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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 $uvw$)

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  • $\begingroup$ Sorry just a typo $\endgroup$
    – nDLynk
    Commented Mar 14, 2019 at 12:39

1 Answer 1

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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!

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    $\begingroup$ 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. $\endgroup$
    – ΑΘΩ
    Commented Mar 14, 2019 at 17:26
  • $\begingroup$ Yes, of course, it's the same. $\endgroup$ Commented Mar 14, 2019 at 17:31
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    $\begingroup$ 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! $\endgroup$
    – ΑΘΩ
    Commented Mar 14, 2019 at 17:36
  • $\begingroup$ @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 $c=\frac{(xy-1)a}{y+1}.$ $\endgroup$ Commented Mar 14, 2019 at 17:47

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