Using the $AM-GM$ inequality we arrive at $LHS \ge 9 $ (from $(x + y + z)^2 \ge 3(xy + yz + zx)$) and $9 \ge RHS$ (from $xy + yz + zx \ge 3 \sqrt{xyz}$).

Equality holds when $x=y=z=3$

$(\sum_{cyc} x )^2 = \sum_{cyc} x^2 + 2\sum_{cyc} xy = \sum_{cyc} x^2 + 2\cdot 27 = \sum_{cyc} x^2 + 54 \implies \sum_{cyc} x^2 = (\sum_{cyc} x)^2 - 54$ Using Cauchy Shwarz $(\sum_{cyc} x^2)(1 + 1 + 1) \ge (\sum_{cyc} x)^2 \implies 3\cdot((\sum_{cyc} x)^2 - 54) \ge (\sum_{cyc} x)^2 \implies 2(\sum_{cyc} x)^2 \ge 3\cdot 54 \implies (\sum_{cyc} x)^2 \ge 81$ $\implies \sum_{cyc} x \ge 9$ Using AM-GM $\frac{\sum xy}{3} \ge \sqrt[3]{x^2y^2z^2} \implies 9 \ge \sqrt[3]{x^2y^2z^2} \implies 9 \cdot 9 \cdot 9 \ge x^2y^2z^2 \implies xyz \le 27$ $\implies 3xyz \le 3\cdot 27 = 81 \implies 3xyz \le 9 \cdot 9 \implies 3xyz \le (\sum_{cyc} x)(\sum_{cyc} x) \implies \sqrt{3xyz} \le \sum_{cyc} x$ Thus $\boxed{x + y + z \ge \sqrt{3xyz}}$ and equality occurs when $\boxed{x = y = z = 3}$ Proved

Lukaluce wrote: Let $x, y,$ and $z$ be positive real numbers such that $xy + yz + zx = 27$. Prove that $x + y + z \ge \sqrt{3xyz}$. When does equality hold? Suppose $(x,y,z)$ are associated to some polynomial in $\mathbb{R}[t]$ then let's define $\sigma_k$ as elementary symmetric polynomials in one variable and $S_k$ as their averages i.e, we've $S_k = \frac{\sigma_j}{\binom nj}$ Then the inequality is equivalent to $\sigma \ge \sqrt{3\sigma_3 }$ Then by Maclaurin's inequalities we've $S_1 \ge \sqrt{S_2} \ge \sqrt[3]{S_3}$ From the first inequality we get $$\frac{\sigma_1}{3} \ge \sqrt{\frac{\sigma_2}{3}}=\sqrt{9} \Leftrightarrow \sigma_1 \ge 9$$Then form the second inequality we get $$\sqrt[3]{\frac{\sigma_{3}}{1}} \le \sqrt{\frac{\sigma_2}{3}} =\sqrt{9} \Leftrightarrow \sigma_3 \le 27 \Leftrightarrow \sqrt{3\sigma_3} \le 9$$From these results the desired inequality follows immediately $\blacksquare$

Squaring both sides we have \[ x^2+y^2+z^2+2(xy+yz+xz)=x^2+y^2+z^2+54 \ge 3xyz \] LHS time: By the rearrangement inequality we have $x^2+y^2+z^2 \ge xy+yz+xz$ so we have \[ x^2+y^2+z^2+2(xy+yz+xz) \ge 3(xy+yz+xz)=81 \] RHS time: By AM-GM we have \[ xy+yz+xz \ge 3\sqrt[3]{x^2y^2z^2} \implies xyz \le 27 \] Thus we also have $3(27)=81 \ge 3xyz$. So we have the inequality \[ x^2+y^2+z^2+2(xy+yz+xz) \ge 81 \ge 81 \ge 3xyz \] Thus our proof is complete. The equality case is $x=y=z=3$. Also 200th post yay

It suffices to show that $x^2+y^2+z^2+2(xy+yz+zx)\ge3xyz$, which follows since $27=xy+yz+zx\ge3(xyz)^{2/3}\Rightarrow xyz\le27$ and $x^2+y^2+z^2\ge xy+yz+zx$. Equality when $x=y=z$, so $x=y=z=3$.

Practicing speed First using AM-GM $$27=xy+yz+zx \ge 3 \cdot (xyz)^{\frac{2}{3}} \implies 27 \ge xyz \implies 9 \ge \sqrt{3xyz}$$Now use $(x+y+z)^2 \ge 3(xy+yz+zx)$ to get $x+y+z \ge 9$ hence $$x+y+z \ge 9 \ge \sqrt{3xyz}$$With equality at $x=y=z=3$, thus we are done

By using a well-known inequality, (a+b+c)²≥3(ab+ac+bc) which gives a+b+c≥9.Now the main thing is just to prove xyz≤27 which is true for AM-GM xy+yz+zx≥∛(xyz)²

