Please Help in this inequality

Question

let $a,b,c$ be positive real numbers such that $a,b,c>0$ and $abc=1$ prove that:

$$\frac{1}{a^{20}+b^{11}+c}+\frac{1}{c^{20}+a^{11}+b}+\frac{1}{b^{20}+c^{11}+a}\le1$$

Any Ideas?

Answer 1

Using Cauchy-Schwarz, we have

$$\frac1{a^{20}+b^{11}+c} = \frac{(1+b^9+c^{19})}{(a^{20}+b^{11}+c)(1+b^9+c^{19})} \le \frac{(1+b^9+c^{19})}{(a^{10}+b^{10}+c^{10})^2}$$

So it is sufficient to show that: $$(a^{10}+b^{10}+c^{10})^2 \ge \sum_{cyc} (1+a^9+b^{19}) \tag{1}$$

$$\iff \sum_{cyc}\left(a^{20}+\frac2{a^{10}} - a^{19}-a^9-1 \right) \ge 0$$

So it is now sufficient to show that for $t > 0$, we have $$f(t) = t^{20} + \frac2{t^{10}} +28 \log t - t^{19} - t^9-1 \ge 0$$ which is true (as shown below).

We note $f'(t) = 20t^{19}- 19t^{18} - 9t^8+\dfrac{28}t -\dfrac{20}{t^{11}}$ which can be written as $$t^{11}f'(t) = 20t^{30}-19t^{29}-9t^{19}+28t^{10}-20 = p(t)$$

Now $p(t)$ has exactly one positive root, $t=1$. Further, $p < 0$ for $0< t< 1$ and $p > 0$ for $t> 1$. Hence $f$ has a minimum at $t=1$, which is $f(1)=0$.

Perhaps $(1)$ can be proved directly by AM-GM or similar, couldn't think of any simple way though.

Answer 2

You have the following:

$$ \frac{1}{f(a, b, c)} + \frac{1}{f(c, a, b)} + \frac{1}{f(b, c, a)} \leq 1\text{, where } abc = 1 $$

We can attempt to find the maximum value (the following assumes a maximum exists):

$$ g(a, b, c) = \frac{1}{f(a, b, c)} + \frac{1}{f(c, a, b)} + \frac{1}{f(b, c, a)}\text{, where } abc = 1 $$

Write away, there is symmetry in this function. It must be the case that:

$$ \max\limits_{a, b, c} g(a, b,c)= \max\limits_{a, b, c} g(c, a, b) = \max\limits_{a, b, c} g(b, c, a) $$

It must be the case $a = b = c$ to maximize this function. A simple proof by contradiction can show this. Let's assume that $a = b \neq c$ maximizes $g(a, b, c)$. This would imply that $\max_{a', b', c'} g(b', c', a') \rightarrow a' =c, b' = a, c' = b$. Since $g(a, b, c) = g(b, c, a)$, finding the maximum of $g(b, c, a)$ should be able to generate the same set of $\{a, b, c\}$ (which maximizes the original function). But my original assumption was that $a = b \neq c$ and yet I was able to generate a set such that $a \neq b = c$ which contradicts my original assumption therefore $a = b = c$ (this was only one case, but the only other case to handle is $a \neq b\neq c$ which follows similarly as the above).

Plugging into the constraint equation, we would find that $a = b = c = 1$ gives a maximum of this function (if a maximum exists!). A maximum should exist if $f(a, b, c)$ has a finite minimum value over the range of $a, b, c, > 0 \wedge abc = 1$. We can see whether or not this is the case by using the method of Lagrange multipliers to minimize the function $f(a, b, c) = a^\alpha + b^\beta + c^\gamma$:

$$ f(a, b, c) = a^\alpha + b^\beta + c^\gamma, g(a, b, c) = abc - 1 = 0 \\ \nabla f = \left\langle\alpha a^{\alpha - 1}, \beta b^{\beta - 1}, \gamma c^{\gamma - 1}\right\rangle = \lambda \nabla g = \lambda \langle bc, ac, ab\rangle \\ \alpha a^\alpha = \lambda abc = \lambda \rightarrow a = \left(\frac{\lambda}{\alpha}\right)^\frac{1}{\alpha} \\ \beta b^\beta = \lambda abc = \lambda \rightarrow b = \left(\frac{\lambda}{\beta}\right)^\frac{1}{\beta}\\ \gamma c^\gamma = \lambda abc = \lambda \rightarrow c = \left(\frac{\lambda}{\gamma}\right)^\frac{1}{\gamma} \\ abc = 1 = \frac{\lambda^{\frac{1}{\alpha} + \frac{1}{\beta} + \frac{1}{\gamma}}}{\alpha^\frac{1}{\alpha}\beta^\frac{1}{\beta}\gamma^\frac{1}{\gamma}} \rightarrow \lambda = \left(\alpha^\frac{1}{\alpha}\beta^\frac{1}{\beta}\gamma^\frac{1}{\gamma}\right)^\frac{\alpha\beta\gamma}{\alpha\beta + \alpha\gamma + \beta\gamma} = \left(\alpha^{\beta\gamma}\beta^{\alpha\gamma}\gamma^{\alpha\beta}\right)^\frac{1}{\alpha\beta + \alpha\gamma + \beta\gamma} $$

This gives:

$$ a^\alpha + b^\beta + c^\gamma = \lambda\left(\frac{1}{\alpha} + \frac{1}{\beta} + \frac{1}{\gamma}\right) $$

I believe as long as $\frac{1}{\alpha}+ \frac{1}{\beta} + \frac{1}{\gamma} > 0$, this function indeed has a finite (and positive) minimum value (if that sum is negative, then the minimum value is a negative number, which means there exist positive values to the function which are very close to $0$ and thus make the original $g(a, b, c)$ possibly infinite). This extremum cannot possibly be a maximum because there cannot be a finite maximum to this function as long as at least one of the $\alpha, \gamma, \beta > 0$ which must be the case by the above constraint (I can choose any of $a$, $b$, or $c$ to be as large as I like--it may make another term very small, but at least one of the terms can be extremely large thus dominating the function).

Finally, using $1$) the maximum of $g(a, b, c)$ (at the top) occurs when $a = b = c, abc = 1 \rightarrow a = b= c = 1$ and $2$) the fact that $f(a, b, c)$ (in this case) attains a finite and positive minimum value means that a maximum value of $g(a, b, c)$ exists and occurs at:

$$ \max_{a, b, c} \frac{1}{f(a, b, c)} + \frac{1}{f(c, a, b)} + \frac{1}{f(b, c, a)} = \frac{3}{f(1, 1, 1)} = \frac{3}{1^\alpha + 1^\beta + 1^\gamma} = 1 \text{, when }\frac{1}{\alpha}+ \frac{1}{\beta} + \frac{1}{\gamma} > 0 $$

Therefore:

$$ \frac{1}{a^\alpha + b^\beta + c^\gamma} + \frac{1}{c^\alpha + a^\beta + b^\gamma} + \frac{1}{b^\alpha + c^\beta + a^\gamma} \leq 1 \text{, when } \frac{1}{\alpha}+ \frac{1}{\beta} + \frac{1}{\gamma} > 0 $$