Claim :

Let $0.5\geq a \geq b \geq 0.25\geq c\geq 0$ such that $a+b+c=1$ then we have :

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

To prove it I have tried Bernoulli's inequality .

For $0\leq x\leq 0.25$ we have :

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

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 variables $a,b$ we want to show :

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

Now by Bernoulli's inequality we have:

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

Remains to show :

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

The function :

$$f(x)=2(1+2x)x^2(1-x)$$ is concave for $x\in [\frac{1}{8}+\frac{\sqrt{\frac{19}{3}}}{8},0.5]$

So we can use Jensen's inequality remains to show :

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

So it reduces to a one variable inequality and using derivatives it's not hard to show that :

$$g(c)=2f\left(\frac{1-c}{2}\right)+2c^2+c\leq 1$$

For $c\in[0,1-2\left(\frac{1}{8}+\frac{\sqrt{\frac{19}{3}}}{8}\right)]$

It shows the equality case $a=b=0.5$ and $c=0$ but inequality $(1)$ is false for the other equality case $a=0.5$ and $b=c=0.25$.

We have also the inequality for $x\in[0.25,0.5]$ (we can prove it using logarithm and then derivative)

$$x^{(2(1-x))}\leq x^22^{-5(x-0.25)(x-0.5)+1}$$

Using Bernoulli's inequality :

$$x^22^{-5(x-0.25)(x-0.5)+1}\leq 2(x^2+x^2(-5(x-0.25)(x-0.5)))$$

So Remains to show :

$$2(a^2+a^2(-5(a-0.25)(a-0.5)))+2(b^2+b^2(-5(b-0.25)(b-0.5)))+2c^2+c\leq 1\quad (2)$$

Question :

Have you a proof ? How to show $(2)$ ?

Thanks in advance !


1 Answer 1


Some thoughts

We first give some auxiliary results (Facts 1 through 4). The proofs are not difficult and thus omitted.

Fact 1: If $\frac{1}{2} \ge x \ge \frac{1}{4}$, then $x^{2(1-x)} \le \frac{528x^2+572x-93}{650}$.

Fact 2: If $0\le x \le \frac{1}{4}$, then $x^{2(1-x)} \le 3x^2 - 2x^3$. (Hint: Use Bernoulli inequality.)

Fact 3: If $\frac{1}{2} \ge x \ge \frac{1}{4}$, then $x^{2(1-x)} \le \frac{752x^2-24x-1}{320}$.

Fact 4: If $0\le x \le \frac{1}{4}$, then $x^{2(1-x)} \le 3x^2 - 4x^3$. (Hint: Use Bernoulli inequality.)

Now, we split into two cases:

  1. $c \le \frac{1}{5}$:

By Facts 1-2, it suffices to prove that $$\frac{528a^2+572a-93}{650} + \frac{528b^2+572b-93}{650} + 3c^2 - 2c^3 + c \le 1$$ or $$650c^3-264a^2-264b^2-975c^2-286a-286b-325c+418 \ge 0.$$ It is verified by Mathematica.

  1. $\frac{1}{5} < c \le \frac{1}{4}$:

By Facts 1, 3, 4, it suffices to prove that $$\frac{528a^2+572a-93}{650} + \frac{752b^2-24b-1}{320} + 3c^2 - 4c^3 + c \le 1$$ or $$83200c^3-16896a^2-48880b^2-62400c^2-18304a+1560b-20800c+23841 \ge 0.$$ It is verified by Mathematica.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.