suppose that:$a \geq b \geq c$ we have $a^2 \geq b^2 \geq c^2$ and $\frac{1}c \geq \frac{1}b \geq \frac{1}a$ using inequalty tchebychev we have : $\frac{a^2}b + \frac{b^2}c + \frac{c^2} a \geq \frac{1}3(a^2+b^2+c^2)(\frac{1}c+ \frac{1}b +\frac{1}a)1 =\frac{1}3(a^2+b^2+c^2)(\frac{1}a+ \frac{1}b +\frac{1}c)(a+b+c)$ ,"because $a+b+c=1$" and $(a+b+c)(\frac{1}a+ \frac{1}b +\frac{1}c) \geq 9$ for $a,b,c>0$

tchebytchev wrote: suppose that:$a \geq b \geq c$ we have $a^2 \geq b^2 \geq c^2$ I think you can't suppose that since the LHS is cyclic.

From $ab^2(b-c)^2+bc^2(c-a)^2+ca^2(a-b)^2\geq 0$ we have $ab^4+bc^4+ca^4+abc(ab+bc+ac)\geq 2abc(a^2+b^2+c^2)$. Now it's enough to prove $a^2b^3+b^2c^3+c^2a^3\geq abc(ab+ac+bc)$.. This is trivial after aplaying CBS on vectors $(\frac{\sqrt{b}}c,\frac{\sqrt{c}}a,\frac{\sqrt{a}}b)$ and $(\frac{1}{\sqrt{b}},\frac{1}{\sqrt{c}},\frac{1}{\sqrt{a}})$

dule_00 wrote: Now it's enough to prove $a^{2}b^{3}+b^{2}c^{3}+c^{2}a^{3}\geq abc(ab+ac+bc)$. It's Muirhead for $(3,2,0)$ and $(2,2,1).$

bodan wrote: dule_00 wrote: Now it's enough to prove $a^2b^3+b^2c^3+c^2a^3\geq abc(ab+ac+bc)$. It's Miurhead for $(3,2,0)$ and $(2,2,1).$ It isn't Miurhead because LHS is cyclic.

$\displaystyle\sum a\sum\frac{a^2}b = \sum a^2+\sum\frac{(a+c)a^2}b\ge 3\sum a^2\Leftrightarrow$ $\Leftrightarrow \displaystyle \sum\frac{(a+c)a^2}b\ge 2\sum a^2$ $\displaystyle\sum\frac{a^3}b+\sum ab\ge 2\sum a^2$ Not it is enough to prove that $\displaystyle\sum \frac cba^2\ge\sum ab$ But from CBS$\displaystyle\sum\frac cba^2\sum bc\ge\left(\sum ab\right)^2$ from wich follows the conclusion.

Oh,nice . And trebusep, $a^2\ge b^2\ge c^2$ , $\frac{1}{c}\ge \frac{1}{b}\ge \frac{1}{a}$. We can't use it.

Needn't any well-known inequalities, just homogenize, the result follows. Sorry, I don't have much time.

view in solution in jpg

Attachments:

This inequality can be broken in $\frac{a^3}{b}+\frac{b^3}{c}+\frac{c^3}{a}\geq\ a^2+\ b^2+\ c^2$ and $\frac{a^2c}{b}+\frac{b^2a}{c}+\frac{c^2b}{a}\geq\ a^2+\ b^2+\ c^2$. The first inequality followes immediately by Cauchy and the second it's trivial applying Schur.

I have a proof: $\displaystyle\sum a\sum\frac{a^2}b = \sum a^2+\sum\frac{(a+c)a^2}b\ge 3\sum a^2$ $\Leftrightarrow \displaystyle \sum\frac{(a+c)a^2}b\ge 2\sum a^2$ apply chauchy ineq$\sum{b(a+c)}\sum\frac{(a+c)a^2}{b^2}\geq (\sum a^2+\sum ab)^2$ so$\sum\frac{(a+c)a^2}{b^2}\geq \frac{(\sum a^2+\sum ab)^2}{2\sum ab}\geq 2\sum a^2$

