Prove that: (a) if $y<\frac12$ and $n\ge3$ is a natural number then $(y+1)^n\ge y^n+(1+2y)^\frac n2$; (b) if $x,y,z$ and $n\ge3$ are natural numbers for which $x^2-1\le2y$ then $x^n+y^n\ne z^n$.
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Tags: number theory, algebra, inequalities
29.06.2024 12:21
i believe y should be more than or equal to -1/2 or else $(1+2y)^{\frac{n}{2}}$ wouldnt be a real number when n is odd in (a) note that $y^2\geq0 \Rightarrow y^2+2y+1\geq 2y+1$ $\Rightarrow y+1\geq \sqrt{2y+1},~~\forall y \in \mathbb{R}^{+}\cup[-\frac{1}{2},0]~$ with equality case holding at $y=0~~~(*)$ let $p(k)$ represent $(y+1)^k\geq y^k+(1+2y)^{\frac{k}{2}},~~\forall y \in \mathbb{R}^{+}\cup[-\frac{1}{2},0]$ base step: from $AM \geq GM;~\frac{3}{4}(1+2y)^2+\frac{1}{4}\geq (1+2y)^{\frac{3}{2}}$ $\Rightarrow3y^2+3y+1\geq (1+2y)^{\frac{3}{2}}\Rightarrow (y+1)^3\geq y^3+(1+2y)^{\frac{3}{2}},~~\forall y \in \mathbb{R}^{+}\cup[-\frac{1}{2},0]$ with equality holding at $y=0$ $\therefore~p(3)$ is true let $m\in\mathbb{N}-\{1,2\}$ which $p(k)$ is true thus $(y+1)^m\geq y^m+(1+2y)^{\frac{m}{2}},~~\forall y \in \mathbb{R}^{+}\cup[-\frac{1}{2},0]$ $\Rightarrow (y+1)^{k+1}\geq y^{k+1}+y^k+(1+2y)^{\frac{k}{2}}(y+1)$ $\Rightarrow (y+1)^{k+1}\geq y^{k+1}+y^k+(1+2y)^{\frac{k+1}{2}}~~;~~from~(*)$ $\Rightarrow (y+1)^{k+1}\geq y^{k+1}+(1+2y)^{\frac{k+1}{2}},~~\forall y \in \mathbb{R}^{+}\cup[-\frac{1}{2},0]$ with equality case holding at $y=0$ $\therefore~p(m+1)$ is true from induction we can conclude that $(y+1)^k\geq y^k+(1+2y)^{\frac{k}{2}},~\forall y \in \mathbb{R}^{+}\cup[-\frac{1}{2},0],k\in\mathbb{N}-\{1,2\}~~~~(1)$ with equality case holding at $y=0$ $\therefore$ (a) is clearly done for (b): assume the contrary that there is $x,y,z\in\mathbb{N}$ which $x^2-1\leq 2y,~x^n+y^n=z^n$ where $n\in\mathbb{N}-\{1,2\}$ thus $x^n\leq (1+2y)^{\frac{n}{2}}$ we can obviosly see that $z\geq y+1$ therefore $x^n+y^n=z^n\geq (y+1)^n$ $\Rightarrow (1+2y)^{\frac{n}{2}}+y^n \geq x^n+y^n \geq (y+1)^n$ $\Rightarrow y^n+(1+2y)^{\frac{n}{2}}\geq (y+1)^n~~~(2)$ from $(1),~(2)$ we'll get that $y^n+(1+2y)^{\frac{n}{2}}=(y+1)^n$ since the equality case holds at $y=0$, it contradicts with $y\in\mathbb{N}$ $\therefore$ we can conclude that if $x,y,z\in\mathbb{N},~x^2-1\leq 2y$ then $x^n+y^n\ne z^n,~\forall n \in\mathbb{N}-\{1,2\}$ lmk if there is any mistakes