By AM-GM inequality we have $\dfrac{2x^k}{1+x^{2k}}\le 1\implies \sum_{k=1}^n \dfrac{kx^k}{1+x^{2k}}\le \dfrac{1}{2}\sum_{k=1}^n k=\dfrac{n(n+1)}{4}.$ Equality occurs when $x=\pm 1$. Let $f(x)=\sum_{k=1}^n \dfrac{kx^k}{1+x^{2k}}$ then $f(-1)<f(1)=\dfrac{n(n+1)}{4}$, so $x=-1$ is not a solution.

I too got \(x=1\). But my solution is different. And rhs can be written as \(1/2(1+2+3+...+n)\). Then I expanded lhs. And since lhs=rhs So I compared each term of them showing it for n that only 1 is only real that satisfies it. Other values don't present in the real set.

Mate007 wrote: I too got \(x=1\). But my solution is different. And rhs can be written as \(1/2(1+2+3+...+n)\). Then I expanded lhs. And since lhs=rhs So I compared each term of them showing it for n that only 1 is only real that satisfies it. Other values don't present in the real set. Same

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MathKnight16 wrote: By AM-GM inequality we have $\dfrac{2x^k}{1+x^{2k}}\le 1\implies \sum_{k=1}^n \dfrac{kx^k}{1+x^{2k}}\le \dfrac{1}{2}\sum_{k=1}^n k=\dfrac{n(n+1)}{4}.$ Equality occurs when $x=1$. This is not enough. We'll have to show that no negative $x$ satisfies the equation. Add-on : Suppose, $\exists~x<0$ satisfying the equation. Then, we have $x^k\leqslant \mid x\mid^k$, but $x^{2k}=\mid x\mid^{2k}$. Therefore, $$\sum_{k=1}^n\frac{kx^k}{1+x^{2k}}\leqslant \sum_{k=1}^n\frac{k\mid x\mid^k}{1+\mid x\mid^{2k}}\leqslant \frac{n(n+1)}{4}$$Thus, equality must hold and for the equality to hold, we must have $$x=\mid x\mid ~~~\implies~~ x>0$$which is a contradiction since we assumed $x<0$. So there exists no solution in $\mathbb{R}^-$.

I used the fact that $$a+\frac{1}{a} \geq 2$$for positive real \(a\). Then I rewrote \(\sum^n_{k=1}\frac{kx^k}{1+x^{2k}}\) as \(\sum^n_{k=1}\frac{k}{\frac{1}{x^{k}}+x^{k}}\). If x is positive then L.H.S is less than equal to R.H.S with equality at \(x=1\) and if x is negative then, for even k the terms on left are non-strictly less than their corresponding rights and for odd k the inequality is strict, so, all in all, if x is negative then equality does not hold. x=0 is easily eliminated. The only solution is x=1.

$1+x^2k \ge 2x^k$ ,$\frac{kx^k}{1+x^2k}\le \frac{k}{2}$ and from summation case if $n=1,2,3,4,5,....$ ,$x=1$ works And from Equality Case in $a^2+b^2\ge 2ab$ $a=b$ .thus in this case $x^2k =1$,$x=1$ $n=2$ starting with first naturals gives $\frac{x}{1+x^2}+\frac{2x^2}{1+x^4}=\frac{3}{2}$ If $x=1$ ,satisfy $x=–1$ not possible Missed that one if $n=1$ $\frac{x}{1+x^2}=\frac{1}{2}$ $x^2–2x+1=0$,$(x–1)^2=0,x=1$

By AM - GM, \[\frac{x^{2k} + 1}{x^k} \ge 2\cdot\frac{\sqrt{x^{2k}\cdot 1}}{x^k} = 2\cdot\frac{|x^k|}{x^k} \ge 2\]So, \[\sum_{k = 1}^n \frac{kx^k}{1 + x^{2k}} \le \sum_{k=1}^n \frac{k}{2} = \frac{n(n+1)}{4}.\]Equality is achieved when $x^{2k} = 1 \implies x \in {-1, 1}$ and $|x^k| = x^k \implies k \equiv 0 \pmod{2}\text{ or } x > 0$. These together imply that $x = 1$ is the only solution.

Simple Am-Gm $\rightarrow \dfrac{kx^{k}}{1+x^{2k}} \le \dfrac{k}{2}$

We have \[\sum^n_{k=1}\frac{kx^k}{1+x^{2k}}=\frac{n(n+1)}4\]or \[\sum^n_{k=1}\frac{2kx^k}{1+x^{2k}}=\frac{n(n+1)}2 = 1+2+\ldots +n.\] Now we have \begin{align*} 0 &= \sum^n_{k=1}\frac{2kx^k}{1+x^{2k}} -k \\ & = \sum_{k=1}^{n} \frac{(x^k-1)^2}{1+(x^k)^2}\geq 0. \end{align*}Thus $x^k -1$ or $x=1.$

