# Help!

(1): Let $$x,y$$ and $$z$$ be real numbers such that $$x+y+z=1$$, find the maximum possible value of $$x(x+y)^2 (y+z)^3 (z+x)^4$$.

(2): Let $$\mathbb R^+$$ be the set of all positive real numbers. Find all the functions $$f: \mathbb R^+ \to \mathbb R^+$$ satisfying $$f(x f(y)) = f(xy) + x$$.

Note by Naitik Sanghavi
2 years ago

MarkdownAppears as
*italics* or _italics_ italics
**bold** or __bold__ bold
- bulleted- list
• bulleted
• list
1. numbered2. list
1. numbered
2. list
Note: you must add a full line of space before and after lists for them to show up correctly
paragraph 1paragraph 2

paragraph 1

paragraph 2

[example link](https://brilliant.org)example link
> This is a quote
This is a quote
    # I indented these lines
# 4 spaces, and now they show
# up as a code block.

print "hello world"
# I indented these lines
# 4 spaces, and now they show
# up as a code block.

print "hello world"
MathAppears as
Remember to wrap math in $$...$$ or $...$ to ensure proper formatting.
2 \times 3 $$2 \times 3$$
2^{34} $$2^{34}$$
a_{i-1} $$a_{i-1}$$
\frac{2}{3} $$\frac{2}{3}$$
\sqrt{2} $$\sqrt{2}$$
\sum_{i=1}^3 $$\sum_{i=1}^3$$
\sin \theta $$\sin \theta$$
\boxed{123} $$\boxed{123}$$

Sort by:

As far as the second problem is concerned: $\text{Let } P(x,y) \text{be the assertion } f(xf(y))=f(xy)+x \quad (*) \quad \text{Then we have: } \\ P(y,x) \Rightarrow f(yf(x))=f(xy)+y \quad (1) \\ P(1,x) \Rightarrow f(f(x))=f(x)+1 \quad (2) \\ P(f(x),y) \Rightarrow f(f(x)f(y))=f(f(x)y)+f(x) \stackrel{(1)}{\Rightarrow}f(f(x)f(y))=f(xy)+y+f(x) \quad (3) \\ P(f(y),x) \Rightarrow f(f(y)f(x))=f(f(y)x)+f(y) \stackrel{(*)}{\Rightarrow} f(f(y)f(x))=f(xy)+x+f(y) \quad (4) \\ (3)-(4) \Rightarrow 0=y-x+f(x)-f(y) \Leftrightarrow f(x)-x=f(y)-y \quad \forall x,y \in \mathbb{R^+} \\ \text{Thus we can say that } (f(x)-x) \text{ is constant for all values of x, hence} \\ f(x)-x=k \Leftrightarrow f(x)=x+k \quad (5) \quad \text{ for some real } k \\ (5) \stackrel{x\rightarrow f(x)}{\Longrightarrow} f(f(x))=f(x)+k \stackrel{(2)}{\Rightarrow} f(x)+1=f(x)+k \Leftrightarrow k=1 \quad (6).\\ \text{Therefore } (5)\stackrel{(6)}{\Rightarrow} \boxed{f(x)=x+1}$

- 2 years ago

In case you are not familiar with this solution I post a solution using that $$f(x)$$ is 1-1. $\text{Let } P(x,y) \text{ be the assertion } f(xf(y))=f(xy)+x \quad (*) \text{ thus:} \\ P(y,x) \Rightarrow f(yf(x))=f(xy)+y \quad (1) \\ P(f(x),y) \Rightarrow f(f(x)f(y))=f(yf(x))+f(x) \stackrel{(1)}{\Rightarrow} f(f(x)f(y))=f(xy)+f(x)+y \\ \stackrel{x=1}{\Rightarrow} f(f(1)f(y))=f(y)+f(1)+y \Leftrightarrow f(f(1)f(x))-f(x)-f(1)=x \quad (2) \\ \text{We will prove that } f(x) \text{ is } 1-1 \text{:} \\ f(x_1)=f(x_2) \Leftrightarrow -f(x_1)-f(1)=-f(x_2)-f(1) \quad (3) \\ f(x_1)=f(x_2) \Leftrightarrow f(1)f(x_1)=f(1)f(x_2) \Leftrightarrow f(f(1)f(x_1))=f(f(1)f(x_2)) \quad (4) \\ (3)+(4) \Rightarrow f(f(1)f(x_1))-f(x_1)-f(1)=f(f(1)f(x_2))-f(x_2)-f(1) \\ \stackrel{(2)}{\Rightarrow} x_1=x_2 \quad \text{Thus } f^{-1}(x) \text{ exists } \\ P(1,x) \Rightarrow f(f(x))=f(x)+1 \quad (5) \\ P(\frac{f(x)}{f(y)},y) \Rightarrow f(f(x))=f(\frac{f(x)}{f(y)}y)+\frac{f(x)}{f(y)} \\ \stackrel{(5)}{\Rightarrow} f(x)+1=f(\frac{f(x)}{f(y)}y)+\frac{f(x)}{f(y)} \\ \stackrel{y=1}{\Rightarrow} f(\frac{f(x)}{f(1)})-f(x)-1=-\frac{f(x)}{f(1)} \\ \stackrel{x\rightarrow f^{-1}(xf(1))}{\Longrightarrow} f(x)-f(1)x-1=-x \\ \Leftrightarrow f(x)=(f(1)-1)x+1 \quad (6) \\ \stackrel{x\rightarrow f(x)}{\Rightarrow} f(f(x))=(f(1)-1)f(x)+1 \stackrel{(5)}{\Rightarrow} f(x)+1=(f(1)-1)f(x)+1 \\ \stackrel{f(x)=w}{\Rightarrow} w=(f(1)-1)w \Leftrightarrow (f(1)-2)w=0 \quad \forall w \in \mathbb{R^+} \\ \text{Thus it should be } f(1)-2=0 \Leftrightarrow f(1)=2 \quad (7) \\ \text{Therefore } (6) \stackrel{(7)}{\Rightarrow} \boxed{f(x)=x+1}$

- 2 years ago

Thanks much!!

- 2 years ago