An inequality with condition

I have a new inequality this is the following :

Let $x,y,z$ be real strictly positive number such as :

$$-2 = - x y z + x + y + z$$

Then we have :

$$\sqrt{\frac{1}{x}}+\sqrt{\frac{1}{y}}+\sqrt{\frac{1}{z}}\geq \frac{9}{\sqrt{15+((x+1)(y+1)(z+1))^{(\frac{1}{3})}}}$$

My try :

I put the following substitution :

$a^3=x$; $b^3=y$; $c^3=z$

And I have tried Holder (a generalized version) to get : $$((a^3+1)(b^3+1)(c^3+1))^{(\frac{1}{3})}\geq abc+1$$

The inequality becomes :

$$\sqrt{\frac{1}{a^3}}+\sqrt{\frac{1}{b^3}}+\sqrt{\frac{1}{c^3}}\geq \frac{9}{\sqrt{15+(abc+1)}}$$ And

$$-2=a^3b^3c^3+a^3+b^3+c^3$$

After that I'm stuck...There is someone to achieve this ?

The condition gives $\sum\limits_{cyc}\frac{1}{x+1}=1.$

Let $\frac{1}{x+1}=\frac{a}{3},$ $\frac{1}{y+1}=\frac{b}{3}$ and $\frac{1}{z+1}=\frac{c}{3}.$

Hence, $a+b+c=3$ and we need to prove that $$\sum_{cyc}\sqrt{\frac{a}{b+c}}\geq\frac{9}{\sqrt{15+\frac{3}{\sqrt[3]{abc}}}}$$ or $$\sum_{cyc}\sqrt{a(a+b)(a+c)}\geq\sqrt{\frac{81\prod\limits_{cyc}(a+b)}{15+\frac{a+b+c}{\sqrt[3]{abc}}}}.$$ Now, by C-S $$\sum_{cyc}\sqrt{a(a+b)(a+c)}=\sqrt{\sum_{cyc}\left(a(a+b)(a+c)+2\sqrt{a(a+b)(a+c)}\sqrt{b(a+b)(b+c)}\right)}=$$ $$=\sqrt{\sum_{cyc}\left(a^3+a^2b+a^2c+abc+2\sqrt{(a^2(a+b+c)+abc)(b^2(a+b+c)+abc)}\right)}\geq$$ $$\geq\sqrt{\sum_{cyc}\left(a^3+a^2b+a^2c+abc+2(ab(a+b+c)+abc)\right)}=$$ $$=\sqrt{\sum_{cyc}\left(a^3+3a^2b+3a^2c+5abc\right)}.$$ Thus, it's enough to prove that $$\sum_{cyc}\left(a^3+3a^2b+3a^2c+5abc\right)\geq\frac{81\prod\limits_{cyc}(a+b)}{15+\frac{a+b+c}{\sqrt[3]{abc}}}.$$ Let $a+b+c=3u$, $ab+ac+bc=3v^2$ and $abc=w^3$.

Thus, we need to prove that $$27u^3+9w^3\geq\frac{81(9uv^2-w^3)}{15+\frac{3u}{w}},$$ which is a linear inequality of $v^2$.

Hence, it's enough to prove the last inequality for an extreme value of $v^2$,

which happens for an equality case of two variables.

Since the last inequality is homogeneous, we can assume that $b=c=1$.

Also, let $a=t^3$.

Id est, we need to prove that $$(t^3+15t+2)(t^9+6t^6+21t^3+8)\geq162t(t^3+1)^2$$ or $$(t-1)^2\left(t^{10}+2t^9+18t^8+42t^7+66t^6+18t^5+3t^4-12t^3-36t^2-10t+16\right)\geq0,$$ which is very strong, but true.

• Many thanks it always a pleasure to collaborate with you Dear Michael . I accept your answer . – user448747 Apr 21 '18 at 12:19
• You are welcome! – Michael Rozenberg Apr 21 '18 at 13:32