Utkirstudios wrote: view in solution in jpg Nice solution

We denote: $a=\frac{x}{r}$, $b=\frac{y}{r}$, $c=\frac{z}{r}$, with $x,y,z,r>0$. From condition $a+b+c=1$, results $r=x+y+z$ and $a=\frac{x}{x+y+z}$, $b=\frac{y}{x+y+z}$, $c=\frac{z}{x+y+z}$. The inequality given is equivalent with: $(x+y+z)(\frac{x^2}{y}+\frac{y^2}{z}+\frac{z^2}{x})\geq 3(x^2+y^2+z^2)$ $\Leftrightarrow$ $(*)$ $x^2(\frac{x}{y}+\frac{z}{y})+y^2(\frac{y}{z}+\frac{x}{z})+z^2(\frac{z}{x}+\frac{y}{x})\geq 2(x^2+y^2+z^2)$ $\Leftrightarrow$ Using $\frac{x}{y}\geq 2-\frac{y}{x}, \forall x,y>0$ and analogues, we obtain: $(*)$ $\Leftrightarrow$ $\frac{x^2 z}{y}+\frac{y^2 x}{z}+\frac{z^2 y}{x}\geq xy+yz+zx$. The last one is obvious from Cauchy-Schwarz.

zhaobin wrote: Utkirstudios wrote: view in solution in jpg Nice solution I agree

One-line proof: \[ \frac{a^2}{b}+\frac{b^2}{c}+\frac{c^2}{a} \ge 3(a^2+b^2+c^2) \\ \iff \sum_{cyc} (a-b)^2 \cdot \frac{c+a}{b(b+c+a)} \ge 0. \]

Lovasz wrote: One-line proof: \[ \frac{a^2}{b}+\frac{b^2}{c}+\frac{c^2}{a} \ge 3(a^2+b^2+c^2) \\ \iff \sum_{cyc} (a-b)^2 \cdot \frac{c+a}{b(b+c+a)} \ge 0. \] Nice solution!

I remember that this problem was posted in this forum. That is \[ \frac{a^2}{b}+\frac{b^2}{c}+\frac{c^2}{a} \ge 3\frac{a^2+b^2+c^2}{a+b+c}. \] (Even it was expended). Solution by S.O.S as Lovasz is the simplest one.

I came up with a solution that I am not 100% sure is not announced, but I considerred it ‘not too standard’ so here it is: (please somebody change it to LateX) The inequality is equivalent to (a^2/b)+(b^2/c)+(c^2/a)>=3(a^2+b^2+c^2)/(a+b+c)_________ (1) which is homogene so we may suppose abc=1 After equilizing denominators and making cross product it reduces to a(b^4)+b(c^4)+c(a^4)+(a^2)(b^3)+(b^2)(c^3)+(c^2)(a^3)>=2(a^2+b^2+c^2)_____________(2) According to CS (a(b^4)+b(c^4)+c(a^4))(1/a+1/b+1/c)>=((a^2+b^2+c^2)^2) (a(b^4)+b(c^4)+c(a^4)>= ((a^2+b^2+c^2)^2)/( (1/a+1/b+1/c)) and (a^2)(b^3)+ab+ac>=3ab according to AM-GM summing inequalities of this type gives (a^2)(b^3)+(b^2)(c^3)+(c^2)(a^3)>=ab+ac+bc Now (2) reduces to (((a^2+b^2+c^2)^2)/( (1/a+1/b+1/c)))+ ab+ac+bc>=2(a^2+b^2+c^2) which is simply AM-GM

Here is my idea: Set $ab=x^{2}.bc=y^{2},ac=z^{2}$ therefore $a=\frac{xy}z,b=\frac{xz}y,c=\frac{yz}x$ and $(a+b+c)^{2}=1=a^{2}+b^{2}+c^{2}+2(x^{2}+y^{2}+z^{2})$ thus $a^{2}+b^{2}+c^{2}=1-2(x^{2}+y^{2}+z^{2})$ therefore the ineq desired is rewritten as: $x\frac{y^{3}}{z^{3}}+y\frac{z^{3}}{x^{3}}+z\frac{x^{3}}{y^{3}}\geq 3-6(x^{2}+y^{2}+z^{2})$ which is rewritten as $\frac{xy^{3}}{z^{3}}+\frac{yz^{3}}{x^{3}}+\frac{zx^{3}}{y^{3}}+6(x^{2}+y^{2}+z^{2})\geq 3$ which is equivalent to $\sum_{cyc}x^{4}y^{6}+6\sum_{cyc}x^{5}y^{3}z^{3}\geq 3x^{3}y^{3}z^{3}$.For the moment my brain is stuck at this one[probably because I am a bit tired].If anyone cares to prove the last one,be my guest. PS:It must be easy,because I haven`t used any inequality so far..Good luck

