prove $2 \geqslant a^{k_ob^2}+b^{k_oc^2}+c^{k_oa^2}$

Question

$a,b,c >0$ and $a+b+c=1$, prove $$2 \geqslant a^{k_ob^2}+b^{k_oc^2}+c^{k_oa^2}$$ where $$k_o = 9 \left( \frac{\ln3-\ln2}{\ln3} \right) \approx 3.32163$$ I don't know if this inequality is true or not. Thousand of Excel calculations do not yield any counter-examples yet. I assumed $c=\frac12$ and prove that the inequality is true. But I have no clue how to solve a general case.

Answer 1

NOT A PROOF

Just a partial answer.

Let $f(a,b,c)$ be the expression on the right of the inequality.

Using Lagrange optimization, we set the Lagrangian as $$L=f(a,b,c)-\lambda(a+b+c-1)$$

Then we will obtain 4 equations. As the process is straightforward but tedious, I am not stating them explicitly(I hate typing Mathjax). You may do it on your own.

You would see that in the 4 equations, $a,b,c$ are kind of symmetric. $a=b=c=\frac13$ is a solution to the system.

For $f(a,b,c)\le 2$ to be true, $$f(\frac13,\frac13,\frac13)\le2$$ $$3\times 3^{-k/9}\le 2$$ $$\color{RED}{k\ge9\left(1-\frac{\ln2}{\ln3}\right)}$$ which is exactly what we are looking for.

Unfortunately, I am unable to prove that the point $(\frac13,\frac13,\frac13)$ is a global maxima.

Answer 2

Use the lemma 3.2 of the following paper particulary the note 3.3 and remark that your inequality is equivalent to :

$$2\geq a^{0.5(\frac{k_0}{2}b)}+b^{0.5(\frac{k_0}{2}c)}+c^{0.5(\frac{k_0}{2}a)}$$

With $\sqrt{a}+\sqrt{b}+\sqrt{c}=1$