Conjecture: $a^{(2(1-a))^{\frac{1}{n}}}+b^{(2(1-b))^{\frac{1}{n}}}+c^{(2(1-c))^{\frac{1}{n}}}+\frac{c}{n}\leq 1$

Question

Hi it's a conjecture inspired by a work of Vasile Cirtoaje :

Claim :

Let $0.5\geq a \geq b \geq c\geq 0$ such that $a+b+c=1$ and $n\geq 1$ a natural number then we have :

$$a^{(2(1-a))^{\frac{1}{n}}}+b^{(2(1-b))^{\frac{1}{n}}}+c^{(2(1-c))^{\frac{1}{n}}}+\frac{c}{n}\leq 1$$

To prove it I have tried Bernoulli's inequality as we have :

$$a^{(2(1-a))^{\frac{1}{n}}}\leq 1-(1-a)(2(1-a))^{\frac{1}{n}}$$

But it's not enough and wrong.

I have tested for $n=1,2,3,4,5$.Moreover the equality case is $a=b=0.5$.

Case $n=1$ :

As in my previous posts we have the inequality $x\in[0,0.5]$ :

$$x^{2(1-x)}\leq 2^{2x+1}x^2(1-x)$$ applying this for each variable we want to show :

$$2^{2a+1}a^2(1-a)+2^{2b+1}b^2(1-b)+2^{2(1-(a+b))+1}(1-(a+b))^2(a+b)+1-(a+b)\leq 1$$

Now by Bernoulli's inequality we have:

$$2^{2x+1}\leq 2(1+2x)$$

Remains to show :

$$2(1+2a)a^2(1-a)+2(1+2b)b^2(1-b)+2(1+2(1-(a+b)))(1-(a+b))^2(a+b)+1-(a+b)\leq 1$$

Wich is wrong because we have an equality case at $b=c=0.25$ and $a=0.5$!

My question :

Have you a counter-example or later (if it's true) a proof ?

Thanks in advance !

Answer 1

I have found a counterexample: $$a=\frac{2}{5},\ b=\frac{3}{10},\ c=\frac{3}{10},\ n=1.$$ This gives: $$a^{(2(1-a))^{\frac{1}{n}}}+b^{(2(1-b))^{\frac{1}{n}}}+c^{(2(1-c))^{\frac{1}{n}}}+\frac{c}{n} = \left(\frac{2}{5}\right)^{\frac{6}{5}}+\left(\frac{3}{10}\right)^{\frac{7}{5}}+\left(\frac{3}{10}\right)^{\frac{7}{5}}+\frac{3}{10}\approx 1.0037 >1.$$

Answer 2

if you pick $a=b=c=\frac{1}{3}$ and $n=1$ then $$a^{(2(1-a))^{\frac{1}{n}}}+b^{(2(1-b))^{\frac{1}{n}}}+c^{(2(1-c))^{\frac{1}{n}}}+\frac{c}{n}=(\frac{1}{3})^{\frac{4}{3}} +(\frac{1}{3})^{\frac{4}{3}}+(\frac{1}{3})^{\frac{4}{3}}+\frac{1}{3}\geq \frac{1}{3}+\frac{1}{3}+\frac{1}{3}+\frac{1}{3}>1$$ And even if you delete the last term $\frac{c}{n}$ in this case it would be also bigger than 1. It is also valid for $n=2$ because you will have: $$a^{(2(1-a))^{\frac{1}{n}}}+b^{(2(1-b))^{\frac{1}{n}}}+c^{(2(1-c))^{\frac{1}{n}}}+\frac{c}{n}=3\cdot(\frac{1}{3})^{\sqrt{\frac{4}{3}}} +\frac{1}{6}=3^{1-\sqrt{\frac{4}{3}}}+\frac{1}{6}=1.010368679... $$