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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 .

Michael Rozenberg
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Miss and Mister cassoulet char
  • 3,511

3 Answers3

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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.

River Li
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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 $x=y=z=1$.

Batominovski
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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!

Michael Rozenberg
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