My teacher has shown me the following problem:
Problem. Let $x,y,z\in \mathbb{R}_+$ with $xyz=1$. Show that:$$x^{10}+y^{10}+z^{10}\ge x^9+y^9+z^9.$$
I think I solved the problem using Muirhead's Inequality (see below). The problem is that we haven't studied inequalities such as Muirhead, Schur, Hölder etc. so I reckon that the "expected" solution doesn't require them.
My (detailed) thought process
Let $$\sum_{\text{sym}}f(x_1,x_2,\dots, x_n)=\sum_{\pi\in S_n}f(x_{\pi(1)}, x_{\pi(2)},\dots,x_{\pi(n)}).$$
I began by rewriting the right hands side as
$$\begin{align}x^9+y^9+z^9&=\frac{x^9y^0z^0+x^9z^0y^0+y^9x^0z^0+y^9z^0x^0+z^9x^0y^0+z^9y^0x^0}{2}\\&=\frac{1}{2}\sum_{\text{sym}}x^9y^0z^0.\end{align}$$
Then, my first instinct was to come up with a sequence $(\alpha, \beta, \gamma)$ that majorizes $(9,0,0)$ in order to apply Muirhead's inequality to obtain $\sum_{\text{sym}}x^{10}y^0z^0$. Clearly, using the fact that $xyz=1$ must come into play now, and I wanted to get rid of two of the variables, so let's set $\gamma=\beta$, and we're left with the equations $\alpha +2\beta=9$ and $\alpha-\beta=10$, which leads to $\alpha=\frac{29}{3}$ and $\beta=-\frac{1}{3}$*. Since $(29/3, -1/3, -1/3) \succ (9, 0, 0)$**, by Muirhead ($x,y,z\neq 0$ since $xyz=1$):
$$x^9+y^9+z^9=\frac{1}{2}\sum_{\text{sym}}x^9y^0z^0\le \frac{1}{2}\sum_{\text{sym}}x^{29/3}y^{-1/3}z^{-1/3}=\frac{1}{2}\sum_{\text{sym}}x^{30/3}(xyz)^{-1/3}=x^{10}+y^{10}+z^{10}$$
Questions
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(*): Can Muirhead's Inequality be used for negative exponents? My question arises from the fact that some sources define the exponents as non-negative, while others just classify them as real numbers.
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(**), closely related to (*): Is majorization defined for sequences that contain negative numbers, or should this term only be used for sequences of non-negative reals? Again, I've seen several sources which disagree on this.
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(***) How would one solve this inequality without making use of Muirhead and other similar inequalities? I feel like it should be very simple but I fail to see any other solution right now. Additionally, is my proof correct? (I didn't get to use Muirhead much so my experience with it is close to zero)
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(****): Is it correct to generalize the problem as I did below? I noticed that a generalization is possible, so this led me to believe that my proof is incorrect, given the fact that only three numbers are given, and thus I may have missed something.
Possible generalization
Claim. Let $(x_i)_{i=1}^n\in \mathbb{R}_+^n$ with $\prod_{i=1}^n x_i=1$. Then$$\sum_{i=1}^nx_i^{10}\ge\sum_{i=1}^nx_i^9$$
Proof. Using a very similar argument,
$$\sum_{i=1}^nx_i^9=\frac{1}{(n-1)!}\sum_{\text{sym}}x_1^9\cdot x_2^0\cdot\dots\cdot x_n^0,$$
and the sequence of exponents that majorizes $(9,0,0,\dots,0)$ is $(\alpha,\beta,\beta,\dots,\beta)$ such that $10=\alpha-\beta$ and $(n-1)\beta+\alpha=9$, which gives $\beta=-1/n$ and $\alpha=10-1/n$. By Muirhead and using $\prod_{i=1}^n x_i=1$:
$$\sum_{i=1}^nx_i^9\le \frac{1}{(n-1)!}\sum_{\text{sym}}x_1^{10-1/n}\cdot(x_2\cdot \dots\cdot x_n)^{-1/n}=\sum_{i=1}^nx_i^{10}.$$
Best Answer
My instinct is to use $xyz=1$ to rewrite the RHS as $$x^9+y^9+z^9=x^{28/3}y^{1/3}z^{1/3}+x^{1/3}y^{28/3}z^{1/3}+x^{1/3}y^{1/3}z^{28/3} $$ and then use AM/GM on each of these terms: $$x^{28/3}y^{1/3}z^{1/3}\le\frac{28x^{10}+y^{10}+z^{10}}{30}$$ via the $n=30$ case of AM/GM. Do this for all three terms and add.
This method should prove your generalisation, and also work for other values of $9$ and $10$.