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It's a new problem that I find interesting :

Let $a,b,c>0$ such that $a+b+c=3$ then we have : $$3\geq (ab)^{bc}+(bc)^{ca}+(ca)^{ab}$$

My try :

If $ab\leq 1$ , $bc\leq 1$ , $ca\leq 1$ the inequality is proved .

So the essential case is $ab\geq 1$ , $bc\leq 1$ , $ca\leq 1$ :

I think that we have (with the condition) and $a\geq 1$ $b\geq 1$:

$a+b-1\geq (ab)^{bc}$

But I'm not sure of this fact .

Furthermore I think that the general case works too .

I like hints so if you have it would be nice .

Thanks in advance .

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  • $\begingroup$ my guess could be taking $x := ab, y := bc, z := ca$ then we have $x,y,z > 0$ and $$ 9 = \left( a+b+c \right) ^{2} = a^2+b^2+c^2 + 2 \left( ab + bc + ca \right) \geq 3 \left( ab + bc + ca \right) = 3 \left( x+y+z \right) $$ or $x+y+z \leq 3$. The problem becomes prove $$ 3 \geq x^y + y^z + z^x . $$ But I don't know how to go further.. Perhaps it would be suppose by contradict then try to prove that we can not have $$ x^y + y^z + z^x > x+y+z . $$ $\endgroup$ – mortal May 17 at 16:21

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