I think this question is easy for P8 and not suitable for Turkey TST.

Matdem26 wrote: I think this question is easy for P8 and not suitable for Turkey TST. I agree. The answer is $6$ and we need to check only that the equality occurs.

$F(x,y,z)\geqslant F(\sqrt{xy},\sqrt{xy},z)$. Hence, it's enough to find the minimum of $F(x,x,z)$. Fixing $t\in (0,1)$, $F(x,x,t/x^2)=\frac{t(2x+t/x^2)+(x^2+2t/x)(1-t)}{t\sqrt{1-t}}:=\frac{g(x)}{t\sqrt{1-t}}$. The minimum of $g(x)$ when $x\in (0,1]$ occurs at either $x=1$ or $g'(x)=0\implies x=\sqrt[3]{t}$. $-t^2+3t+1=g(1)\geqslant g(\sqrt[3]{t})=-3t^{5/3}+3t^{4/3}+3t^{2/3}\iff (t^{2/3}-1)^2(1+2t^{1/3}+t-t^{4/3})\geqslant 0$ which is true. Hence, it's enough to find the minimum of $F(k,k,k)=\frac{-3k^3+3k^2+3}{k\sqrt{1-k^3}}:=h(k)$ when $k\in (0,1)$. $h'(k)=0$ has exactly one root $r\in (0,1)$ that $r^3+r^2=1$. $F(r,r,r)=6$.

arqady wrote: Matdem26 wrote: I think this question is easy for P8 and not suitable for Turkey TST. I agree. The answer is $6$ and we need to check only that the equality occurs. How?????

AM-GM: Let $xyz = t^3$ then $LHS \geq \frac{3t^4 + 3t^2(1-t^3)}{t^3 \sqrt{1-t^3}} = 3 \frac{1+t^2-t^3}{t \sqrt{1-t^3}}$ But this just am-gm because $t \sqrt{1-t^3} = \sqrt{t^2(1-t^3)} \leq \frac{1}{2} (1+t^2-t^3)$ Thus $LHS \geq 6$ Equality $t^3+t^2=1$ and $x=y=z=t$

Let $x,y,z$ be real numbers such that $0<x,y,z<1$. Then$$\frac {xyz(x+y+z)+(xy+yz+zx)(1-xyz)}{xyz\sqrt {1-xyz}}\geq\frac{3(1-xyz+\sqrt[3]{x^2y^2z^2})}{\sqrt{(1-xyz)\sqrt[3]{x^2y^2z^2}}}\geq 6$$Equality holds when $x=y=z=t $ and $t^3+t^2=1.$

$ AM-GM $ $ x+y+z\geq 3(xyz)^{\frac{1}{3}} $ $ xy+xz+yz\geq 3(xyz)^{\frac{2}{3}} $ $$\frac {xyz(x+y+z)+(xy+yz+zx)(1-xyz)}{xyz\sqrt {1-xyz}}=\frac{x+y+z}{\sqrt{1-xyz}}+\frac{(xy+xz+yz)(\sqrt{1-xyz})}{xyz}\geq \frac{3(xyz)^{\frac{1}{3}}}{(1-xyz)^{\frac{1}{2}}}+\frac{3(xyz)^{\frac{2}{3}}(1-xyz)^{\frac{1}{2}}}{xyz}\geq 2\sqrt{(\frac{3(xyz)^{\frac{1}{3}}}{(1-xyz)^{\frac{1}{2}}})*(\frac{3(xyz)^{\frac{2}{3}}(1-xyz)^{\frac{1}{2}}}{xyz})}=6 $$

Guys, I somewhat agree that this indeed is simple for P8, but I think the equality case still worth some emphasis. As above letting $xyz\triangleq t^3$, and using AM-GM for lower bounding $x+y+z$ and $xy+yz+zx$, we arrive at the expression is greater than or equal to $6$, with equality for $x=y=z=t$, where $1-t^3=t^2$. Now let $\varphi(t)=t^3+t^2-1$. To establish $\varphi(\cdot)$ has a root in $(0,1)$ (note that $0<x,y,z<1$ forces $t\in(0,1)$); we use intermediate value property. $\varphi(0)=-1<0$ and $\varphi(1)=1>0$. Since $\varphi(\cdot)$ is continuous, we conclude there indeed is a $t_0\in(0,1)$ such that $t_0^3+t_0^2=1$.

