# Is it true that $\sum_{i=1}^n ( nGx_i^{G} + G^{x_i}) \ge n^2G + G^2n$, for all $x_i>0$, where $G=\prod_{j=1}^nx_j$?

Prove or disprove that, for all $$x_1,x_2,\ldots,x_n>0$$, it holds that $$\sum_{i=1}^n ( nGx_i^{G} + G^{x_i}) \ge n^2G + G^2n, \space \space \space \text{where} \space \space \space G=\prod_{j=1}^nx_j.$$

The case $$n=2$$ is equlvalent to $$2\sqrt{xy}\left(x^{\sqrt{xy}}+y^{\sqrt{xy}}\right)+\left(\sqrt{xy}^{x}+\sqrt{xy}^y\right)\geq 4\sqrt{xy}+2xy\,.$$ We will show that the inequality above is true at least when $$xy\geq 1$$. By AM-GM, $$x^{\sqrt{xy}}+y^{\sqrt{xy}}\geq 2\,\sqrt{x^{\sqrt{xy}}y^{\sqrt{xy}}}=2\,\sqrt{xy}^{\sqrt{xy}}\geq 2\,\sqrt{xy}\,.$$ Also by AM-GM, $$\sqrt{xy}^x+\sqrt{xy}^y\geq 2\,\sqrt{\sqrt{xy}^x\sqrt{xy}^y}=2\,\sqrt{xy}^{\frac{x+y}{2}}\geq 2\,\sqrt{xy}^{\sqrt{xy}}\geq 2\,\sqrt{xy}\,.$$ Thus, \begin{align}2\sqrt{xy}\left(x^{\sqrt{xy}}+y^{\sqrt{xy}}\right)+\left(\sqrt{xy}^{x}+\sqrt{xy}^y\right)&\geq 2\sqrt{xy}(2\sqrt{xy})+(2\sqrt{xy})\\&=4xy+2\sqrt{xy}\geq 4\sqrt{xy}+2xy\,.\end{align}

I did not copy the answer by River Li here because the user wants to work on the problem a little bit more.

The inequality is true for $$G \ge 1$$. For simplicity, rewrite it $$\sum_{i = 1}^n \left(x_i^G + \frac{G^{x_i - 1}}n\right) \overset ?\ge n + G$$ As mentioned by @The.old.boy, $$x \mapsto x^G + \frac{G^{x - 1}}n$$ is a convex function. Hence, Jensen's Inequality gives $$\sum_{i = 1}^n \left(x_i^G + \frac{G^{x_i - 1}}n\right) \ge nm^G + G^{m - 1}$$ where $$m = \frac{x_1 + \dots + x_n}n$$ is the arithmetic mean of the $$x_i$$. We need to check that $$nm^G + G^{m - 1} \overset ?\ge n + G$$ knowing that $$m \ge \sqrt[n]G \ge 1$$ by AM-GM. As $$nx^G + G^{x - 1}$$ is strictly increasing, we only need to show that $$nm^{m^n} + m^{n(m - 1)} \ge n + m^n$$ for all $$x \ge 1$$. However the derivative of $$nx^{x^n} + x^{n(x - 1)} - x^n$$ is $$nx^{n - 1}\left(x^{x^n}\left(n\ln x + 1\right) + x^{n(x - 2)}(x + \ln x - 1) - 1\right)$$ and is negative on $$]0, 1[$$ and positive on $$]1, \infty[$$ (because the inside is strictly increasing). Hence $$nm^{m^n} + m^{n(m - 1)} - m^n \ge n\cdot 1^{1^n} + 1^{n(1 - 1)} - 1^n = n$$

The case $$G < 1$$ is substantially harder because you can't rely anymore on Jensen. However, the Tangent Line Trick might do the job. I'll update my answer should I get anywhere.

EDIT : Looking at the function $$f : x \mapsto e^{Gx} + \frac{G^{e^x - 1}}n$$ is much more relevant because your inequality becomes $$f(a_1) + \dots + f(a_n) \ge n + G$$ for all $$a_1 + \dots + a_n = \ln G$$ (by setting $$x_i = e^{a_i}$$). If $$f$$ had exactly one inflexion point, an olympiad brutal technique called n - 1 EV (see here) would imply that the minimum value of $$f(a_1) + \dots + f(a_n)$$ is reached when $$n - 1$$ of the $$a_i$$s are equal. However $$f$$ has either $$0$$ (in which case $$f$$ is convex and the same Jensen trick concludes) or $$2$$ inflexion points. The technique is adaptable and leaves a simpler inequality to prove:

