# Prove the inequality $\sum x+6\ge 2(\sum\sqrt{xy})$

Let $$x;y;z\in R^+$$ such that $$x+y+z+2=xyz$$. Prove that $$x+y+z+6\ge 2(\sqrt{xy}+\sqrt{yz}+\sqrt{xz})$$

This inequality is not homogeneous and look at the condition i thought that i would substitute the variables $$x;y;z$$ such as:

+)If i need to solve $$x^2+y^2+z^2+2xyz=1$$, i will let $$x=\frac{2a}{\sqrt{\left(a+b\right)\left(a+c\right)}}$$

+)If i need to solve $$xy+yz+xz+xyz=4$$,i will let $$x=\frac{2a}{b+c}$$

but failed. Please explain for me how can i get this substitution (if have a solution by substitution)

I also tried to solve it by $$u,v,w$$.Let $$\sum_{cyc} x=3u;\sum_{cyc} xy;\Pi_{cyc}a=w^3(3u+2=w^3;u,v,w>0 )$$ then $$u\le w^3-3u$$ or $$4u\le w^3$$ but stuck (I am really bad at $$uvw$$)

• Sorry just a typo Commented Mar 14, 2019 at 12:39

The condition gives $$\sum_{cyc}\frac{1}{x+1}=1.$$ Now, let $$x=\frac{b+c}{a}$$ and $$y=\frac{a+c}{b},$$ where $$a$$, $$b$$ and $$c$$ are positives.

Thus, $$z=\frac{a+b}{c}$$ and we need to prove that: $$\sum_{cyc}\frac{b+c}{a}+6\geq2\sum_{cyc}\sqrt{\frac{(b+c)(a+c)}{ab}}$$ or $$\sum_{cyc}(a^2b+a^2c+2abc)\geq2\sum_{cyc}a\sqrt{bc(a+b)(a+c)},$$ which is true by AM-GM.

Indeed, $$2\sum_{cyc}a\sqrt{bc(a+b)(a+c)}=2\sum_{cyc}a\sqrt{(ac+bc)(ab+bc)}\leq$$ $$\leq\sum_{cyc}a(ac+bc+ab+bc)=\sum_{cyc}(a^2b+a^2c+2abc).$$ Done!

• To be more to the point in the crucial substitution step, one can say ''let $a=\frac{1}{x+1}$ and the analogues''; then $a,b,c>0, a+b+c=1$ and it will turn out that $x=\frac{1}{a}-1=\frac{b+c}{a}$ etc.
– ΑΘΩ
Commented Mar 14, 2019 at 17:26
• Yes, of course, it's the same. Commented Mar 14, 2019 at 17:31
• At the risk of being pedantic, I would like to respectfully disagree. Phrasing things like ''let $x=\frac{b+c}{a}$ etc immediately raises the question ''do such $a, b, c$ affording the desired expressions actually exist?''. Of course this question can be tacitly answered in the background but a presentation such as the one suggested above clarifies everything from the onset and precludes any worries about such questions. Nevertheless, with all my fussing over matters, a very elegant solution!
– ΑΘΩ
Commented Mar 14, 2019 at 17:36
• @ΑΘΩ Nice question! Since $z(xy-1)=x+y+2>0,$ we obtain $xy>1$ and we can take $b=\frac{(x+1)a}{y+1}$ and $c=\frac{(xy-1)a}{y+1}.$ Commented Mar 14, 2019 at 17:47