$$A=\frac {xyz(x+y+z)+(xy+yz+zx)(1-xyz)}{xyz\sqrt {1-xyz}}=\frac{x+y+z}{\sqrt{1-xyz}}+\frac{(xy+yz+zx)\sqrt{1-xyz}}{xyz}\geq 2\sqrt{\frac{(x+y+z)(xy+yz+zx)}{xyz}}$$ Cauchy$\Rightarrow (x+y+z)(yz+xz+xy)\geq (3\sqrt{xyz})^2=9xyz$. $\Rightarrow$ $$A\geq2\sqrt{\frac{(x+y+z)(xy+yz+zx)}{xyz}}\geq2\sqrt{\frac{9xyz}{xyz}}=6$$ $A=6$ if $(i)$ $x=y=z=a$ for $0<a<1$. $(ii)$ $\frac{x+y+z}{\sqrt{1-xyz}}=\frac{(xy+yz+zx)\sqrt{1-xyz}}{xyz}\Rightarrow \frac{3a}{\sqrt{1-a^3}}=\frac{(3a^2)\sqrt{1-a^3}}{a^3}\Rightarrow a^2=1-a^3$. So if we select $a\in(0,1)$ as the root of the polynomial $P(x)=x^3+x^2-1$, we are done. Luckily, this polynomial has a root between $0$ and $1$, because $P(0)=-1$, $P(1)=1$, so for a $r\in(0,1)$, we have $P(r)=0$. Let $x=y=z=r$. Done.

CinarArslan wrote: Let $x,y,z$ be real numbers such that $0<x,y,z<1$. Find the minimum value of: $$\frac {xyz(x+y+z)+(xy+yz+zx)(1-xyz)}{xyz\sqrt {1-xyz}}$$

CinarArslan wrote: Let $x,y,z$ be real numbers such that $0<x,y,z<1$. Find the minimum value of: $$\frac {xyz(x+y+z)+(xy+yz+zx)(1-xyz)}{xyz\sqrt {1-xyz}}$$ P=$\frac{xyz(x+y+z)+(xy+yz+zx)(1-xyz)}{xyz\sqrt{1-xyz}}$ Let $xyz=a^3$ By AM-GM $x+y+z\geq 3\sqrt[3]{xyz}$ = $3a$ $xy+yz+zx\geq 3\sqrt[3]{(xyz)^2}$ = $3a^2$ P=$\frac{a^3.3a + 3a^2.(1-a^3)}{a^3\sqrt{1-a^3}}$ = $\frac{3a^4+3a^2(1-a^3)}{a^3\sqrt{1-a^3}}$ =$\frac{3a}{\sqrt{1-a^3}}+\frac{3a^2(1-a^3)}{a^3\sqrt{1-a^3}}$ =$\frac{3a}{\sqrt{1-a^3}}+\frac{3\sqrt{1-a^3}}{a}$ By again AM-GM $\frac{3a}{\sqrt{1-a^3}}+\frac{3\sqrt{1-a^3}}{a}\geq 2\sqrt{9}$ =$6$ Afterward algebraic prove is not included cause every message has it.

Generalization 1 Let $a,b,c$ be reels such that $0<x,y,z<p$. Then find the minimum value of

Generalization 2 Let $a_{1},a_{2},\cdots,a_{n}$ positive reels and $0<a_{1},a_{2},\cdots,a_{n}<p$ . Then prove that $$\dfrac{\prod{a_{1}}.\left(\sum\limits_{cyc}{a_{1}}\right)+\left(\sum\limits_{sym}{a_{1}a_{2}}\right).\left(p^n-\prod{a_{1}}\right)}{\prod{a_{1}}\sqrt{p^n-\prod{a_{1}}}}\geq 2n$$

Let $ x,y,z>0 $.Prove that $$ \frac{xyz(x+y+z+\sqrt{x^2+y^2+z^2})}{(xy+yz+zx)^2}\leq\frac{3+\sqrt{3}}{9} $$

ehuseyinyigit wrote: Generalization 2 Let $a_{1},a_{2},\cdots,a_{n}$ positive reels and $0<a_{1},a_{2},\cdots,a_{n}<p$ . Then prove that $$\dfrac{\prod{a_{1}}.\left(\sum\limits_{cyc}{a_{1}}\right)+\left(\sum\limits_{sym}{a_{1}a_{2}}\right).\left(p^n-\prod{a_{1}}\right)}{\prod{a_{1}}\sqrt{p^n-\prod{a_{1}}}}\geq 2n$$ Very nice!

crezk wrote: Very nice! Sağ ol abi