11
$\begingroup$

Let $a$, $b$ and $c$ be positive numbers such that $abc=1$. Prove that: $$\sqrt{\frac{a}{4a+2b+3}}+\sqrt{\frac{b}{4b+2c+3}}+\sqrt{\frac{c}{4c+2a+3}}\leq1$$

The equality "occurs" also for $a\rightarrow+\infty$, $b\rightarrow+\infty$ and $a>>b$.

I tried AM-GM, C-S and more, but without any success.

$\endgroup$
  • $\begingroup$ This kind of cyclic but asymmetric inequalities is usually tackled through mixing variables or similar techniques. $\endgroup$ – Jack D'Aurizio Jan 27 '17 at 13:50
  • $\begingroup$ @Jack D'Aurizio MV works for symmetric inequalities. I think it can not help here. $\endgroup$ – Michael Rozenberg Jan 27 '17 at 13:52
  • 1
    $\begingroup$ Are you sure about the "equality" for $a=b \to \infty$? Wouldn't the limit be $1/\sqrt 6 + 1/2 + 0 < 1$? $\endgroup$ – Martin R Jan 27 '17 at 13:55
  • 2
    $\begingroup$ Equality will occur when $a=b=c=1$ $\endgroup$ – Joseph Quarcoo Jan 27 '17 at 14:05
  • 1
    $\begingroup$ Hello I can transform your inequality in a cyclic inequality . I will post my answer tomorrow .It's a nice inequality thanks for that ! $\endgroup$ – max8128 Jun 3 '17 at 18:51
2
$\begingroup$

With the substitutions $a = \frac{x}{y}, \ b = \frac{y}{z}$ and $c = \frac{z}{x}$, it suffices to prove that $$\sum_{\mathrm{cyc}} \sqrt{\frac{zx}{4zx + 2y^2 + 3yz}} \le 1.$$ It suffices to prove that (The desired result follows by summing cyclically.) $$\sqrt{\frac{zx}{4zx + 2y^2 + 3yz}} \le \frac{9x^2+9z^2 + 16xy + 4yz + 34zx}{18x^2+18y^2+18z^2+54xy+54yz+54zx}.$$ Squaring both sides, it suffices to prove that $f(x, y, z) \ge 0$ where \begin{align} f(x,y,z) &= 162 x^4 y^2-549 x^4 y z+504 x^4 z^2+576 x^3 y^3-452 x^3 y^2 z-1300 x^3 y z^2+1708 x^3 z^3+512 x^2 y^4\nonumber\\ &\quad +1144 x^2 y^3 z-1148 x^2 y^2 z^2-1582 x^2 y z^3+504 x^2 z^4-68 x y^4 z-440 x y^3 z^2-596 x y^2 z^3\nonumber\\ &\quad+180 x y z^4+32 y^4 z^2+192 y^3 z^3+378 y^2 z^4+243 y z^5. \end{align} We use the Buffalo Way. There are three possible cases:

1) $z = \min(x,y,z)$: Let $y = z + s, \ x = z+ t; \ s, t\ge 0$. We have $$f(z+t, z+s, z) = a_4z^4 + a_3z^3 + a_2z^2 + a_1z + a_0$$ where \begin{align} a_4 &= 5616 s^2-1728 s t+1728 t^2, \\ a_3 &= 3376 s^3+12864 s^2 t-1176 s t^2+1000 t^3, \\ a_2 &= 476 s^4+7400 s^3 t+10156 s^2 t^2-1376 s t^3+117 t^4, \\ a_1 &= 956 s^4 t+4920 s^3 t^2+1924 s^2 t^3-225 s t^4, \\ a_0 &= 512 s^4 t^2+576 s^3 t^3+162 s^2 t^4. \end{align} It is easy to prove that $a_4\ge 0, \ a_3\ge 0, \ a_2 \ge 0, \ a_0 \ge 0$ and $4a_2a_0 \ge a_1^2$. Thus, $f(z+t, z+s, z) \ge 0$.

2) $y = \min(x,y,z)$: Let $z = y+s, \ x = y+t; \ s, t\ge 0$. We have \begin{align} f(y+t, y, y+s) &= (5616 s^2-1728 s t+1728 t^2) y^4+(6400 s^3+6600 s^2 t+5088 s t^2+1000 t^3) y^3\nonumber\\ &\quad +(2277 s^4+6116 s^3 t+11626 s^2 t^2+3908 s t^3+117 t^4) y^2\nonumber\\ &\quad +(243 s^5+1188 s^4 t+5558 s^3 t^2+5840 s^2 t^3+459 s t^4) y+504 s^4 t^2+1708 s^3 t^3+504 s^2 t^4. \end{align} Clearly, $f(y+t, y, y+s)\ge 0$.

3) $x = \min(x,y,z)$: Similar.