I will solve this problem with $uvw$ and calculus. ( Please check my solution ) let $3u=x+y+z,3v^2=xy+yz+zx,xyz=w^3$ then we have $$3v^2=27$$$$3u \ge \sqrt{3w^3}$$ square them and write the inequality as $$9u^2-3w^3 \ge 0$$ Note that $3v^2$ is already fixed ( it is constant), now fix the value of $w^3$ then we have $$f(u)=9u^2-3w^3 \ge 0$$ Note that by Tejs theorem (UVW Lemma), the function $f(u)$ is minimized when $x=y$, with that condition, we have: $x^2+2xz=27$ ,$ z=\frac{27-x^2}{2x}$ note that $$3v^2=27$$$$3u=2x+\frac{27-x^2}{2x}$$$$w^3=\frac{x(27-x^2)}{2}$$we have to Prove $$f(x)=\left( 2x+\frac{27-x^2}{2x} \right)^2 -3 \left( \frac{ x(27-x^2)}{2} \right) \ge 0 $$ , we have to find all critial points of f(x), note that we have to find $$f^{'}(x)=\frac{9(x^2-9)(x^3+x^2+9)}{2x^3}=0$$ it is not hard to see that $(x^3+x^2+9)$ can't be achieved for positive reals, hence we have $x^2-9=0$ which means $x=3$ so $f(x)$ has only one critial point which is $x=3$, checking we can see that this indeed holds, and the eqaulity holds at $x=y=3$ and $z=\frac{27-x^2}{2x}=3$ We are done

KhayalAliyev wrote: By using a well-known inequality, (a+b+c)²≥3(ab+ac+bc) which gives a+b+c≥9.Now the main thing is just to prove xyz≤27 which is true for AM-GM xy+yz+zx≥∛(xyz)² You are very bad at MATH @khayal Aliyev

$(x+y+z)^2 \ge 3(xy+yz+zx)$ ,so we get $x+y+z \ge 9$ By AM GM,we have $xy+yz+xz \ge 3 \sqrt[3]{(xyz)^2}$,thus $xyz \le 27.$ So we have $\sqrt{3xyz} \le 9$. Hence we have $x+y+z \ge 3xyz$ Equality holds for x=y=z=3.

Upon squaring both sides, it suffices to show $$ x^2+y^2+z^2+2(xy+yz+xz)\ge 3xyz .$$Note that $x^2+y^2+z^2 \ge xy+yz+xz$, so $$ x^2+y^2+z^2+2(xy+yz+xz)\ge 3(xy+yz+xz)=81\ge 3xyz $$so we only need to verify $27\ge xyz$ which is true since $27=xy+yz+xz\ge 3\sqrt[3]{xy\cdot yz\cdot xz}\Rightarrow \left(\frac{27}{3}\right)^\frac32 =27\ge xyz$ by AM-GM.

@above how’d u get this step: franzliszt wrote: Note that $x^2+y^2+z^2 \ge xy+yz+xz$

adatta0517 wrote: @above how’d u get this step: franzliszt wrote: Note that $x^2+y^2+z^2 \ge xy+yz+xz$ Rearrangement.

adatta0517 wrote: @above how’d u get this step: franzliszt wrote: Note that $x^2+y^2+z^2 \ge xy+yz+xz$ Overkill: Rearrangement. Easy way: Observe that $\frac{(x-y)^2+(y-z)^2+(z-x)^2}{2}\geq 0$, then expand everything.

For AM-GM: $x^{2}$ + $y^{2}$ + $z^{2}$ $\ge xy+yz+zx$ $\Rightarrow$ $(x+y+z)^{2}$ $\ge 3xy+3yz+3xz$ $\Rightarrow$ $(x+y+z)^{2}$ $\ge 81$ $\Rightarrow$ $x+y+z \ge 9$ We know that $xy+yz+zx=27$ $\Rightarrow$ For AM-GM: $9$ $\ge$ $\sqrt{3xyz}$ $\Rightarrow$ $x+y+z$ $\ge$ $\sqrt{3xyz}$ what we wanted to show How we use AM-GM inequality is satisfied with the equality $x=y=z$

Hello mates that is my solution Check it pls The first We know that $(x+y+z)^2 \ge 3(xy+yz+zx)$ By $(x+y+z)^2=x^2+y^2+z^2+2(xy+yz+xz)$ And $x^2+y^2+z^2+2*(xy+yz+xz) \ge 3(xy+xz+yz)$ And $x^2+y^2 + z^2+ \ge xy+yz+zx $ It prove with $x^2 + y^2 \ge 2xy$ By the equation We get $x+y+z \ge 9$ So by well known lemma $xy+yz+xz \ge \sqrt{3xyz(x+y+z)}$ We get $27 \ge xyz$ So that we have to prove that $xy+yz+zx \ge \sqrt{3xyz}$ $81 \ge 3xyz$ And we are done And note that Equality case is $x=y=z=3$

FriIzi wrote: Hello mates that is my solution Check it pls The first We know that $(x+y+z)^2 \ge 3(xy+yz+zx)$ By $(x+y+z)^2=x^2+y^2+z^2+2(xy+yz+xz)$ And $x^2+y^2+z^2+2*(xy+yz+xz) \ge 3(xy+xz+yz)$ And $x^2+y^2 + z^2+ \ge xy+yz+zx $ It prove with $x^2 + y^2 \ge 2xy$ By the equation We get $x+y+z \ge 9$ So by well known lemma $xy+yz+xz \ge \sqrt{3xyz(x+y+z)}$ We get $27 \ge xyz$ So that we have to prove that $xy+yz+xz \ge \sqrt{3xyz}$ $81 \ge 3xyz$ And we are done And note that Equality case is $x=y=z=3$ one nice!