5
$\begingroup$

I have this to solve :

Let $x,y,z>0$ such that $x+y+z=3$ then we have : $$\frac{xy^2}{4y^3+3}+\frac{yz^2}{4z^3+3}+\frac{zx^2}{4x^3+3}\leq \frac{3}{7}$$

I try to use Jensen's inequality but the function $f(x)=\frac{x^2}{4x^3+3}$ is neither concave or convex on the interval $[0,3]$

I can't use Karamata's inequality too .

Maybe brute force is the only way to solve it .

I try also to use the derivative but it becomes a little bit difficult .

In fact my idea was to use rearrangment inequality we have :

$$\frac{xy^2}{4y^3+3}+\frac{yz^2}{4z^3+3}+\frac{zx^2}{4x^3+3}\leq \frac{x^3}{4x^3+3}+\frac{y^3}{4y^3+3}+\frac{z^3}{4z^3+3}$$

An use the inequality of Jensen's on $[0.8,1.2]$ with $f(x)=\frac{x^3}{4x^3+3}$

So it's a partial answer .

My question is how to complete my answer or can you provide an other answer ?

Thanks a lot for sharing your knowledge and your time .

$\endgroup$
  • 2
    $\begingroup$ Maybe some of the answer to this question could be useful. $\endgroup$ – Arnaud D. Sep 27 at 13:06
  • 3
    $\begingroup$ Sometimes I wonder how do they come up with questions like these te begin with, let alone finding an answer. $\endgroup$ – imranfat Sep 27 at 13:12
4
$\begingroup$

Note that (tangent line trick) $$\frac{5+2y}{49} - \frac{y^2}{4y^3+3} = \frac{(8y^2+36y+15)(y-1)^2}{49(4y^3+3)}.$$ Thus, we have $$\frac{y^2}{4y^3+3} \le \frac{5+2y}{49}, \quad \forall y \ge 0.$$ Thus, we have $$\sum_{\mathrm{cyc}} \frac{xy^2}{4y^3+3} \le \sum_{\mathrm{cyc}} \frac{x(5+2y)}{49} = \frac{5(x+y+z) + 2(xy+yz+zx)}{49} \le \frac{3}{7}$$ where we have used the fact that $xy+yz+zx \le \frac{(x+y+z)^2}{3}$. We are done.

$\endgroup$
  • $\begingroup$ By AM-GM $(4y^3+3)(2y+5)\geq49\sqrt[7]{\left(y^3\right)^4y^2}=49y^2.$ $\endgroup$ – Michael Rozenberg Sep 28 at 0:54
  • 1
    $\begingroup$ @Michael Rozenberg Nice. Thank you. $\endgroup$ – River Li Sep 28 at 1:08
2
$\begingroup$

Using AM-GM, we have $$4y^3+3=y^3+y^3+y^3+y^3+1+1+1\geq 7\left((y^3)^41^3\right)^{1/7}=7y^{12/7}.$$ So $$\frac{xy^2}{4y^3+3}\leq\frac{xy^2}{7y^{12/7}}=\frac{xy^{2/7}}{7}.$$ Now, note by Holder's inequality that $$\sum_{cyc}xy^{2/7}\leq (x+y+z)^{5/7}(xy+yz+zx)^{2/7}.$$ By Cauchy, $xy+yz+zx\leq \frac{(x+y+z)^2}{3}$. Thus, $$\sum_{cyc}xy^{2/7}\leq \frac{(x+y+z)^{9/7}}{3^{2/7}}.$$ So $$\sum_{cyc}\frac{xy^2}{4y^3+3}\leq \frac{\sum_{cyc}xy^{2/7}}{7}\leq \frac{(x+y+z)^{9/7}}{7\cdot 3^{2/7}}.$$ Thus, $\sum_{cyc}\frac{xy^2}{4y^3+3}\le\frac{(x+y+z)^{9/7}}{7\cdot 3^{2/7}}$ for all $x,y,z\ge 0$, where the equality case is $x=y=z=1$. When $x+y+z=3$, we get $$\sum_{cyc}\frac{xy^2}{4y^3+3}\leq \frac{3^{9/7}}{7\cdot 3^{2/7}}=\frac{3}{7}.$$ The equality holds iff $x=y=z=1$.

More generally, for non-negative real numbers $x,y,z$, for parameters $a,b>0$, and for real exponents $m,n$ such that $$(m-1)(a+b)\le an\le m(a+b),$$ we have $$\sum_{cyc}\frac{xy^m}{ay^n+b}\leq \frac{\sum_{cyc}xy^{\frac{m(a+b)-an}{a+b}}}{a+b}\le \frac{\left(\sum_{cyc}x\right)^{\frac{an-(m-1)(a+b)}{a+b}}\left(\sum_{cyc}xy\right)^{\frac{m(a+b)-an}{a+b}}}{a+b}\leq \frac{\left(\sum_{cyc}x\right)^{\frac{(m+1)(a+b)-an}{a+b}}}{3^{\frac{m(a+b)-an}{a+b}}(a+b)}.$$ The equality case is $x=y=z=1$. In particular if, in addition, $x+y+z=3$, we get $$\sum_{cyc}\frac{xy^m}{ay^n+b}\leq \frac{3}{a+b}.$$ The equality holds iff $x=y=z=1$.

$\endgroup$
2
$\begingroup$

Let $\{x,y,z\}=\{a^2,b^2,c^2\},$ where $a\geq b\geq c>0$.

Thus, $a^2+b^2+c^2=3$ and by AM-GM, C-S, Rearrangement and AM-GM again we obtain: $$\sum_{cyc}\frac{xy^2}{4y^3+3}=\sum_{cyc}\frac{xy^2}{2y^3+1+2y^3+2}\leq\sum_{cyc}\frac{xy^2}{3y^2+4\sqrt{y^3}}\leq$$ $$\leq\frac{1}{(3+4)^2}\sum_{cyc}xy^2\left(\frac{3^2}{3y^2}+\frac{4^2}{4\sqrt{y^3}}\right)=\frac{9}{49}+\frac{4}{49}\sum_{cyc}x\sqrt{y}=$$ $$=\frac{9}{49}+\frac{4}{49}\left(\sqrt{x}\sqrt{xy}+\sqrt{y}\sqrt{yz}+\sqrt{z}\sqrt{zx}\right)\leq\frac{9}{49}+\frac{4}{49}\left(a\cdot ab+b\cdot ac+c\cdot bc\right)=$$ $$=\frac{9}{49}+\frac{4b}{49}\left(a^2+ac+c^2\right)=\frac{9}{49}+\frac{4abc}{49}+\frac{4b(3-b^2)}{49}\leq\frac{9}{49}+\frac{4\cdot1}{49}+\frac{4\cdot2}{49}=\frac{3}{7}$$ and we are done!

$\endgroup$

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.