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Let $x;y;z>0$ such that: $xy+yz+zx=1$. Prove that:

$\frac{x}{\sqrt{yz}+\sqrt{3}}+\frac{y}{\sqrt{xz}+\sqrt{3}}+\frac{z}{\sqrt{yx}+\sqrt{3}}\leq \frac{1}{4\sqrt{3}xyz}$

I think:

$1=xy+yz+zx\geq 3\sqrt[3]{x^2y^2z^2}\Rightarrow xyz\leq \frac{1}{3\sqrt{3}}\Rightarrow \frac{1}{4\sqrt{3}xyz}\geq \frac{3}{4}$

Then, we need to prove that:

$\frac{x}{\sqrt{yz}+\sqrt{3}}+\frac{y}{\sqrt{xz}+\sqrt{3}}+\frac{z}{\sqrt{yx}+\sqrt{3}}\leq \frac{3}{4}$

But i don't know how :(

Thanks :)

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  • $\begingroup$ Take $x=y \to 1, z\to 0$ and you can see that the last inequality does not always hold. So you need something tighter. $\endgroup$ – Macavity May 27 '15 at 5:19
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Let $a^2=3xy, b^2=yz, c^2=3zx$, then $a^2+b^2+c^2=3$ and we need to show $$\sum_{cyc} \frac{bc}{a(3+a)} \le \frac3{4abc} \iff \sum_{cyc} \frac{b^2c^2}{3+a} \le \frac34$$

Elementary methods seem difficult on that, though we can use the $uvw$ trick. Rationalising denominators and using $3u=a+b+c, \; 3v^2=ab+bc+ca \implies 2v^2=3u^2-1,\; w^3=abc$, you can eventually get the following equivalent: $$27u^2(2-3u^2)(3u^2+6u+5)+2(54u^3+90u^2+54u-5)w^3-8w^6 \ge 0 \tag{$\dagger$}$$

The relevant aspect about $(\dagger)$ is that considered as a function of $w^3$, LHS is concave (a quadratic facing down), hence for any possible $u$, LHS achieves minimum only when $w^3$ takes an extreme value in the allowable interval. This occurs when two among $a, b, c$ are equal or $abc=0$.

Case 1: WLOG it is enough to check the inequality for the case $b=c=t$. Then $a^2+2t^2=3 \implies a = \sqrt{3-2t^2}$ and we need to just show for $t \in (0, \sqrt{3/2}]$: $$\frac{t^4}{3+\sqrt{3-2t^2}}+\frac{2(3-2t^2)t^2}{3+t} \le \frac34$$ which while cumbersome is verifiable.

Case 2: WLOG let $c=0$. Then we have $a^2+b^2=3$ and need to show $\dfrac{a^2b^2}3 \le \dfrac34$ which is an easy AM-GM.

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I can slightly simplify the question as follows.

Put $a=\sqrt{yz}$, $b=\sqrt{xz}$, and $c=\sqrt{xy}$. Then $x=\frac{bc}a$, $y=\frac{ac}b$, and $z=\frac{ab}c$, $a^2+b^2+c^2=1$ and we have to prove

$$\frac{bc}{a^2+a\sqrt{3}}+\frac{ac}{b^2+b\sqrt{3}}+\frac{ab}{c^2+c\sqrt{3}}\leq \frac{1}{4\sqrt{3}abc}.$$

$$\frac{1}{a^3+a^2\sqrt{3}}+\frac{1}{b^3+b^2\sqrt{3}}+\frac{1}{c^3+c^2\sqrt{3}}\leq \frac{1}{4\sqrt{3}a^2b^2c^2}.$$

Remark. Let $f(t)=\frac{1}{t^3+\sqrt{3}t^2}$. Unfortunately, $f’’(t)=\frac{2(6t^2+8\sqrt{3}t+9)}{t^4(t+\sqrt{3})^3}>0$ provided $t>0$, so we cannot apply Jensen inequality.

If we wish to use a common denominator then we have to prove .

$$(a^3b^3+a^3c^3+b^3c^3)+\sqrt{3}(a^3b^2+b^3a^2+a^3c^2+c^3a^2+b^3c^2+c^3b^2)+3(a^2+b^2+c^2)\le$$ $$\frac{1}{4\sqrt{3}}(abc+\sqrt{3}(ab+bc+ac)+3(a+b+c)+3\sqrt{3})$$

Since $a^2+b^2+c^2=1$

$$(a^3b^3+a^3c^3+b^3c^3)+\sqrt{3}(a^3b^2+b^3a^2+a^3c^2+c^3a^2+b^3c^2+c^3b^2)+\frac{9}{4}\le$$ $$ \frac{1}{4\sqrt{3}}(abc+\sqrt{3}(ab+bc+ac)+3(a+b+c)).$$

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