Tetration of cos : An original Inequality .

It's a little bit crazy for me the problem I propose to you today :

Let $$a,b,c\geq 0$$ such that $$a+b+c=1$$ then we have : $$3\cos(\frac{1}{3})^{\cos(\frac{1}{3})^{\cos(\frac{1}{3})}}\geq\cos(a)^{\cos(b)^{\cos(c)}}+\cos(c)^{\cos(a)^{\cos(b)}}+\cos(b)^{\cos(c)^{\cos(a)}}\geq 2+\cos(1)$$

It's nice because we have an upper bound and a lower bound (try with $$\sin(x)$$ it doesn't works) Furthermore I don't see an inequality of this kind on the website .

I have tested numerically until it becomes obvious for me .

To prove this I would like to use Am-Gm for the RHS but it partial and it becomes very complicated . I'm new with tetration and I haven't the technics to go further . As it's cyclic we can't use Jensen's inequality so it's make the problem harder .

Maybe an angel can prove this...

Many thanks .

• Any smooth function $f(a,b,c)$ which is symmetric in $a, b, c$ is going to have $a = b = c$ as either a local maximum or a local minimum, because if increasing one variable causes the value to rise, then so will increasing either of the others. – Paul Sinclair Jul 21 at 4:43
• @PaulSinclair but $f(a,b,c)\neq f(c,b,a)$ ? – user674646 Jul 23 at 13:37
• So its symmetry is cyclic, not anti-cyclic. That is still sufficient for the argument I gave to work. – Paul Sinclair Jul 23 at 14:38