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Find all functions $f:\mathbb{R}^+ \to \mathbb{R}^+$ such that $$(x + y)f(f(x)y) = x^2 f(f(x) + f(y)) \mbox{, for all } x,y\in\mathbb{R}^+.$$ I tried out various substitutions such as $x=y$, $x=1$, $x+y=x^2$, and nothin' works.

Thanks in advance for your reply.

Pathbreaker
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1 Answers1

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$$\frac{f(f(x)y)}{x^2} = \frac{f(f(x) + f(y))}{x + y }\tag{1}$$ Using the symmetry of the right side of $(1)$: $$\frac{f(f(x)y)}{x^2} = \frac{f(f(x) + f(y))}{x + y }= \frac{f(f(y) + f(x))}{y + x }=\frac{f(f(y)x)}{y^2} \implies\\ $$ $$ \frac{f(f(x)y)}{x^2} =\frac{f(f(y)x)}{y^2} \tag{2} $$ With $ x=y $ in $ (1) $ we have : $$ \frac{f(xf(x))}{x} = \frac{f(2f(x))}{2} \tag{3}\\ $$ One solution to $ (2) $ is : $ f(x)=x^2 $. A solution to $ (3) $ is: $ f(x) = x $.

Lemma 1 : if $ f(x) $ exists it must be injective.

Let $ f(a)=f(b) $: $$ A: \enspace \frac{f(af(b))}{a}=\frac{f(af(a))}{a} = (3) = \frac{f(2f(a))}{2} = \frac{f(2f(b))}{2} = (3) = \frac{f(bf(b))}{b}=\frac{f(bf(a))}{b}\\ B: \enspace (2) \implies \frac{f(af(b))}{b^2} = \frac{f(bf(a))}{a^2} \\ A \enspace and \enspace B: \enspace \implies \frac{a}{b} \cdot \frac{f(bf(a))}{b^2} = \frac{f(af(b))}{b^2} = \frac{f(bf(a))}{a^2} \implies a^3=b^3 \implies a = b \enspace \square \\ $$

Now we prove $ (1) $ has no solutions because $ f(x) <0 $ for some $ x $ .

Let: $ x > 1 $. $$ (1) \implies \frac{f(f(x)(x^2-x))}{x^2} = \frac{f(f(x) + f(x^2-x))}{x+ x^2 -x}\\ f(f(x)(x^2-x)) = f(f(x) + f(x^2-x)) \implies \text{ with injectivity Lemma 1 } \implies \\ f(x)(x^2-x) = f(x) + f(x^2-x) \implies x^2-x=\frac{f(x)}{f(x)} + \frac{ f(x^2-x)}{f(x)} \implies \\ x^2-x -1 = \frac{f(x^2-x)}{ f(x)} $$ We see that $ x^2-x -1 < 0 $ for $ 1 <x < \frac{1+\sqrt{5}}{2} $ ( $ \implies f(x^2-x) < 0 \lor f(x) < 0 $ ) and conclude $ (1) $ has no solutions in the range $f:\mathbb{R}^+ \to \mathbb{R}^+$ $ \enspace \square $
(Note that : $ \frac{1+\sqrt{5}}{2} $ is the famous golden ratio ).

Rutger Moody
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