mornik wrote: $a,b,c>0$ $a+b+c=1$ Prove: \[\frac{a^{2}}b+\frac{b^{2}}c+\frac{c^{2}}a \ge 3(a^{2}+b^{2}+c^{2}) \] $\frac{a^{2}}b+\frac{b^{2}}c+\frac{c^{2}}a \ge 3(a^{2}+b^{2}+c^{2})$ $\leftrightarrow \frac{a^{2}}{b}+\frac{b^{2}}{c}+\frac{c^{2}}{a}\ge \frac{3(a^{2}+b^{2}+c^{2})}{a+b+c}$ $\leftrightarrow M(a-b)^{2}+N(a-c)(b-c) \ge 0$ With: $M=\frac{a+b}{ab}-\frac{2}{a+b+c}\ge 0$ $N=\frac{b+c}{ac}-\frac{2}{a+b+c}\ge 0$ Done.

$\sum\frac{a^{2}}{b}=\sum\frac{a^{4}}{a^{2}b}$ Cauchy gives: $\sum\frac{a^{4}}{a^{2}b}\geq\frac{(\sum a^{2})^{2}}{\sum a^{2}b}$ So it suffices to show that $\frac{(\sum a^{2})^{2}}{\sum a^{2}b}\geq\ 3(\sum a^{2})\leftrightarrow\sum a^{2}\geq\ 3\sum a^{2}b\leftrightarrow\sum a(a-b)^{2}\geq 0$ Done. 10maths_tp0609, how do you come up with such transformations? Is there some kind of algorithm or what?

Behemont, can you explain the last step to me? I understand that we should prove $a^{2}+b^{2}+c^{2}\ge 3a^{2}b+3b^{2}c+3c^{2}a$, but I don't understand how is this equivalent to $a(a-b)^{2}+b(b-c)^{2}+c(c-a)^{2}\ge 0$. Also, where did you use $a+b+c=1$?

$a^{2}+b^{2}+c^{2}-3(a^{2}b+b^{2}+c^{2}a)=(a^{2}+b^{2}+c^{2})(a+b+c)-3(a^{2}b+b^{2}+c^{2}a)=$ $a^{3}-2a^{2}b+ab^{2}+b^{3}-2b^{2}c+bc^{2}+c^{3}-2c^{2}a+ca^{2}=a(a-b)^{2}+b(b-a)^{2}+c(c-a)^{2}$ practice melkac, practice

Thank you. I would never think of something like that.

http://www.mathlinks.ro/viewtopic.php?p=515092#p515092 this problem was last year in romanian JBTST

We will prove that $ (\frac{x^{2}}{y}+\frac{y^{2}}{z}+\frac{z^{2}}{x})(x+y+z)\geq3(x^{2}+y^{2}+z^{2})$ $ \Leftrightarrow\frac{x^{2}(x+z)}{y}+\frac{y^{2}(y+x)}{z}+\frac{z^{2}(z+y)}{x}\geq 2\cdot (x^{2}+y^{2}+z^{2})$ $ \Leftrightarrow \frac{x^{2}(x+z)^{2}}{y(x+z)}+\frac{y^{2}(y+x)^{2}}{z(y+x)}+\frac{z^{2}(z+y)^{2}}{x(z+y)}\geq 2\cdot (x^{2}+y^{2}+z^{2})$. But by CBS we have, after we note $ a=x^{2}+y^{2}+z^{2}$ and $ b=xy+yz+zx$ we get $ \frac{x^{2}(x+z)^{2}}{y(x+z)}+\frac{y^{2}(y+x)^{2}}{z(y+x)}+\frac{z^{2}(z+y)^{2}}{x(z+y)}\geq \frac{(a+b)^{2}}{2b}\geq2a$ qed EDIT: equlity when a=b=c=...

$ \sum \frac{a^{2}}{b}\geq 3\sum a^{2}$ $ \Leftrightarrow \sum(\frac{a^{3}}{b}+\frac{a^{2}c}{b})\geq 2\sum a^{2}$ $ \Leftrightarrow \frac{1}{2}\sum\frac{a^{3}}{b}+\frac{1}{2}\sum(\frac{a^{3}}{b}+\frac{a^{2}c}{b}+\frac{b^{2}a}{c})\geq 2\sum a^{2}$ And this is true because... $ \frac{1}{2}\sum\frac{a^{3}}{b}=\frac{1}{2}\sum\frac{a^{4}}{ab}\geq \frac{1}{2}\frac{(\sum a^{2})^{2}}{\sum ab}\geq \frac{1}{2}\sum a^{2}$ $ \frac{1}{2}\sum(\frac{a^{3}}{b}+\frac{a^{2}c}{b}+\frac{b^{2}a}{c})\geq\frac{3}{2}\sum\sqrt[3]{\frac{a^{3}}{b}\cdot\frac{a^{2}c}{b}\cdot\frac{b^{2}a}{c}}=\frac{3}{2}\sum a^{2}$

