After using Cauchy-Schwarz we'll get something obvious. Also SOS helps here. This inequality is very easy.

arqady wrote: After using Cauchy-Schwarz we'll get something obvious. Also SOS helps here. This inequality is very easy. Yes, it is obvious .

$\sum \frac{a^4+5b^4}{a(a+2b)} =\sum \frac{a^4}{a(a+2b)} + 5\sum \frac{b^4}{a(a+2b)} \geq \frac {(\sum a^2)^2} {\sum a(a+2b)} + 5\frac {(\sum a^2)^2} {\sum a(a+2b)}$. Now, $\sum a(a+2b) = \sum a^2 + 2\sum ab = (\sum a)^2 = 1$, and $3\sum a^2 \geq (\sum a)^2 = 1$, so the above continues $ = 6(\sum a^2)^2 \geq 2\sum a^2$. Write $ 2\sum a^2 \geq 1 - \sum ab$ as $ 4\sum a^2 + 2\sum ab \geq 2$, and indeed, $ 4\sum a^2 + 2\sum ab = 3\sum a^2 + (\sum a)^2 \geq 1+1=2$.

For all $t> 0$ one has $\displaystyle\frac{t^4+5}{t(t+2)} \geq \frac{1}{3}(3t^2-10t+13)$ because this is (for $t>0$!) equivalent to $(t-1)^2(4t+15) \geq 0$. For $t = a/b$, $t = b/c$ and $t = c/a$ this implies that \[\begin{aligned}\sum \frac{a^4+5b^4}{a(a+2b)} & \geq \sum \frac{1}{3}(3a^2-10ab+13b^2) \\ & = \frac{1}{3}\left(16(a^2+b^2+c^2) - 10(ab+bc+ca)\right).\end{aligned}\]So it suffices to show that $16(a^2+b^2+c^2) \geq 3 + 7(ab+bc+ca)$. Using $a+b+c = 1$, we write $3 = 3(a+b+c)^2$ to get the equivalent $a^2+b^2+c^2\geq ab+bc+ca$, which is true by $\sum (a-b)^2\geq 0$.

For all positive real numbers $a, b, c$, prove that \[ \frac{a^4+5b^4}{a(a+2b)} + \frac{b^4+5c^4}{b(b+2c)} + \frac{c^4+5a^4}{c(c+2a)} \geq 2(a^2+b^2+c^2). \]

Kurt Gödel wrote: For all $t> 0$ one has $\displaystyle\frac{t^4+5}{t(t+2)} \geq \frac{1}{3}(3t^2-10t+13)$ because this is (for $t>0$!) equivalent to $(t-1)^2(4t+15) \geq 0$. For $t = a/b$, $t = b/c$ and $t = c/a$ this implies that \[\begin{aligned}\sum \frac{a^4+5b^4}{a(a+2b)} & \geq \sum \frac{1}{3}(3a^2-10ab+13b^2) \\ & = \frac{1}{3}\left(16(a^2+b^2+c^2) - 10(ab+bc+ca)\right).\end{aligned}\]So it suffices to show that $16(a^2+b^2+c^2) \geq 3 + 7(ab+bc+ca)$. Using $a+b+c = 1$, we write $3 = 3(a+b+c)^2$ to get the equivalent $a^2+b^2+c^2\geq ab+bc+ca$, which is true by $\sum (a-b)^2\geq 0$. How could you find this inequality $\displaystyle\frac{t^4+5}{t(t+2)} \geq \frac{1}{3}(3t^2-10t+13)$ ??

sqing wrote: For all positive real numbers $a, b, c$, prove that \[ \frac{a^4+5b^4}{a(a+2b)} + \frac{b^4+5c^4}{b(b+2c)} + \frac{c^4+5a^4}{c(c+2a)} \geq 2(a^2+b^2+c^2). \] ___________________ We have: $ \frac{a^4+5b^4}{a(a+2b)}+\frac{b^4+5c^4}{b(b+2c)}+\frac{c^4+5a^4}{c(c+2a)}\geq 2(a^2+b^2+c^2). $ 1) $ \Leftrightarrow\sum\frac{(a^2)^2}{a(a+2b)}+5\sum\frac{(b^2)^2}{b(b+2c)}\geq 2(a^2+b^2+c^2). $ We apply the Cauchy-Schwarz inequality, we get: $ LHS\ge\frac{6(a^2+b^2+c^2)^2}{(a+b+c)^2}=\frac{2(a^2+b^2+c^2)3(a^2+b^2+c^2)}{(a+b+c)^2} \ge 2(a^2+b^2+c^2)$ since $ 3(a^2+b^2+c^2)\ge (a+b+c)^2 $ The proof is ended! _________ Sandu Marin

