Let $a$, $b$, $c$ and $d$ be positive numbers such that $abcd=1$. Prove that: $$\frac{a^2}{a^3+2}+\frac{b^2}{b^3+2}+\frac{c^2}{c^3+2}+\frac{d^2}{d^3+2}\leq\frac{4}{3}$$ Vasc's LCF Theorem does not help here. Also I tried MV method, but without success.

  • $\begingroup$ :I have an idea, that is to prove $F(a,b,c,d)\leqF(\sqrt{ab},\sqrt{ab},\sart{cd},\sqrt{cd}), and then prove $F(\sqrt{ab},\sqrt{ab},\sart{cd},\sqrt{cd})\leq 0$, but there are still some arguments I did not complete, if you have not tried and interested, maybe it works! $\endgroup$ – Zuo Jan 5 '17 at 7:18

Have not found a contest-type solution, but I think one can brute force Lagrange multiplier here: $$x^2(x^3-4) = -\lambda(x^3+2)^2$$,

Since, $(a,b,c,d)$ that gives the maximum value must satisfy the above equation. I will try to work out a solution.

  • $\begingroup$ I checked your reasoning. For $a$ and $b$ we get also $a^5b^5-4a^2b^2(a^3+b^3)-8a^3b^3(a+b)-4(a^4+b^4)-20a^2b^2+16(a+b)=0$ and what is the rest? $\endgroup$ – Michael Rozenberg Dec 17 '16 at 9:17

Let : $a= e^{x_1}\ , \ b=e^{x_2}\ , \ c=e^{x_3} \ , \ d =e^{x_4}$

So we must to prove:

$$f(x_1)+f(x_2)+f(x_3)+f(x_4) \le \dfrac{4}{3}$$

for $x_1+x_2+x_3+x_4=0 $

$$f(x)= \dfrac{e^{2x}}{e^{3x}+2}$$

Since : $f'(x)= \dfrac{e^{2x}(4-e^{3x})}{(2+e^{3x})^2}$ and $f''(x)=\dfrac{e^{2x}(e^{6x}-26e^{3x}+16)}{(2+e^{3x})^3}$

We only need to consider the inequality in case : $x_1=x_2=x_3=t\ , \ x_4= -4t$

$\Leftrightarrow a=b=c=x \ ,\ d=\dfrac{1}{x^3}$

$g(x)=\dfrac{3x^2}{x^3+2}+\dfrac{x^3}{1+2x^9} \ , x>0$


Maximum is attained at $x=1\ ,\ g(1)=\dfrac{4}{3}$

Equality holdes for : $(a=b=c=d=1)$

  • $\begingroup$ No! The Vasc's theorems don't help here. $\endgroup$ – Michael Rozenberg Feb 25 '17 at 10:55
  • $\begingroup$ it is not Vasc. $\endgroup$ – Sergic Primazon Feb 25 '17 at 11:03
  • $\begingroup$ If so, explain please why we can assume $a=b=c$? $\endgroup$ – Michael Rozenberg Feb 25 '17 at 11:16

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