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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 !

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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.

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