Kurt Gödel wrote: \[\begin{aligned}\sum \frac{a^4+5b^4}{a(a+2b)} & \geq \frac{1}{3}\left(16(a^2+b^2+c^2) - 10(ab+bc+ca)\right).\end{aligned}\] It is true the following \[ \frac{a^4+5b^4}{a(a+2b)}+\frac{b^4+5c^4}{b(b+2c)}+\frac{c^4+5a^4}{c(c+2a)}\geq \frac{31}{3}(a^2+b^2+c^2)-\frac{25}{3}(ab+bc+ca). \]

using by AM-GM inequality: ... The equality holds: $a=b=c=\frac{1}{3}.$

sqing wrote: For all positive real numbers $a, b, c$, prove that \[ \frac{a^4+5b^4}{a(a+2b)} + \frac{b^4+5c^4}{b(b+2c)} + \frac{c^4+5a^4}{c(c+2a)} \geq 2(a^2+b^2+c^2). \] By my proof it suffices to show that $\frac{1}{3}\left(16(a^2+b^2+c^2)-10(ab+bc+ca)\right)\geq 2(a^2+b^2+c^2)$ which is obviously true. shinichiman wrote: Kurt Gödel wrote: For all $t> 0$ one has $\displaystyle\frac{t^4+5}{t(t+2)} \geq \frac{1}{3}(3t^2-10t+13)$ because this is (for $t>0$!) equivalent to $(t-1)^2(4t+15) \geq 0$. For $t = a/b$, $t = b/c$ and $t = c/a$ this implies that \[\begin{aligned}\sum \frac{a^4+5b^4}{a(a+2b)} & \geq \sum \frac{1}{3}(3a^2-10ab+13b^2) \\ & = \frac{1}{3}\left(16(a^2+b^2+c^2) - 10(ab+bc+ca)\right).\end{aligned}\]So it suffices to show that $16(a^2+b^2+c^2) \geq 3 + 7(ab+bc+ca)$. Using $a+b+c = 1$, we write $3 = 3(a+b+c)^2$ to get the equivalent $a^2+b^2+c^2\geq ab+bc+ca$, which is true by $\sum (a-b)^2\geq 0$. How could you find this inequality $\displaystyle\frac{t^4+5}{t(t+2)} \geq \frac{1}{3}(3t^2-10t+13)$ ?? The right hand side of the original inequality is quadratic in disguise, so we want to bound each term $(a^4+5b^4)/(a(a+2b))$ by something quadratic ("isolated fudging"). By dehomogenizing with respect to $b = 1$, this means that you want to find $\alpha, \beta, \gamma$ such that $t^4 + 5 \geq t(t+2)(\alpha t^2 + \beta t + \gamma)$ with equality in $t = 1$. This means that the polynomial $f(t) = t^4+5 - t(t+2)(\alpha t^2+\beta t + \gamma)$ should have a double zero in $t = 1$: this already gives two equations for $\alpha,\beta,\gamma$. The third equation comes from the quadratic quotient $f(t)/(t-1)^2$: it should have a nonnegative leading coefficient and only negative roots (or none). Different people handle the quadratic term differently, so I guess that's why MariusStanean got another bound. He actually got a better bound, and I think he did this by taking $\frac{1}{2}g''(1)(t-1)^2 + g'(1)(t-1) + g(1)$ as bound, where $g(t) = (t^4+5)/t(t+2)$.

Thank you very much!!!

Really, isn't it too easy, especially for Turkey? It can be solved in a minute...

@above really.1 min solve by T2