Because it will allow us to wipe out terms more easily, look at the continuous version:

For all $$k$$, $$\lambda_1, \dots \lambda_n > 0$$ and $$a_1, \dots, a_k \in \mathbb R$$ with $$\lambda_1 + \dots + \lambda_k = n$$ and $$\lambda_1a_1 + \dots + \lambda_ka_k = \ln G$$, we have $$\lambda_1f(a_1) + \dots + \lambda_kf(a_k) \ge n + G$$

First establish the following lemma:

If $$\lambda_1f(a_1) + \dots + \lambda_kf(a_k)$$ is minimal, then $$f'(a_1) = \dots = f'(a_k)$$ and $$f''(a_1), \dots, f''(a_k) \ge 0$$.

Proof

$$\bullet$$ Suppose that $$f'(a_i) \ne f'(a_j)$$ and $$\lambda_i = \lambda_j$$ (by breaking down $$\max(\lambda_i, \lambda_j)$$ if needed). Then we can replace $$a_i, a_j$$ by $$a_i + x, a_j - x$$. This doesn't change $$\lambda_1a_1 + \dots + \lambda_ka_k$$ and Taylor's interpolation gives $$f(a_i + x) + f(a_j - x) - f(a_i) - f(a_j) \underset{x \rightarrow 0}\sim x(f'(a_i) - f'(a_j))$$ In particular, we can choose $$x$$ to make this difference negative, which shows that we weren't on a minima.

$$\bullet$$ Suppose that $$f''(a_i) < 0$$. Then we can replace $$a_i$$ by $$a_i - x$$ and $$a_i + x$$ with respective $$\lambda$$s being both $$\frac{\lambda_i}2$$. This doesn't change $$\lambda_1a_1 + \dots + \lambda_ka_k$$ and Taylor's interpolation gives $$f(a_i + x) + f(a_i - x) - 2f(a_i) \underset{x \rightarrow 0}\sim \frac{x^2}2 f''(a_i) < 0$$ That shows we weren't on a minima.

Then this lemma:

If $$\lambda_1f(a_1) + \dots + \lambda_kf(a_k)$$ is minimal, then $$\{a_1, \dots, a_k\} \le 2$$. That is, we can assume that $$k = 2$$.

Proof: $$f$$ has at most $$2$$ inflexion points, which means it has at most $$2$$ convex parts. On each of these, $$f'' > 0$$ which implies that $$f'$$ is injective. As the previous lemma says that all $$f'(a_i)$$ must be equal, there is space for only one in each of the convex parts of $$f$$.

Thus we can restrict ourselves to the case $$k = 2$$, needing to prove $$\lambda\left(e^{Ga} + \frac{G^{e^a - 1}}n\right) + (1 - \lambda)\left(e^{G\frac{\ln G - \lambda a}{n - \lambda}} + \frac{G^{e^{\frac{\ln G - \lambda a}{n - \lambda}} - 1}}n\right) \ge n + G$$ for all $$a$$ and all $$\lambda \in [0, n]$$.

Assume that $$G=Constant\geq 1$$ and $$\sum_{i=1}^{n}x_i\geq 2n$$

it's not hard to see that following function is convex on $$(0,\infty)$$: $$f(x)=nGx^G+G^x$$

As the sum of two convex function .

So we can apply Jensen's inequality :

$$\sum_{i=1}^n ( nGx_i^{G} + G^{x_i}) \ge ( n^2Ga^{G} + nG^{a})$$

Where $$a=\frac{\sum_{i=1}^{n}x_i}{n}$$

But with the assumptions we have $$a^G\geq 2^G$$ and $$G^a\geq G^2$$

So :$$\sum_{i=1}^n ( nGx_i^{G} + G^{x_i}) \ge ( n^2G2^{G} + nG^{2})> n^2G+G^2n$$

Update the case $$x_i\leq 1$$:

This is an observation by River Li. Here is the quote.

I didn't find a counterexample. By the way, for $$x_i\le 1, \forall i$$, I have a proof as follows. By AM-GM, we have $$\sum x_i^G \ge n (x_1x_2\cdots x_n)^{G/n} = nG^{G/n} = n \mathrm{e}^{(G\ln G)/n} \ge n (1 + (G\ln G)/n)$$ and $$\sum G^{x_i} \ge n G^{(x_1+x_2+\cdots + x_n)/n} \ge nG\,.$$ It suffices to prove that $$nG \cdot n (1 + (G\ln G)/n) + nG \ge n^2G + G^2n$$ or $$1 - G + G\ln G \ge 0$$ which is true.