mentioned in 4th post, what is CBS inequality?

It`s Cauchy-Schwartz B stands for Buniakovski

You can use just "simple" AM-GM Homogenising gives $\sum\left(a^{4}c+a^{2}b^{3}\right)\geq 2\sum a^{3}bc$ Since $a^{4}c+\frac{4}{7}a^{2}b^{3}+\frac{2}{7}c^{2}a^{3}+\frac{1}{7}b^{2}c^{3}\geq a^{4}c+a^{2}b^{2}c \geq 2a^{3}bc$, we are done

Makaveli wrote: I came up with a solution that I am not 100% sure is not announced, but I considerred it ‘not too standard’ so here it is: (please somebody change it to ) The inequality is equivalent to \[\frac{a^{2}}{b}+\frac{b^{2}}{c}+\frac{c^{2}}{a}\geq 3\frac{a^{2}+b^{2}+c^{2}}{a+b+c}\;\;\;\;\; (1)\] which is homogene so we may suppose $abc=1.$ After equilizing denominators and making cross product it reduces to \[ab^{4}+bc^{4}+ca^{4}+a^{2}b^{3}+b^{2}c^{3}+c^{2}a^{3}\geq2(a^{2}+b^{2}+c^{2}).\;\;\;\;\;(2)\] According to CS \[(ab^{4}+bc^{4}+ca^{4})\left(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}\right)\geq(a^{2}+b^{2}+c^{2})^{2}(ab^{4}+bc^{4}+ca^{4})\geq\frac{(a^{2}+b^{2}+c^{2})^{2}}{\frac{1}{a}+\frac{1}{b}+\frac{1}{c}}\] and \[a^{2}b^{3}+ab+ac\geq3ab\] according to AM-GM. Summing inequalities of this type gives \[a^{2}b^{3}+b^{2}c^{3}+c^{2}a^{3}\geq ab+ac+bc.\] Now (2) reduces to \[\frac{(a^{2}+b^{2}+c^{2})^{2}}{\frac{1}{a}+\frac{1}{b}+\frac{1}{c}}+ab+ac+bc\geq 2(a^{2}+b^{2}+c^{2})\] which is simply AM-GM

is 3 the largest constant? how we prove it?

$\frac{a^2}{b}+\frac{b^2}{c}+\frac{c^2}{a}-3\frac{a^2+b^2+c^2}{a+b+c}=\frac{1}{abc(a+b+c)(ab+bc+ca)}((a^3+abc)(b^2-ac)^2+(b^3+abc)(c^2-ab)^2+(c^3+abc)(a^2-bc)^2)$ also one line

red3 wrote: $\frac{a^2}{b}+\frac{b^2}{c}+\frac{c^2}{a}-3\frac{a^2+b^2+c^2}{a+b+c}=\frac{1}{abc(a+b+c)(ab+bc+ca)}((a^3+abc)(b^2-ac)^2+(b^3+abc)(c^2-ab)^2+(c^3+abc)(a^2-bc)^2)$ also one line is also true because $LHS = \sum{(a-b)^2(\frac{1}{b}-1)}$ and $a,b,c \le 1$ (because $a+b+c = 1$ and $a,b,c \in R^+$ )

hungkhtn wrote: I remember that this problem was posted in this forum. That is \[ \frac{a^2}{b}+\frac{b^2}{c}+\frac{c^2}{a} \ge \frac{3(a^2+b^2+c^2)}{a+b+c}. \](Even it was expended). Solution by S.O.S as Lovasz is the simplest one. Let $a,b,c>0$.Prove: $$\frac{a^2}{b}+\frac{b^2}{c}+\frac{c^2}{a} \ge \frac{2(a^2+b^2+c^2)}{a+b+c}+\sqrt[6]{\frac{a^6+b^6+c^6}{3}} $$

