^{40}$ in order to obtain that $40x^{60}-39x^{53}-20x^{30}+39x^{14}-20=0$. This is the part where I used Desmos.


As you can see, the only real root of the polynomial is 1, so this means that $x=y=z=1$ yet again. It's obvious that $f(1,1,1)=0$, but now we have to prove that this is the minimal value, because we wanted to prove that $f(x,y,z)≥0$ when $g(x,y,z)=1$. Since this is a global extreme, we only have to find a triplet $x,y,z$ for which the conditions apply. By taking $z$ very large, towards infinity, and with $x$ and $y$ taken accordingly, we obtain an $f$ that approaches infinity, and, thus, is bigger than 0. This concludes the proof.

Perhaps I should have tried finding the roots of that polynomial without computer help. I am not sure how I could have done it, but I asked one of the teachers at Oxford during the interview about that and he suggested a method of approximating the roots. Even so, calculating the roots without Desmos would have taken a really long while to write, and I am forever thankful to computers. I really liked this inequality, it really tests the solver's capacity of manipulating functions...and graphing applications. Tune in next time for a combinatorics problem taken from a college contest, or something completely random!

Cris.