<s>[INST] Prove that $a^ab^b+a^bb^a\le 1$. 
Let $a$ and $b$ be positive real numbers such that $a+b=1$. Prove that $a^ab^b+a^bb^a\le 1$.

This is RMO 2012.
I got a solution, but I think it's flawed.
Consider $a+\dots +a (a$ $ \text{times})$ and $b+\dots +b(b$ $ \text{times})$. We get $a\cdot a$+ $b\cdot b \\ $.
Applying AM-GM , we get  $$\sqrt[a+b]{a^a\cdot b^b}=a^a\cdot b^b\le a\cdot a+b\cdot b.$$
Considering $a+\dots +a (b$ $\text{times})$ and $b+\dots+b(a$ $\text{times})$, we get
$$\sqrt[a+b]  {a^bb^a}=a^bb^a \le 2\cdot ab \Rightarrow a^a\cdot b^b +a^b\cdot b^a\le (a-b)^2=1$$
I have a lot of flaws. Like $a$ is real so we can't actually consider this $a+\dots +a (b$ $\text{times})$ . Can this be corrected somehow?
 [/INST] Your idea is correct however its representation has to be fixed as you gussed the representation is only valid when $a,b$ are natural
Actually we  have as a subcase the weighted AM-GM inequality as follows $$xa+yb\ge (x+y)\sqrt[x+y]{a^xb^y}$$ where $a,b,x,y$ are positive reals.
For more reference see here.
So instead of writing $$a+a+a...(a \space \text{times})+b+b+.....(b \space \text{times})=a^2+b^2\ge \sqrt[a+b]{a^ab^b}$$ i would just write it as by weighted AM-GM $$a\cdot a+b\cdot b\ge \sqrt[a+b]{a^ab^b}$$
Similarly for the other...:)
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