UM Jensen is like a instasolve

I don't remember it being instant (the derivatives take some time to muck through), but yes, Jensen does happen to work...

Yes once the derivative is calculated. I was just trying to say that the problem is not exact require any insight, just some elementary calculation.

Does anybody have solutions?

http://www.artofproblemsolving.com/Forum/viewtopic.php?f=52&p=3513124: $\sum_{i=1}^{n}x_i=k$,and $0<x_i\le\sqrt{2+\sqrt5}$,Show that\[\prod_{k=1}^{n}(x_i+\dfrac{1}{x_i})\ge(\dfrac{k}{n}+\dfrac{n}{k})^n\]

I dont think sqing give the solution Sonce in the link that given there is range And the provlems donnot have any range

By using Jensen's inequality, it is clear that this problem boils down to proving that $ f(x) = \text{ln}\left(x^k + x^{-k}\right) $ is a convex function on the interval $ [0, 1]. $ Note that $ f''(x) = \frac{n(4nx^{2n} + 1 - x^{4n})}{x^2(x^{2n} + 1)^2} > 0 $ for all $ x \in [0, 1] $ which implies the desired result.

I think clearing the denominators and applying Lagrange Multipliers works fine.

A bit cleaner, IMHO Taiwan 2014 Quizzes wrote: Let $a_i > 0$ for $i=1,2,\dots,n$ and suppose $a_1 + a_2 + \dots + a_n = 1$. Prove that for any positive integer $k$, \[ \left( a_1^k + \frac{1}{a_1^k} \right) \left( a_2^k + \frac{1}{a_2^k} \right) \dots \left( a_n^k + \frac{1}{a_n^k} \right) \ge \left( n^k + \frac{1}{n^k} \right)^n. \] My Solution: Claim: $f(x)=\log(x^k + x^{-k})$ is convex for $k \geqslant 1.$ Proof. $$f''(x) = -\dfrac{k\left(x^{4k}-4kx^{2k}-1\right)}{x^2\left(x^{2k}+1\right)^2} > 0$$for all $x\in(0,1).\quad\square$ Now, by Jensen we get: \begin{align*} \frac{\log\left(a_1^k+\frac{1}{a_1^k}\right)+\cdots+\log\left(a_n^k+\frac{1}{a_n^k}\right)}{n} &\geqslant \log\left(\frac{a_1^k+\frac{1}{a_1^k}+\cdots+a_n^k+\frac{1}{a_n^k}}{n}\right) \\&=\log\left(\frac{a_1^k+a_2^k\cdots+a_n^k}{n}+\frac{\frac{1}{a_1^k}+\frac{1}{a_2^k}+\cdots+\frac{1}{a_n^k}}{n}\right). \end{align*} Next, by Power Mean inequality applied to positive reals $\{a_1, a_2, \ldots, a_n\}$ with weights $\left\{\frac 1n, \frac 1n, \cdots, \frac 1n\right\}$, we have $\mathcal{P}(k) \geqslant \mathcal{P}(1)$, i.e.: $$\left(\frac{a_1^k+a_2^k\cdots+a_n^k}{n}\right)^{\frac 1k} \geqslant \left(\frac{a_1^1+a_2^1+\cdots+a_n^1}{n}\right)^{\frac 11} \implies \frac{a_1^k+a_2^k\cdots+a_n^k}{n} \geqslant \frac{1}{n^k}$$ Since $k>-1$, again by Power Mean applied to positive reals $\left\{\frac{1}{a_1}, \frac{1}{a_2}, \cdots, \frac{1}{a_n}\right\}$ with weights $\left\{\frac 1n, \frac 1n, \cdots, \frac 1n\right\}$, we have $\mathcal{P}(k) \geqslant \mathcal{P}(-1)$ i.e.: $$\left(\frac{\frac{1}{a_1^k}+\frac{1}{a_2^k}+\cdots+\frac{1}{a_n^k}}{n}\right)^{\frac 1k} \geqslant \left(\frac{\frac{1}{a_1^{-1}}+\frac{1}{a_2^{-1}}+\cdots+\frac{1}{a_n^{-1}}}{n}\right)^{\frac {1}{-1}} \implies \frac{\frac{1}{a_1^k}+\frac{1}{a_2^k}+\cdots+\frac{1}{a_n^k}}{n} \geqslant n^k$$So, we have $$\frac{a_1^k+a_2^k\cdots+a_n^k}{n} + \frac{\frac{1}{a_1^k}+\frac{1}{a_2^k}+\cdots+\frac{1}{a_n^k}}{n} \geqslant \left(n^k + \frac{1}{n^k}\right)$$$$\implies \log\left(\left(a_1^k+\frac{1}{a_1^k}\right)\left(a_2^k+\frac{1}{a_2^k}\right)\cdots\left(a_n^k+\frac{1}{a_n^k}\right)\right) \geqslant n\log\left(n^k+\frac{1}{n^k}\right) = \log\left(n^k+\frac{1}{n^k}\right)^n.\quad\blacksquare$$

