Let $x,y,z \in \mathbb{R}^{+}$ and $x^2+y^2+z^2=x+y+z$. Prove that $$x+y+z+1 \ge 4\sqrt[3]{\frac{xy+yz+zx}{3}}$$$$x+y+z+k \ge (k+3) \sqrt[3]{\frac{xy+yz+zx}{3}}$$Where $k\geq 1.$

Sketch of my solution in the contest ( It can be wrong)

Jeez I can't believe I fake solved this easy problem

Let $x,y,z \in \mathbb{R}^{+}$ and $x^2+y^2+z^2=x+y+z$. Prove that $$x+y+z+3 \ge 6 \sqrt[4]{\frac{xy+yz+zx}{3}}$$

This is true too: ( I think?) Let $x,y,z \in \mathbb{R}^{+}$ and $x^2+y^2+z^2=x+y+z$. Prove that $$x+y+z+3 \ge 6 \sqrt[k]{\frac{xy+yz+zx}{3}}$$Where $k$ is any real number

Iora wrote: Let $x,y,z \in \mathbb{R}^{+}$ and $x^2+y^2+z^2=x+y+z$. Prove that $$x+y+z+3 \ge 6 \sqrt[3]{\frac{xy+yz+zx}{3}}$$ By AM-GM, $$x+y+z+3=\frac{x+y+z}{3}+\frac{x+y+z}{3}+\frac{x+y+z}{3}+1+1+1 \ge 6 \sqrt[6]{\frac{(x+y+z)^3}{27}}$$$$\ge 6 \sqrt[6]{\frac{(xy+yz+zx)(x^2+y^2+z^2)}{9}} \ge 6 \sqrt[3]{\frac{xy+yz+zx}{3}}$$

Iora wrote: Sketch of my solution in the contest ( It can be wrong)

I don't think so.

Iora wrote: Let $x,y,z \in \mathbb{R}^{+}$ and $x^2+y^2+z^2=x+y+z$. Prove that $$x+y+z+3 \ge 6 \sqrt[3]{\frac{xy+yz+zx}{3}}$$ $AM-GM$ gives:$2(\frac{xy+yz+zx}{3}+2)\geqslant RHS$ So we want to prove that:$LHS\geqslant 2(\frac{xy+yz+zx}{3}+2)\Rightarrow 3(x+y+z)\geqslant 2(xy+yz+zx)+3$ Using the condition we have: $ 3(x+y+z)^2\geqslant 2(xy+yz+zx)(x+y+z)+3(x^2+y^2+z^2)\Rightarrow 6 (xy+yz+zx)\geqslant 2(xy+yz+zx)(x+y+z)\Rightarrow 3\geqslant x+y+z$ which is true becayse $ 3(x^2+y^2+z^2)\geqslant (x+y+z)^2$

From $C-S$ we have: $$x^2+y^2+z^2 \ge \frac{(x+y+z)^2}{3} \Longleftrightarrow 3 \ge x+y+z$$And by $AM-GM$ $$xy+yz+zx \le x^2+y^2+z^2=x+y+z$$Replacing $LHS$ with these 2 expressions we get: $$x+y+z +3\ge 6\sqrt[3]{\frac{x+y+z}{x+y+z}}=6 \Longleftrightarrow x+y+z \ge 3$$which is contraction to the first expression we get, After doing this I concluded somehow $x+y+z=3$ is true, eh wrong solution

Here's an alternative approach to prove $3(x+y+z)\geq 2(xy+yz+zx)+3$. Let $x+y+z=x^2+y^2+z^2=a$. We need to prove that $$3a\geq (a^2-a)+3\iff 0\geq (a-3)(a-1).$$Of course, $a\leq 3$ by Cauchy. Note that $$a^2-a=(x+y+z)^2-(x^2+y^2+z^2)>0. $$Thus, $a>1$, and we're done.

Let $x+y+z=x^2+y^2+z^2=a$ and $\frac{xy+yz+zx}{3}=b$. Then we have $a^2=a+6b$. We need to prove: $a+3\ge 6\sqrt[3]{b}$ Taking cube of both sides, we get: $a^3+9a^2+27a+27 \ge 216b$ After replacing $6b=a^2-a$ into last equation, we get: $a^3+9a^2+27a+27 \ge 36a^2-36a$ $\implies$ $a^3-27a^2+63a+27 \ge 0$ $\implies$ $(a-3)(a^2-24a-9) \ge 0$ This is true since we have: 1) From Cauchy-Schwarz: $3(x+y+z) = 3(x^2+y^2+z^2) \ge (x+y+z)^2 \implies 3 \ge x+y+z=a$ thus, $(a-3) \leq 0$ 2) $a^2 \leq 3a \implies a^2-24a-9 \leq 3a-24a-9 = -21a-9 <0 \implies (a^2-24a-9)<0$ Multiplying these two inequalities gives the result. Equality is when $a=3 \implies$ From C-S in 1) and the condition given in the question, we get $x=y=z=1$

Cauchy+Am-Gm

Iora wrote: From $C-S$ we have: $$x^2+y^2+z^2 \ge \frac{(x+y+z)^2}{3} \Longleftrightarrow 3 \ge x+y+z$$And by $AM-GM$ $$xy+yz+zx \le x^2+y^2+z^2=x+y+z$$Replacing $LHS$ with these 2 expressions we get: $$x+y+z +3\ge 6\sqrt[3]{\frac{x+y+z}{x+y+z}}=6 \Longleftrightarrow x+y+z \ge 3$$which is contraction to the first expression we get, After doing this I concluded somehow $x+y+z=3$ is true, eh wrong solution I think urs first expression is wrong cause if they are equal hence we have to keep the equality case