integrated_JRC wrote: MathKnight16 wrote: By AM-GM inequality we have $\dfrac{2x^k}{1+x^{2k}}\le 1\implies \sum_{k=1}^n \dfrac{kx^k}{1+x^{2k}}\le \dfrac{1}{2}\sum_{k=1}^n k=\dfrac{n(n+1)}{4}.$ Equality occurs when $x=1$. This is not enough. We'll have to show that no negative $x$ satisfies the equation. Add-on : Suppose, $\exists~x<0$ satisfying the equation. Then, we have $x^k\leqslant \mid x\mid^k$, but $x^{2k}=\mid x\mid^{2k}$. Therefore, $$\sum_{k=1}^n\frac{kx^k}{1+x^{2k}}\leqslant \sum_{k=1}^n\frac{k\mid x\mid^k}{1+\mid x\mid^{2k}}\leqslant \frac{n(n+1)}{4}$$Thus, equality must hold and for the equality to hold, we must have $$x=\mid x\mid ~~~\implies~~ x>0$$which is a contradiction since we assumed $x<0$. So there exists no solution in $\mathbb{R}^-$. My solution for $x \in \mathbb{R^+}$ was the same. For $x<0$, I took $x=-y$, where $y>0$. Let $$S=\sum^n_{k=1}\frac{ky^k}{1+y^{2k}} \text{ AND } S'(x)=\frac{1}{1+y^2}+\frac{3y^3}{1+y^6}+\dots$$Then $LHS=S-2S'$. By AM-GM (like we did for $x>0$), we have $S \leq \frac{n(n+1)}4$. Also $S'>0$. This gives $LHS<\frac{n(n+1)}4-0 \Rightarrow LHS<RHS$. Thus, no solution exists if $x<0$.

$$0 = \frac{n(n+1)}{4} - \sum_{k=1}^n \frac{kx^k}{1+x^{2k}} = \sum_{k=1}^n \left ( \frac{k}{2} - \frac{kx^k}{1+x^{2k}} \right ) = \sum_{k=1}^n \frac{k(x^k-1)^2}{2(1+x^{2k})} \geq 0$$ So equality must hold everywhere, that is, $(x^k-1) = 0$ for all $1 \leq k \leq n$. Thus $x=1$. Indeed, $x=1$ works.

By am gm u can get given expresion >=k/2 and for each n equality holds solve the equation forx^k U will obtain x^k=1 where one is a root -1 also is root where k is odd but but it is cleary not possible

https://olympiads.hbcse.tifr.res.in/olympiads/wp-content/uploads/2018/05/solutions-crmo-18.pdf

$$\sum_{k=1}^{n} \frac{kx^k}{1+x^{2k}}= \frac{x}{1+x^2}+\frac{2x^2}{1+x^2}+\frac{3x^3}{1+x^6}+ \dots =\frac{n(n+1)}{4}$$If $x>0$ $$\implies \frac{1}{x^{-1}+x^1}+\frac{2}{x^{-2}+x^2}+\frac{3}{x^{-3}+x^3}+\dots \leq \frac12 +\frac22+\frac32 +\dots +\frac{n}{2}=\frac{n(n+1)}{4}$$Because $x^{-1}+x^1\geq 2, x^{-2}+x^2\geq 2, \dots$ by AM-GM. Now, it is given that $$LHS=\frac{n(n+1)}{4}$$$$\implies AM=GM \implies \text{every term has to be it's maximum}\implies \boxed{x=1}.$$Now, the maximum value of LHS will only occur when each term is positive, so, we can omit the negatives $(x<0)$. If $x=0 \implies LHS=0$, but that is clearly not a solution. So, only solution is $x=1$.

DynamoBlaze wrote: Let $n$ be a natural number. Find all real numbers $x$ satisfying the equation $$\sum^n_{k=1}\frac{kx^k}{1+x^{2k}}=\frac{n(n+1)}4.$$ Let $P(n) \implies \sum^n_{k=1}\frac{kx^k}{1+x^{2k}}=\frac{n(n+1)}{4}$. $P(n+1) - P(n) = \frac{(n+1)x^{n+1}}{1+ x^{2n+2}} = \frac{n+1}{2}$. But observe that since $n \in \mathbb{N}$, $n+1 > 0$ and so $\frac{x^{n+1}}{1+x^{2n+2}} = \frac{1}{2}$. Now, using AM-GM, we obtain that $\frac{1 + x^{2n+2}}{2} \geq 2\sqrt{x^{2n+2}} = 2x^{n+1}$ which means that equality must hold or $1 = x^{2n+2}$ or $x = \pm 1$ (sinee $x$ is a real number). We see that $x = 1$ definitely satisfies the equation. But, see that whenever $k$ is odd, $\frac{k(-1)^k}{1 + (-1)^{2k}} < \frac{k(1)^k}{1 + (1)^{2k}}$ and hence $\sum^n_{k=1}\frac{k(-1)^k}{1+(-1)^{2k}} < \frac{n(n+1)}{4}$. Hence, $\boxed{x = 1}$ is the only real number satisfying the given equation

$$1 + x^{2k} \ge 2x^k$$$$\implies \frac{x^k}{1 + x^{2k}} \le \frac 12 \implies \frac{kx^k}{1 + x^{2k}} \le \frac k2$$$$\frac{kx^k}{1 + x^{2k}} \le \frac{k}{2}$$$$\implies \sum^n_{k=1}\frac{kx^k}{1+x^{2k}} \le \sum_{k=1}^n \frac k2$$But we know that $\sum^n_{k=1}\frac{kx^k}{1+x^{2k}}=\frac{n(n+1)}4$ Hence $x^{2k} = 1 \implies x = \pm 1$ But if $x = -1$ then $1 + x^{2k} > 2x^k$ for odd $k$ Hence the sum will not be equal. Thus the only answer is $x=1 \ \blacksquare$

My solution is a bit different and does not use AM-GM.