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$a,b,c >0$ 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.

Rhys Hughes
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HN_NH
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  • I would try setting $a,b>0, a<b, a<\frac 12, c=1-a-b$ to simplify the problem. – Rhys Hughes Jun 10 '18 at 11:20
  • Work-intensive possibility, but it would probably work: how about simply finding the minimum of $f(a,b,c)$ with partial derivatives? – peterh Jun 16 '18 at 11:17
  • proofing this is harder since you're not sure of the inequality, $a+b+c=1$, say $ c = 1-a-b$ $$2 \geqslant a^{k_ob^2}+b^{k_oc^2}+c^{k_oa^2}$$ where $$k_o = 9 \left( \frac{\ln3-\ln2}{\ln3} \right) $$ $$ a^{k_ob^2}+b^{k_o(1-a-b)^2}+(1-a-b)^{k_oa^2} = 2 + \alpha$$ let's say $\alpha $ is some constant, which we have to find.... so we would choose values for it now plot the function, say $a=x$ and $b=y$, to visualize the inequality $$ x^{k_oy^2}+y^{k_o(1-x-y)^2}+(1-x-y)^{k_ox^2} \le 2 $$ $$ x^{k_oy^2}+y^{k_o(1-x-y)^2}+(1-x-y)^{k_ox^2} = 2 +\alpha $$ – Aderinsola Joshua Mar 16 '21 at 14:04

2 Answers2

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

Community
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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.

Szeto
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  • You’re relying on symmetry which can mislead often in such problems. All that is being said above is that if the problem is true then the value of $k$ is consistent for one case, viz. $a=b=c=\frac13$. Far from a proof. – Macavity Jun 10 '18 at 06:29
  • For e.g. if the problem were $a+b+c=3$ for non-negatives, then $2^{-a^2}+2^{-b^2}+2^{-c^2}\geqslant 1+2^{-5/4}$ is correct, but in this case the problem is symmetrical though the particular solution for minimum is not. – Macavity Jun 10 '18 at 06:33
  • @Macavity I agree. This answer is not a proof. However, the $k$ lower bound I obtained coincides with the value given, so I posted it. – Szeto Jun 10 '18 at 08:51
  • @Szeto why couldn't you prove $\left(\frac13,\frac13,\frac13\right)$ is the global minima? Shouldn't the solution to the 4 Lagrangian be the proof? – John Glenn Jun 10 '18 at 13:09
  • @JohnGlenn Well, according to the Wiki page, the solution is guaranteed to be a stationary point, but not necessarily an extrema. – Szeto Jun 10 '18 at 13:15
  • If it's the only stationary point, values proximal and orthogonal would indicate if it was a minimum, maximum or saddle. Because of the symmetry, one proximal value would indicate that it's min or max (not a saddle because of the symmetry). Not as pretty as algebra, but certainly easier than a second derivative. Really clever to use the $\mathcal L$-multipliers. – user121330 Mar 16 '21 at 22:40