Iora wrote: Let $x,y,z \in \mathbb{R}^{+}$ and $x^2+y^2+z^2=x+y+z$. Prove that $$x+y+z+3 \ge 6 \sqrt[3]{\frac{xy+yz+zx}{3}}$$ (pqr method) Let $p = x + y + z, q = xy + yz + zx$. We have $p^2 - 2q = p$ which results in $q = \frac{p^2 - p}{2}$. Using $p^2 \ge 3q$, we have $p^2 \ge 3\cdot \frac{p^2 - p}{2}$ which results in $p \le 3$. We need to prove that $$(p + 3)^3 \ge 216 \cdot \frac{p^2 - p}{6}$$or $$(3-p)(-p^2 + 24p + 9) \ge 0$$which is true.

sqing wrote: Let $x,y,z \in \mathbb{R}^{+}$ and $x^2+y^2+z^2=x+y+z$. Prove that $$x+y+z+3 \ge 6 \sqrt[4]{\frac{xy+yz+zx}{3}}$$ \[x + y + z + 3 \ge 6\sqrt[5]{{\frac{{xy + yz + zx}}{3}}}\]

Find the real constants $k$ for which $$x+y+z+3\geq6\left(\frac{xy+yz+zx}{3}\right)^k$$overall $x,y,z \in \mathbb{R}^{+}$ and $x^2+y^2+z^2=x+y+z.$

mihaig wrote: Find the real constants $k$ for which $$x+y+z+3\geq6\left(\frac{xy+yz+zx}{3}\right)^k$$overall $x,y,z \in \mathbb{R}^{+}$ and $x^2+y^2+z^2=x+y+z.$ $k \geq \dfrac{1}{5}$ is a necessary and sufficient condition

anh1110004 wrote: mihaig wrote: Find the real constants $k$ for which $$x+y+z+3\geq6\left(\frac{xy+yz+zx}{3}\right)^k$$overall $x,y,z \in \mathbb{R}^{+}$ and $x^2+y^2+z^2=x+y+z.$ $k \geq \dfrac{1}{5}$ is a necessary and sufficient condition Of course, it remains the eternal "why"

mihaig wrote: Find the real constants $k$ for which $$x+y+z+3\geq6\left(\frac{xy+yz+zx}{3}\right)^k$$overall $x,y,z \in \mathbb{R}^{+}$ and $x^2+y^2+z^2=x+y+z.$ Bump

Is my solution wrong? Quote: $$\sum{x^2 + 1} \geq 2\sum{x} = 2\sum{x^2} \geq 6\sqrt[3]{\frac{xy+yz+xz}{3}}$$Which is $$x^2 + y^2 + z^2 \geq 3\sqrt[3]{\frac{xy+yz+xz}{3}}$$apply C.S. to LHS and get $$\frac{(x+ y + z)^2}{3} \geq 3\sqrt[3]{\frac{xy+yz+xz}{3}} \implies \frac{(x+ y + z)^6}{3^3} \geq (3^3)\frac{xy+yz+xz}{3} \implies (xy + yz + zx)^3 \geq 9(xy + yz + zx) \implies (xy + yz + zx) \geq 3$$We will prove this with AM-GM $$x^2+y^2+z^2 \geq xy + yz + zx$$$$(x+y+z)^2 \geq 3(xy + yz + zx)$$Applying our condition and combining these we get $$xy + yz + zx \geq 3$$Which ends our solution

Equality holds when $x=y=z=1$, I think, using AM-GM, $x+y+z+3 \ge 6 \sqrt[6]{\frac{(x+y+z)^3}{3}}$ that's all I could get :skull:

$x+y+z+3=x+y+z/3+x+y+z/3+x+y+z/3+1+1+1$ Then AM-GM $x+y+z+3=x+y+z/3+x+y+z/3+x+y+z/3+1+1+1 \ge 6 \sqrt[6]{\frac{(x+y+z)^3}{27}} \ge 6 \sqrt[6]{\frac{(xy+yz+xz)(x^2+y^2+z^2)}{9}} $\ge$ 6 \sqrt[6]{\frac{(xy+xz+yz)(x^2+y^2+z^2)}{9}}$ Quote: We got this from Chebyshev's theorem, after that we'll use Muirhead (2,0,0) \ge (1,1,0)

Ferum_2710 wrote: Is my solution wrong? Quote: $$\sum{x^2 + 1} \geq 2\sum{x} = 2\sum{x^2} \geq 6\sqrt[3]{\frac{xy+yz+xz}{3}}$$Which is $$x^2 + y^2 + z^2 \geq 3\sqrt[3]{\frac{xy+yz+xz}{3}}$$apply C.S. to LHS and get $$\frac{(x+ y + z)^2}{3} \geq 3\sqrt[3]{\frac{xy+yz+xz}{3}} \implies \frac{(x+ y + z)^6}{3^3} \geq (3^3)\frac{xy+yz+xz}{3} \implies (xy + yz + zx)^3 \geq 9(xy + yz + zx) \implies (xy + yz + zx) \geq 3$$We will prove this with AM-GM $$x^2+y^2+z^2 \geq xy + yz + zx$$$$(x+y+z)^2 \geq 3(xy + yz + zx)$$Applying our condition and combining these we get $$xy + yz + zx \geq 3$$Which ends our solution Yes, it is. Try $(1,0,0).$