Prove $2(xy+yz+zx)+3\ge \sqrt{5x^2+4}+\sqrt{5y^2+4}+\sqrt{5z^2+4}$ when $x,y,z>0: x+y+z=3xyz.$

Question

Question

Let $x,y,z>0: x+y+z=3xyz.$ Prove that$$2(xy+yz+zx)+3\ge \sqrt{5x^2+4}+\sqrt{5y^2+4}+\sqrt{5z^2+4}$$ I tried some classical inequality as AM-GM, Cauchy-Schwarz, etc but non of them work.

For example, by Cauchy-Schwarz we will prove $$2(xy+yz+zx)+3\ge \sqrt{3}.\sqrt{5(x^2+y^2+z^2)+12}$$ which is already wrong.

The $\sqrt{5x^2+4}$ should be a long with a good upper bound. I am stuck to find it.

I need some advices. Thanks.

Answer 1

Let $\frac{1}{xy}=x$, $\frac{1}{xz}=b$,$\frac{1}{yz}=a$, $a+b+c=3u$, $ab+ac+bc=3v^2$ and $abc=w^3$.

Thus, $a+b+c=3$, $u=1$ and we need to prove that: $$\frac{2(ab+ac+bc)}{abc}+3\geq\sum_{cyc}\sqrt{\frac{5a}{bc}+4}$$ or $$2(ab+ac+bc)+3abc\geq\sum_{cyc}\sqrt{a^2bc(5a+4bc)}.$$ Now, by C-S $$\sum_{cyc}\sqrt{a^2bc(5a+4bc)}\leq\sqrt{(a+b+c)\sum_{cyc}abc(5a+4bc)}=\sqrt{3u(15u^2+12v^2)w^3}$$ and it's enough to prove that: $$6uv^2+3w^3\geq3\sqrt{(5u^3+4uv^2)w^3}$$ or $$(2uv^2+w^3)^2\geq(5u^2+4uv^2)w^3$$ or $$4u^2v^4+w^6-5u^3w^3\geq0.$$ Now, let $f(w^3)=4u^2v^4+w^6-5u^3w^3.$

Thus, $$f'(w^3)=2w^3-5u^3<0,$$ which says that it's enough to prove $f(w^3)\geq0$ for a maximal value of $w^3$, which by $uvw$ happens for equality case of two variables.

Let $b=c$.

Since our inequality is homogeneous and for $b=c=0$ it's obvious, we can assume $b=c=1$, which gives $$\frac{4(a+2)^2(2a+1)^2}{81}+a^2\geq\frac{5(a+2)^3a}{27}$$ or $$(a-1)^2(a-4)^2\geq0$$ and we are done!

About $uvw$ see here: https://artofproblemsolving.com/community/c6h278791

Answer 2

Proof

By denoting $a=\dfrac{1}{x};b=\dfrac{1}{y};c=\dfrac{1}{z}.$ Now, we prove the following inequality

Problem. Let $a,b,c>0: ab+bc+ca=3.$ Prove that$$\frac{2(a+b+c)}{abc}+3\ge \sqrt{4+\frac{5}{a^2}}+\sqrt{4+\frac{5}{b^2}}+\sqrt{4+\frac{5}{c^2}}.$$ Set $p=a+b+c\ge 3; q=ab+bc+ca=3; r=abc\le 1.$ By AM-GM $$2\sqrt{4+\frac{5}{a^2}}\le \frac{9}{2a}+5-\frac{21}{2(a+2)}.$$ Similarly, we obtain $$LHS\le \frac{27}{4r}-\frac{21(4p+15)}{4(r+4p+14)}+\frac{15}{2}.$$It remains to prove$$\color{blue}{f(r)=(8p-27)(2p+7)+r(10p+18-9r)\ge 0. }\tag{*}$$ It is easy $f^{'}(r)=10p+18(1-r) >0$ hence $f(r)$ is increasing function.

We split $(*)$ into two cases.

$\bullet p\ge \dfrac{27}{8}$ : $f(r)\ge f(0)=(8p-27)(2p+7)\ge 0.$

$\bullet 3\le p\le \dfrac{27}{8}:$ By Schur of third degree \begin{align*} f(r)&\ge f\left(\frac{12p-p^3}{9}\right)\\&=\frac{(p-3)\left[(7-2p)(16p^4+104p^3+284p^2+1042p+1871)+5047\right]}{288}\ge 0. \end{align*} The proof is done. Equality holds at $a=b=c=1.$