$\frac{1}{x}+\frac{1}{y}+\frac{1}{z}=3$ $x^2y+\frac{1}{y}+y^2z+\frac{1}{z}+z^2x+\frac{1}{x} \geq 2(x+y+z)$

It is from a $BMO$.

GeniusF wrote: It is from a $BMO$. Where? Thanks.

https://artofproblemsolving.com/community/c6h588115p3481492

GeniusF wrote: https://artofproblemsolving.com/community/c6h588115p3481492 Balkan Mathematics Olympiad 2014 - Problem-1 Thank you very much.

What is AIG???

Did2 wrote: What is AIG??? You should be talking about the AM - GM inequality

For condition: $\frac{1}{y}+\frac{1}{z}+\frac{1}{x}=3$. Using Cauchy-Schwarz, we done that: $(x^2 y + y^2 z + z^2 x)\left( \frac{1}{y}+\frac{1}{z}+\frac{1}{x} \right)\geq (x+y+z)^{2}\Rightarrow x^2 y + y^2 z + z^2 x\geq \frac{(x+y+z)^2}{3}$ Finally, $((x+y+z)-3)^2 \geq 0\Rightarrow (x+y+z)^{2}\geq 6(x+y+z)-9\Rightarrow \frac{(x+y+z)^2}{3}\geq 2(x+y+z)-3$ $x^2 y + y^2 z + z^2 x\geq 2(x+y+z)-3 \blacksquare$

GeniusF wrote: It is from a $BMO$. indeed

tchebytchev wrote: Let $x,y$ and $z$ be positive real numbers such that $xy+yz+zx=3xyz$. Prove that $$x^2y+y^2z+z^2x \geq 2(x+y+z)-3$$. In which cases do we have equality ? By computer,we have $$LHS-RHS=\sum_{cyc}{\frac{(xz-1)^2}{x}}\ge{0}$$

xy+yz+xz=3xyz ................(1) we know that by AM-GM inequality, 3xyz≤ x3+y3+z3 Applying Cbebyshev’s inequality on both sides we get that , x+y+z≥3 ..................(2) AM-GM applied to LHS, in (1) 3xyz ∧ 2/3≤3xyz This gives ; xyz∧1/3≥1 and xyz≥1 ...............(3) By Chebyshevs inequality; X2y+y2x+z2x≥((xy+yz+xz)(x+y+z))/3 =(3xyz(x+y+z))/3 X2y+y2z+z2x≥xyz(x+y+z) X2y+y2z+z2x-2(x+y+z)≥(xyz-2)(x+y+z) ≥(1-2)3= -1•3= -3 X2y+y2z+z2x -2(x+y+z) ≥-3 Therefore x2y+y2z+z2x≥ 2(x+y+z)-3. Obviously equality holds for x=y=z=1(sorry x2y implies x square y thanks)

szl6208 wrote: tchebytchev wrote: Let $x,y$ and $z$ be positive real numbers such that $xy+yz+zx=3xyz$. Prove that $$x^2y+y^2z+z^2x \geq 2(x+y+z)-3$$. In which cases do we have equality ? By computer,we have $$LHS-RHS=\sum_{cyc}{\frac{(xz-1)^2}{x}}\ge{0}$$ What is computer?

tchebytchev wrote: Let $x,y$ and $z$ be positive real numbers such that $xy+yz+zx=3xyz$. Prove that $$x^2y+y^2z+z^2x \geq 2(x+y+z)-3$$. In which cases do we have equality ? For equality: $$\sum \frac{1}{x} = 3 \Rightarrow \sum x^2y -2y - \frac{1}{y} = 0$$$$\sum y(x^2-2-\frac{1}{y^2}) = 0 \Rightarrow (x+\frac{1}{y})(x-\frac{1}{y})=2$$It should be more clear.

Let $x,y$, and $z$ be positive real numbers such that $xy+yz+zx=3xyz$. Prove that $$x^2y+y^2z+z^2x \geq \frac{5}{2}(x+y+z)-7$$$$x^2y+y^2z+z^2x \geq \frac{14}{5}(x+y+z)-\frac{41}{5}$$

sqing wrote: Let $x,y$, and $z$ be positive real numbers such that $xy+yz+zx=3xyz$. Prove that $$x^2y+y^2z+z^2x \geq \frac{5}{2}(x+y+z)-7$$$$x^2y+y^2z+z^2x \geq \frac{14}{5}(x+y+z)-\frac{41}{5}$$ Let $x,y$, and $z$ be positive real numbers such that $xy+yz+zx=3xyz$. Prove that $$x^2y+y^2z+z^2x \geq \frac{5}{2}(x+y+z)-7$$ For the 1) question-- let $p=x+y+z,q=xy+yz+zx,r=xyz$ where $q=3r$ and by AM-GM $$x^2y+y^2z+z^2x\geq 3xyz =q$$, it suffises to prove that $$q\geq \frac{5}{2}p-7 $$or $$2q-5p+14\geq 0$$ Fix the value of $q=q_0$ then r is fixed 'cuz of condition, the inequality becomes $$f(p)=-5p+2q+14 \geq0$$ The function f(p) is at minimum when p is minimum, by PQR Lemma p is minimum when x=y, using this at condition we can get $x=y$ and $z=\frac{x}{3x-2}$ then $$q=\frac{3x^3}{3x-2}$$, $$p=\frac{3x(2x-1)}{3x-2}$$ Using these variables at function f(p) it is sufficient to prove $$f(p)=\frac{-15x(2x-1)}{3x-2}+\frac{6x^3}{3x-2}+14=\frac{6x^3-30x^2+57x-28}{3x-2}\geq 0$$ To find the minimum value of this polynomial, using derivatives and long calculations $$f^{'}(p)=\frac{6x^3-30x^3+57x-28}{3x-2}$$ is equal to 0 at x=1 or at $x=\frac{5}{4} + \frac{\sqrt{\frac{35}{4}}}{4}$ putting them into the inequality we can see both of them holds and the equality case is at x=y=z=1 Same calculations can be done to the original BMO question and your second question

Thanks you

The following generalizations #20,#21 were posted on Balkan MO 2014 thread too.

Generalization 5 For $x,y,z,k\in \mathbf{R^+}$ such that $xy+yz+zx=kxyz$. Prove the following $$x^{n+1}y^n+y^{n+1}z^n+z^{n+1}x^n\geq (n+1)(x+y+z)-kn$$ Determine equality case.

Generalization 6 For $a_{1},a_{2},\cdots,a_{p},k\in \mathbf{R^+}$ such that $\sum_{i=1-> p}{\frac{a_{1}a_{2}\cdots a_{n}}{a_{i}}}=k(a_{1}a_{2}\cdots a_{p})$ Prove the following $$a_{1}^{n+1}a_{2}^{n}+a_{2}^{n+1}a_{3}^{n}+\cdots+a_{p-1}^{n+1}a_{p}^{n}+a_{p}^{n+1}a_{1}^{n}\geq (n+1)(a_{1}+a_{2}+\cdots+a_{p})-kn$$ Determine the equality case.

tchebytchev wrote: Let $x,y$, and $z$ be positive real numbers such that $xy+yz+zx=3xyz$. Prove that $$x^2y+y^2z+z^2x \geq 2(x+y+z)-3.$$In which cases do we have equality? $$\sum_{cyc}\frac{1}{x}=3\implies\sum_{cyc}x^2y\cdot\frac{1}{y}\ge2(x+y+z)-3$$