$\endgroup$
  • $\begingroup$ Maybe there is something human's? $\endgroup$ – Michael Rozenberg Jul 23 at 15:08
  • $\begingroup$ I hope to see it. $\endgroup$ – River Li Jul 23 at 15:50
0
$\begingroup$

We begin with a theorem :

Theorem :

Let $a,b,c,d,e,f$ be positive real number , with $a\geq b \geq c$ , $d\geq e \geq f $ under the three following conditions :

$a\geq d$ , $ab\geq de$ , $abc\geq def$ so we have :

$$a+b+c\geq d+e+f$$

Here we suppose that we have :

$a\geq b \geq 1 \geq c $

So to get the majorization we prove this :

$\sqrt{\frac{a}{4a+2b+3}}\geq \sqrt{\frac{c}{4c+2a+3}}$

Wich is equivalent to :

$\frac{a}{4a+2b+3}\geq \frac{c}{4c+2a+3}$

Or :

$a(4c+2a+3)\geq c(4a+2b+3)$

Wich is obvious under the previous conditions.

With the same reasoning we can prove that we have :

$\sqrt{\frac{b}{4b+2c+3}}\geq \sqrt{\frac{c}{4c+2a+3}}$

Now we study the case :

$\sqrt{\frac{a}{4a+2b+3}}\geq\sqrt{\frac{b}{4b+2c+3}}\geq \sqrt{\frac{c}{4c+2a+3}}$ wich corresponding to $a\geq b \geq c$ in the initial theorem

And

$0.5-\frac{1}{8.2a}\geq 0.5-\frac{1}{8.2b}\geq \frac{1}{8.2a}+\frac{1}{8.2b} $ wich corresponding to $d\geq e \geq f $ in the initial theorem

It's clear that we have :

$\sqrt{\frac{a}{4a+2b+3}}\leq \sqrt{\frac{a}{4a+3}}\leq 0.5-\frac{1}{8.2a}$

And

$\sqrt{\frac{b}{4b+2c+3}}\leq \sqrt{\frac{b}{4b+3}}\leq 0.5-\frac{1}{8.2b}$

So we have :

$\sqrt{\frac{a}{4a+2b+3}}\sqrt{\frac{b}{4b+2c+3}}\leq (0.5-\frac{1}{8.2a})(0.5-\frac{1}{8.2b})$

And

$\sqrt{\frac{a}{4a+2b+3}}\sqrt{\frac{b}{4b+2c+3}}\sqrt{\frac{c}{4c+2a+3}}\leq (0.5-\frac{1}{8.2a})(0.5-\frac{1}{8.2b})(\frac{1}{8.2a}+\frac{1}{8.2b})$

Wich is true because we have with the condition $abc=1$

$$27\leq \prod_{cyc}\sqrt{4a+2b+3}$$

So now you just have to apply the theorem with this majorization .

The case $\sqrt{\frac{b}{4b+2c+3}}\geq \sqrt{\frac{a}{4a+2b+3}} \geq \sqrt{\frac{c}{4c+2a+3}}$ is the same.

And for the case $a\geq 1 \geq b \geq c$ you just have to make the following substitution $B=\frac{1}{b}$ to find the previous case $a\geq b \geq 1 \geq c $

Edit :

With the previous substitution the original inequality becomes with $a\geq b \geq 1 \geq c$ and $ac=b$:

$$\sqrt{\frac{ab}{4ab+2+3b}}+\sqrt{\frac{1}{4+2cb+3b}}+\sqrt{\frac{c}{4c+2a+3}}$$

We can briefly prove that we have :

$\sqrt{\frac{ab}{4ab+2+3b}}\geq \sqrt{\frac{1}{4+2cb+3b}}$

And

$\sqrt{\frac{ab}{4ab+2+3b}}\geq \sqrt{\frac{c}{4c+2a+3}}$

Now we study the case :

$\sqrt{\frac{ab}{4ab+2+3b}}\geq \sqrt{\frac{1}{4+2cb+3b}}\geq \sqrt{\frac{c}{4c+2a+3}}$ wich corresponding to $a\geq b \geq c$ in the initial theorem

And

$0.5-\frac{1}{11(ab)^2}\geq \frac{1}{3} \geq 1-\frac{1}{3}-(0.5-\frac{1}{11(ab)^2})$

wich corresponding to $d\geq e \geq f $ in the initial theorem

Now you just have to apply the theorem with this majorization .

The case $\sqrt{\frac{ab}{4ab+2+3b}}\geq \sqrt{\frac{c}{4c+2a+3}}\geq \sqrt{\frac{1}{4+2cb+3b}}$ works this the same majorization.

$\endgroup$
  • $\begingroup$ In the last your step the inequality changes after substitution $B=\frac{1}{b}$. $\endgroup$ – Michael Rozenberg Aug 11 '17 at 11:39
  • $\begingroup$ Yes I will edit I realize now my mistake $\endgroup$ – user448747 Aug 11 '17 at 12:02
  • $\begingroup$ There are many mistakes. $\endgroup$ – Michael Rozenberg Aug 11 '17 at 12:12
  • $\begingroup$ Can you say me where please ? $\endgroup$ – user448747 Aug 11 '17 at 12:28
  • $\begingroup$ I edit the proof . $\endgroup$ – user448747 Aug 11 '17 at 13:06

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.