Update the case $$G\leq 1$$ and $$n=2k+1$$:

Put : $$x_i=y_i^{\frac{G+1}{G}}$$ such that $$|y_{i+1}-y_i|=\epsilon$$ $$\epsilon>0$$ and $$y_{n+1}=y_1$$ and finally $$y_{\frac{n+1}{2}}=1$$

We have for the LHS:

$$\sum_{i=1}^{n}(nG(y_i)^{G+1}+G^{x_i})$$

Here I use the Hermite-Hadamard inequality .

The following functions are convex on $$(0,\infty)$$ (with the notation of the OP):

$$h(x)=nGx^{G+1}\quad r(x)=G^x$$

We have $$x_n\geq x_{n-1}\geq \cdots\geq x_2\geq x_1$$ and $$y_n\geq y_{n-1}\geq \cdots\geq y_2\geq y_1$$ and $$y_n\geq 1$$:

$$\sum_{i=1}^{n}(nG(y_i)^{G+1})\geq nG\Bigg(\frac{1}{(y_2-y_1)}\int_{y_1}^{y_2}h(x)dx+\frac{1}{(y_3-y_2)}\int_{y_2}^{y_3}h(x)dx+\cdots+\frac{1}{(y_n-y_1)}\int_{y_1}^{y_n}h(x)dx\Bigg)$$

Summing and using the additivity of integration on intervals we get :

$$\Bigg(\frac{1}{(y_2-y_1)}\int_{y_1}^{y_n}h(x)dx+\frac{1}{(y_n-y_1)}\int_{y_1}^{y_n}h(x)dx\Bigg)$$

But a primitive of $$h(x)$$ is :

$$H(x)=nG\frac{x^{G+2}}{G+2}$$

So :

$$\Bigg(\frac{1}{(y_2-y_1)}\int_{ y_1}^{ y_n}h(x)dx+\frac{1}{(y_n-y_1)}\int_{y_1}^{y_n}h(x)dx\Bigg)=\frac{nG}{(y_2-y_1)}\Bigg(\frac{(y_n)^{G+2}}{G+2}-\frac{(y_1)^{G+2}}{G+2}\Bigg)+\frac{nG}{(y_n-y_1)}\Bigg(\frac{(y_n)^{G+2}}{G+2}-\frac{(y_1)^{G+2}}{G+2}\Bigg)$$

Now we have by the Hermite-Hadamard inequality : $$\frac{\frac{(y_n)^{G+2}}{G+2}-\frac{(y_1)^{G+2}}{G+2}}{y_n-y_1}\geq\Big(\frac{y_n+y_1}{2}\Big)^{G+1}= 1$$

And as we have $$|y_{i+1}-y_i|=\epsilon$$ we get :

$$\frac{nG}{(y_2-y_1)}\Bigg(\frac{(y_n)^{G+2}}{G+2}-\frac{(y_1)^{G+2}}{G+2}\Bigg)+\frac{nG}{(y_n-y_1)}\Bigg(\frac{(y_n)^{G+2}}{G+2}-\frac{(y_1)^{G+2}}{G+2}\Bigg)= \frac{n^2G}{y_n-y_1}\Bigg(\frac{(y_n)^{G+2}}{G+2}-\frac{(y_1)^{G+2}}{G+2}\Bigg)\geq n^2G$$

On the other hand we have with Jensen's inequality:

$$\sum_{i=1}^{n}G^{x_i}\geq nG^{\frac{\sum_{i=1}^{n}x_i}{n}}$$

Assuming that $$\sum_{i=1}^{n}x_i\leq 2n$$ we have : $$\sum_{i=1}^{n}G^{x_i}\geq nG^{\frac{\sum_{i=1}^{n}x_i}{n}}\geq nG^2$$

Summing the two result we get the desired inequality .

Hope it helps !

Update:

We can apply the same reasoning to $$y_i^{\frac{G+\alpha}{G}}=x_i$$ instead of $$y_i^{\frac{G+1}{G}}=x_i$$ with $$\alpha> 1-G$$ or $$\alpha<-G$$ it generalize considerably the proof. The proof is still valid if $$y_n+y_1\geq 2$$ so without the restriction $$y_{\frac{n+1}{2}}=1$$