Let $f(x)=\ln\left(x^k+\frac1{x^k}\right)$. As a result, we need to prove that for any $n$ and $\sum_{i=1}^na_i=1$ that $$\prod_{i=1}^ne^{f(a_i)} \ge e^{f(n) \cdot n} = e^{f\left(\frac1n\right)\cdot n}=e^{f\left(\frac{\sum_{i=1}^na_i}{n}\right) \cdot n}$$which hints towards taking the $\ln$ of both sides (credits to Eyed for telling me this smart way of Jensen-ing a bunch of products of a function). Doing that and dividing both sides by $n$ gets we need to prove $$\frac{\sum_{i=1}^n f(a_i)}{n} \ge f\left(\frac{\sum_{i=1}^na_i}{n}\right)$$so if $f$ is convex, we are done. Taking the derivative of $f$ gets us that for a constant $k$, $$\frac{d}{dx}\ln\left(x^k+\frac{1}{x^k}\right) = \frac{k (-1 + x^{2 k})}{x + x^{1 + 2 k}}$$so the second derivative would be $$\frac{d}{dx}\left(\frac{k (-1 + x^{2 k})}{x + x^{1 + 2 k}}\right) = \frac{k (1 + 4 k x^{2 k} - x^{4 k})}{x^2 (1 + x^{2 k})^2}$$meaning since the bottom side is positive by trivial inequality and the top side is positive since $x^{4k} < 1$ if $0 < x < 1$ then the second derivative is positive meaning $f$ is convex, so as a result, $$\frac{\sum_{i=1}^n f(a_i)}{n} \ge f\left(\frac{\sum_{i=1}^na_i}{n}\right)$$which means that taking $e$ as the base and each side as an exponent gets $$\prod_{i=1}^n \left(a_i^k+\frac{1}{a_i^k} \right) \ge \left(n^k+\frac{1}{n^k} \right)^n$$for all positive real numbers $\sum_{i=1}^n = 1$ and positive integer $k$.

We use a smoothing argument. Lemma: For $0 < a_p < a_q \le 1$ and $a_p + \varepsilon \le a_q - \varepsilon$, we have $$\left((a_p + \varepsilon)^k + \frac1{(a_p+\varepsilon)^k} \right) \left((a_q-\varepsilon)^k + \frac1{(a_q-\varepsilon)^k}\right) \le \left(a_p^k + \frac1{a_p^k} \right) \left(a_q^k + \frac1{a_q^k}\right).$$Proof: Note that if $0 < x < y \le 1$, we have $y^k + \frac1{y^k} \le x^k + \frac1{x^k}$. With this fact, the inequality follows by expansion since $0 < a_pa_q < (a_p + \varepsilon)(a_q - \varepsilon) \le 1$ and $0 < \frac{a_p}{a_q} < \frac{a_p + \varepsilon}{a_q - \varepsilon} \le 1$. Thus, through finitely many steps we can make all $a_i = \frac1n$ while the expression $\prod_{i=1}^n \left(a_i^k + \frac1{a_i^k}\right)$ monotonically decreases. At the state where all $a_i = \frac1n$, the desired inequality achieves equality, so we're done.