It is true the strogner one - $ \frac{a^2}{b}+\frac{b^2}{c}+\frac{c^2}{a}\ge \frac{\left(a+b+c\right)\left(a^2+b^2+c^2\right)}{ab+ac+bc}$. After clearing the denominators and putting all the stuff on the left, the rest is equivalent to $ \frac{\left(bc^2-a^2c\right)^2+\left(ab^2-bc^2\right)^2+\left(a^2c-ab^2\right)^2}{2}\ge 0$.

only from Chebyshev inequality we have $ (\frac{x^{2}}{y}+\frac{y^{2}}{z}+\frac{z^{2}}{x})(y+z+x)\geq3(x^{2}+y^{2}+z^{2}) $

The following inequality is also true. Let $a$, $b$ and $c$ be positive numbers such that $a+b+c=1$. Prove that: \[ \frac{a^{2}}{b}+\frac{b^{2}}{c}+\frac{c^{2}}{a}\ge \frac{93(a^2+b^2+c^2)-19}{12} \]

Assume $b$ between $a$ and $c$ We have $\sum \frac{a^2}{b}=\sum (\frac{a^2}{b}-2a+b)+(a+b+c)^2=\sum \frac{(a-b)^2}{b}+(a+b+c)^2\geq \frac{(a-b+b-c+a-c)^2}{a+b+c}+(a+b+c)^2=4(a-c)^2+(a+b+c)^2\geq 3(a^2+b^2+c^2)\Leftrightarrow (a-c)^2+(b-a)(c-b)\geq 0$(True with assume)

$ \left( {\frac {{x}^{2}}{y}}+{\frac {{y}^{2}}{z}}+{\frac {{z}^{2}}{x}} \right) \left( x+y+z \right) -3\,{x}^{2}-3\,{y}^{2}-3\,{z}^{2}={ \frac {{\it \sum} \left( \left( x-y \right) ^{2}{x}^{2}z \right) +{ \it \sum} \left( \left( y-z \right) ^{2}{x}^{2}y \right) }{xyz}}$

mornik wrote: Let $a,b,c>0$ such that $a+b+c=1$. Prove: \[\frac{a^{2}}b+\frac{b^{2}}c+\frac{c^{2}}a \ge 3(a^{2}+b^{2}+c^{2}) \] $x,y,z>0$,prove \[\left( {\frac {{x}^{2}}{y}}+{\frac {{y}^{2}}{z}}+{\frac {{z}^{2}}{x}} \right) \sqrt {xy+zx+yz}\geq \left( {x}^{2}+{y}^{2}+{z}^{2} \right) \sqrt {3}\]

xzlbq wrote: $x,y,z>0$,prove \[\left( {\frac {{x}^{2}}{y}}+{\frac {{y}^{2}}{z}}+{\frac {{z}^{2}}{x}} \right) \sqrt {xy+zx+yz}\geq \left( {x}^{2}+{y}^{2}+{z}^{2} \right) \sqrt {3}\] best is this: Let $x,y,z>0,xy+yz+zx=1$,prove that \[\frac{x^2}{y}+\frac{y^2}{z}+\frac{z^2}{x}\geq (x+y+z)(x^2+y^2+z^2)\] If who C-S can do?

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mornik wrote: Let $a,b,c>0$ such that $a+b+c=1$. Prove: \[\frac{a^{2}}b+\frac{b^{2}}c+\frac{c^{2}}a \ge 3(a^{2}+b^{2}+c^{2}) \] By computer,we have $$LHS-RHS=\sum_{cyc}{\frac{(c+a)(a-b)^2}{b}}\ge{0}$$

Clearing denominators and homogenizing, it suffices to prove that \[(a^3c+b^3a+c^3b)(a+b+c)\geq 3a^3bc+3ab^3c+3abc^3\iff a^4c+b^4a+c^4b+a^2b^3+b^2c^3+c^2a^3\geq 2(a^3bc+ab^3c+abc^3)\iff \sum c(a^2-bc)^2\geq 0,\]which is true, so we're done.

AlgebraFC wrote: $a^4c+b^4a+c^4b+a^2b^3+b^2c^3+c^2a^3\geq 2(a^3bc+ab^3c+abc^3)\iff \sum c(a^2-bc)^2\geq 0$ which is true, so we're done. Why? $\sum c(a^2-bc)^2\geq 0\iff a^4c+b^4a+c^4b+a^2b^3+b^2c^3+c^2a^3\geq 2(a^2b^2c+a^2bc^2+ab^2c^2)$

Clearing the denominator and homogenizing, we get $a^4c+b^4a+c^4b+a^2b^3+b^2c^3+c^2a^3\geq 2(a^3bc+ab^3c+abc^3)$ By AM-GM inequality, $\frac{1}{2}(\sum_{cyc}a^4c)+(\sum_{cyc}a^3c^2)\geq \frac{3}{2}(\sum_{cyc}a^3bc)$ So it is enough to show that $\sum_{cyc}a^4c\geq \sum_{cyc}a^3bc\iff \sum_{cyc}a^3c(a-b)\geq 0$ WLOG, we suppose that $a\geq b\geq c$ or $a\geq c\geq b$. In either case, it is easy to show that this inequality holds