# Solve the inequality $a^{ab}c+b^{bc}a+c^{ca}b\leq 1-2abc$

Let $$a,b,c>0$$ such that $$a+b+c=1$$ then we have : $$a^{ab}c+b^{bc}a+c^{ca}b\leq 1-2abc\quad (1)$$

I have a proof :

By Bernoulli's inequality we have : $$a^{ab}c+b^{bc}a+c^{ca}b\leq (1+(a-1)ab)c+(1+(b-1)bc)a+(1+(c-1)ac)b=a+b+c+abc(a+b+c-3)=1-2abc$$

I was thinking for an alternative proof considering by example Young's inequality or someting like that .

Question :

Have you an alternative proof for $$(1)$$ ?

Thanks in advance !

Regards Max.