Hehe headsolved. I don't understand why everyone is taking derivatives, but okay Claim: If $x+y<1$ then $(x^k+x^{-k})(y^k+y^{-k}) \ge ((\frac{x+y}{2})^k + (\frac{2}{x+y})^k)^2$ Proof: If we expand, we get $(xy)^k + (xy)^{-k} \ge ((\frac{x+y}{2})^{2})^k + ((\frac{x+y}{2})^{2})^{-k}$ which is true because $xy \le (\frac{x+y}{2})^{2} \le \frac 14$. Also, $(\frac xy)^k + (\frac xy)^{-k} \ge 2$ by AM-GM. The conclusion follows by smoothing.

Trivial by Jensen is a thing Ok so we need $\ln \left(x+\frac{1}{x} \right)$ to be convex, so do the cancer operation on ur head (not writing it for the sake of my sanity) $$\frac{\text{d}^2}{\text{dx}^2} \ln \left(x+\frac{1}{x} \right)=\frac{-x^4+4x^2+1}{x^2(x^2+1)^2} \implies \; \text{that thing is convex on} \; [0,1]$$Since "that thing" is convex on $[0,1]$ we can use jensen to get $$\ln \left(a_1^k+\frac{1}{a_1^k} \right)+\ln \left(a_2^k+\frac{1}{a_2^k} \right)+ \cdots + \ln \left(a_n^k+\frac{1}{a_n^k} \right) \ge n \cdot \ln \left(n^k+\frac{1}{n^k} \right)$$And by elevating by $e$ we are done, equality if $a_1=a_2= \cdots =a_n=\frac{1}{n}$ thus we did it

It suffices to show that $$\sum_{i=1}^n \log\left(a_i^k + \frac 1{a_i^k}\right) \geq n\log\left(n^k + \frac 1{n^k}\right).$$Notice that the second derivative of $\log\left(x^k+\frac 1{x^k}\right)$ is $$\frac{k(4kx^{2k} - x^{4k} + 1)}{x^{2k}(x^{2k}+1)^2} > 0\ \forall x \in (0, 1).$$Thus Jensen works.

Since the two sides of the inequality are strictly positive, it suffices to show that : $$\sum_{i=1}^n \ln \left(a_i^k + \frac 1{a_i^k}\right) \geq n\ln\left(n^k + \frac 1{n^k}\right).$$ But this is trivial by Jensen's inequality since the second derivative of $f(x)=\ln ( x^k+\frac{1}{x^k})$ is equal to : $$f"(x)=\frac{k(4kx^{2k} - x^{4k} + 1)}{x^{2k}(x^{2k}+1)^2}$$which is positive for all $0<x<1$ $$\mathbb{Q.E.D.}$$

We take $\ln$ both sides. Hence, it is enough to show that $$\sum_{i=1}^n \ln\left(a_i^k+\frac{1}{a_i^k}\right)\ge n\cdot\ln\left(n^k+\frac{1}{n^k}\right).$$Let $f(x)=\ln\left(x^k+\frac{1}{x^k}\right).$ We will now prove that $f(x)$ is convex. Indeed, we have $$f'(x)=\frac{1}{x^k+\frac{1}{x^k}}\cdot\frac{2k\cdot x^{2k-1}\cdot x^k-(x^{2k}+1)k\cdot x^{k-1}}{x^{2k}} = \frac{k(x^{2k}-1)}{x(x^{2k}+1)}$$and hence $$f''(x)=\frac{k(-x^{4k}+4kx^{2k}+1)}{x^2(x^{2k}+1)^2}.$$ We will prove that $x^{4k}-4kx^{2k}-1\le 0$ for every positive integer $k$ and $x\in(0,1).$ Indeed, let $t=x^{2k}.$ Hence, we get $$t^2-4kt-1\le0\Leftrightarrow 2k-\sqrt{4k^2+1}\le t=x^{2k}\le 2k+\sqrt{4k^2+1}.$$However, $2k-\sqrt{4k^2+1}<0$ and $2k+\sqrt{4k^2+1}>1>x^{2k}>0$ since $x\in(0,1)$, so we are done. It now follows that $f''(x)\ge 0$ for every positive integer $k$ and $x\in(0,1)$. Hence, $f(x)$ is convex and we can apply Jensen. We have $$f(a_1)+f(a_2)+\dots+f(a_n)\ge n\cdot f\left(\frac{1}{n}\right)=n\cdot \ln\left(\frac{1}{n^k}+n^k\right),$$so we are done. Equality occurs when $a_1=a_2=\dots=a_n=\frac{1}{n}.$