# Inequality for $a+b+c=3$ $3\geq (ab)^{bc}+(bc)^{ca}+(ca)^{ab}$

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 .

• 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 $$x^y + y^z + z^x > x+y+z .$$ – mortal May 17 at 16:21