hungkhtn wrote: I remember that this problem was posted in this forum. That is \[ \frac{a^2}{b}+\frac{b^2}{c}+\frac{c^2}{a} \ge 3\frac{a^2+b^2+c^2}{a+b+c}. \](Even it was expended). Solution by S.O.S as Lovasz is the simplest one. <=> $x,y,z>0,$ \[M={x}^{4}z-2\,{x}^{3}zy+{z}^{2}{x}^{3}+{x}^{2}{y}^{3}+x{y}^{4}-2\,x{y}^{ 3}z-2\,y{z}^{3}x+{y}^{2}{z}^{3}+y{z}^{4}\geq 0\] $x$=min($x,y,z$),then \[M=z \left( {z}^{2}+yz+2\,zx-xy+{x}^{2} \right) \left( x-y \right) \left( x-z \right) +x \left( {y}^{2}+xy+zx+{z}^{2} \right) \left( y- z \right) ^{2}\geq 0\] easy do.

Valentin Vornicu wrote: Let $a,b,c>0$ be real numbers with sum 1. Prove that \[ \frac{a^2}b + \frac{b^2}c + \frac{c^2} a \geq 3(a^2+b^2+c^2) . \] NaPrai: \begin{align*}LHS \ge RHS &\Leftrightarrow \frac{a^2}{b}+\frac{b^2}{c}+\frac{c^2}{a}-1\ge 3(a^2+b^2+c^2)-1\\&\Leftrightarrow \left(\frac{a^2}{b}-2a+b\right)+\left(\frac{b^2}{c}-2b+c\right)+\left(\frac{c^2}{a}-2c+a\right)\ge 3(a^2+b^2+c^2)-(a+b+c)^2 \\&\Leftrightarrow \frac{1}{b}(a-b)^2+\frac{1}{c}(b-c)^2+\frac{1}{a}(c-a)^2 \ge (a-b)^2+(b-c)^2+(c-a)^2\\&\Leftrightarrow \left(\frac{1}{b}-1\right)(a-b)^2+\left(\frac{1}{c}-1\right)(b-c)^2+\left(\frac{1}{a}-1\right)(c-a)^2\ge 0\end{align*}which is obviously true because $a,b,c<1$ here here

hungkhtn wrote: I remember that this problem was posted in this forum. That is \[ \frac{a^2}{b}+\frac{b^2}{c}+\frac{c^2}{a} \ge \frac{3(a^2+b^2+c^2)}{a+b+c}. \](Even it was expended). Solution by S.O.S as Lovasz is the simplest one. Let $a,b,c$ be positive real numbers . Prove that $$\frac{b}{a^2}+\frac{c}{b^2}+\frac{a} {c^2}\ge \frac{3(a+b+c)}{a^2+b^2+c^2},$$ܽ$$ \frac{a^3}{b}+\frac{b^3}{c}+\frac{c^3}{a} \ge \frac{3(a^3+b^3+c^3)}{a+b+c}. $$(Nguyen Viet Hung) Let $a,b,c$ be positive real numbers . For $2\leq n\in N,$ prove or disprove ܽ$$ \frac{a^n}{b}+\frac{b^n}{c}+\frac{c^n}{a} \ge \frac{3(a^n+b^n+c^n)}{a+b+c},$$$$ \frac{b}{a^n}+\frac{c}{b^n}+\frac{a}{c^n}\ge \frac{3(a+b+c)}{a^n+b^n+c^n}. $$

Let $a,b,c$ be positive real numbers .Prove that \[\frac{a^2}{b}+\frac{b^2}{c}+\frac{c^2}{a} \geqslant a+b+c+\frac{(a-b)^2}{a+\frac{6}{25}b}.\]\[\frac{a^2}{b}+\frac{b^2}{c}+\frac{c^2}{a} \geqslant a+b+c+\frac{(a-b)^2}{a+kb}.\]where $k_{\min}=0.2133885325$ h