Inequality with condition $2(x^2+y^2+z^2)\leq 3(x+y+z-1)$

Question

If $x,y,z>0$ such that $2(x^2+y^2+z^2)\leq 3(x+y+z-1)$, then find the minimum value of:

$$S=(x+y+z)\bigg(\frac{1}{\sqrt{2x^3+x}}+\frac{1}{\sqrt{2y^3+y}}+\frac{1}{\sqrt{2z^3+z}}\bigg)$$

What I have tried so far is the following: I think the minimum value is $3\sqrt{3}$ when $x=y=z=1$. This made me think to apply the means inequality in the following way:

$$S=(x+y+z)\bigg(\frac{\sqrt{3}}{\sqrt{3(2x^3+x)}}+\frac{\sqrt{3}}{\sqrt{3(2y^3+y)}}+\frac{\sqrt{3}}{\sqrt{3(2z^3+z)}}\bigg)\ge $$

$$(x+y+z)\bigg(\frac{2\sqrt{3}}{3+2x^3+x}+\frac{2\sqrt{3}}{3+2y^3+y}+\frac{2\sqrt{3}}{3+2z^3+z}\bigg)$$

Now, given we want to find the minimum, applying Titu's inequality:

$$S\ge \frac{18\sqrt{3}(x+y+z)}{2(x^3+y^3+z^3)+x+y+z+9}$$

At which point I got stuck and I am not sure whether I started the right way, because I cannot see how to apply the condition here.

I would be thankful for any suggestions.

Answer 1

Since $$(x+y+z)\geq\frac{1}{3}\sum_{cyc}(2x^2+1),$$ by Holder we obtain: $$S^2=(x+y+z)^2\left(\sum_{cyc}\frac{1}{\sqrt{x(2x^2+1)}}\right)^2\geq$$ $$\geq\frac{1}{3}\sum_{cyc}x\sum_{cyc}(2x^2+1)\left(\sum_{cyc}\frac{1}{\sqrt{x(2x^2+1)}}\right)^2\geq\frac{1}{3}(1+1+1)^4=27.$$ Thus, $$S\geq3\sqrt3.$$ The equality occurs for $x=y=z=1,$ which says that we got a minimal value.

Answer 2

OP's idea also works if we tweak it at the start. The problem is that here:

$$\frac{\sqrt{3}}{\sqrt{3(2x^3+x)}} \geq \frac{2\sqrt{3}}{3+2x^3+x}$$

equality occurs only if $x=1$ and the question has two equality cases $(1,1,1)$ and $\left(\frac{1}{2},\frac{1}{2},\frac{1}{2}\right)$. We're good though, if we use AM-GM like this:

$$\frac{\sqrt{3}}{\sqrt{3x(2x^2+1)}}\geq \frac{2\sqrt{3}}{3x+2x^2+1}$$

and from Cauchy-Schwarz:

$$\sum_{cyc}\frac{2\sqrt{3}}{3x+2x^2+1} \geq \frac{18\sqrt{3}}{2(x^2+y^2+z^2)+3(x+y+z)+3}\geq \frac{18\sqrt{3}}{6(x+y+z)}$$

It follows that $S\geq 3\sqrt{3}$.