socrates wrote: a) Find at least two functions $f: \mathbb{R}^+ \rightarrow \mathbb{R}^+$ such that $$\displaystyle{2f(x^2)\geq xf(x) + x,}$$for all $x \in \mathbb{R}^+.$ Choose for example $f(x)=\max(ax,a)$ which is a solution whatever is $a\in[1,+\infty)$

socrates wrote: b) Let $f: \mathbb{R}^+ \rightarrow \mathbb{R}^+$ be a function such that $$\displaystyle{2f(x^2)\geq xf(x) + x,}$$for all $x \in \mathbb{R}^+.$ Show that $ f(x^3)\geq x^2,$ for all $x \in \mathbb{R}^+.$ We have $f(x)\geq \frac{\sqrt{x}}{2}$ and , by induction, $$2^nf(x)\geq 2^{n-1}\sqrt{x}+2^{n-2}x^{3/4}+...+x^{\frac{2^n-1}{2^n}}.$$ Using weighted AM-GM inequality (with weights $\frac{2^{n-1}}{2^n-1},\frac{2^{n-2}}{2^n-1},...,\frac{1}{2^n-1}$) and letting $n\to \infty$ we get $f(x)\geq x^{2/3}.$

socrates wrote: Let $f: \mathbb{R}^+ \rightarrow \mathbb{R}^+$ be a function such that $$\displaystyle{2f(x^2)\geq xf(x) + x,}$$for all $x \in \mathbb{R}^+.$ Can we find the best constant $a\in \Bbb{R}$ such that $f(x)\geq x^a,$ for all $x \in \mathbb{R}^+?$ Any ideas?

socrates wrote: socrates wrote: b) Let $f: \mathbb{R}^+ \rightarrow \mathbb{R}^+$ be a function such that $$\displaystyle{2f(x^2)\geq xf(x) + x,}$$for all $x \in \mathbb{R}^+.$ Show that $ f(x^3)\geq x^2,$ for all $x \in \mathbb{R}^+.$ letting $n\to \infty$ we get $f(x)\geq x^{2/3}.$ can somebody explain this part pleas ? I dont understand why

^By Weighted A.M.-G.M., the right-hand side is more than or equal to $(2^n-1)x^{\frac{2-2^{-n}}{3}}$. socrates wrote: Can we find the best constant $a\in \Bbb{R}$ such that $f(x)\geq x^a,$ for all $x \in \mathbb{R}^+?$ Since $f(x)=\frac{1+2x}{3}$ works, the only possible value of $a$ is $\frac{2}{3}$.

socrates wrote: socrates wrote: b) Let $f: \mathbb{R}^+ \rightarrow \mathbb{R}^+$ be a function such that $$\displaystyle{2f(x^2)\geq xf(x) + x,}$$for all $x \in \mathbb{R}^+.$ Show that $ f(x^3)\geq x^2,$ for all $x \in \mathbb{R}^+.$ We have $f(x)\geq \frac{\sqrt{x}}{2}$ and , by induction, $$2^nf(x)\geq 2^{n-1}\sqrt{x}+2^{n-2}x^{3/4}+...+x^{\frac{2^n-1}{2^n}}.$$ Using weighted AM-GM inequality (with weights $\frac{2^{n-1}}{2^n-1},\frac{2^{n-2}}{2^n-1},...,\frac{1}{2^n-1}$) and letting $n\to \infty$ we get $f(x)\geq x^{2/3}.$ Someone can explain why we have f(x)>=sqrt(x)/2?

kent2207 wrote: Someone can explain why we have f(x)>=sqrt(x)/2? Well, $2f(x^2) \ge xf(x)+x \ge x$, hence $f(x^2) \ge \frac{x}{2}$. Now replace $x$ by $\sqrt{x}$...

Tintarn wrote: kent2207 wrote: Someone can explain why we have f(x)>=sqrt(x)/2? Well, $2f(x^2) \ge xf(x)+x \ge x$, hence $f(x^2) \ge \frac{x}{2}$. Now replace $x$ by $\sqrt{x}$... Oh it's ok, thanks for your idea, but it's R+, i'm not sure that x can be zero.

kent2207 wrote: Oh it's ok, thanks for your idea, but it's R+, i'm not sure that x can be zero. I don't understand. I know that it is $\mathbb{R}_+$. How is this a problem? I never put $x=0$...

Tintarn wrote: kent2207 wrote: Oh it's ok, thanks for your idea, but it's R+, i'm not sure that x can be zero. I don't understand. I know that it is $\mathbb{R}_+$. How is this a problem? I never put $x=0$... Oh ok don't worry about that

socrates wrote: socrates wrote: b) Let $f: \mathbb{R}^+ \rightarrow \mathbb{R}^+$ be a function such that $$\displaystyle{2f(x^2)\geq xf(x) + x,}$$for all $x \in \mathbb{R}^+.$ Show that $ f(x^3)\geq x^2,$ for all $x \in \mathbb{R}^+.$ We have $f(x)\geq \frac{\sqrt{x}}{2}$ and , by induction, $$2^nf(x)\geq 2^{n-1}\sqrt{x}+2^{n-2}x^{3/4}+...+x^{\frac{2^n-1}{2^n}}.$$ Using weighted AM-GM inequality (with weights $\frac{2^{n-1}}{2^n-1},\frac{2^{n-2}}{2^n-1},...,\frac{1}{2^n-1}$) and letting $n\to \infty$ we get $f(x)\geq x^{2/3}.$ How can we make the induction to this inequality? Someone help me!!!. And what is weighted AM-GM inequality? How to use that?

socrates wrote: a) Find at least two functions $f: \mathbb{R}^+ \rightarrow \mathbb{R}^+$ such that $$\displaystyle{2f(x^2)\geq xf(x) + x,}$$for all $x \in \mathbb{R}^+.$ b) Let $f: \mathbb{R}^+ \rightarrow \mathbb{R}^+$ be a function such that $$\displaystyle{2f(x^2)\geq xf(x) + x,}$$for all $x \in \mathbb{R}^+.$ Show that $ f(x^3)\geq x^2,$ for all $x \in \mathbb{R}^+.$ Can we find the best constant $a\in \Bbb{R}$ such that $f(x)\geq x^a,$ for all $x \in \mathbb{R}^+?$ $a) 1.-$ If $x\ge 1 \Rightarrow f(x)=x,$ If $x\le 1 \Rightarrow f(x)=\frac{1}{x}$ $\text{ }\text{ }\text{ } 2.-$ If $x\ge 1 \Rightarrow f(x)=x,$ If $x\le 1 \Rightarrow f(x)=1$