we know $ a^5 + b^5 >= a^2b^2(a+b)$(prove easy) so the inequality is $ >= ab + bc+ ca$ now we have to prove $ ab + bc + ca +2>= 3(ab + bc + ca)$ or $ 2(a^2 + b^2 + c^2) >= 2(ab + bc + ca)$,which is true.

Since $ y = x^5$ is an increasing and downwards convex function for $ x > 0$, by Jensen's inequality we have $ \frac {a^5 + b^5}{2}\geq \left(\frac {a + b}{2}\right)^5\Longleftrightarrow \frac {a^5 + b^5}{ab(a + b)}\geq \frac {1}{16}\cdot \frac {(a + b)^4}{ab} = \frac {1}{16}(a + b)^2\cdot \frac {(a + b)^2}{ab}$ $ \geq \frac {1}{16}(a + b)^2\cdot 4\ \because (a + b)^2\geq 4ab$ for $ a > 0,\ b > 0$. Thus for $ a > 0,\ b > 0,\ c > 0$, $ \frac {a^5 + b^5}{ab(a + b)} + \frac {b^5 + c^5}{bc(b + c)} + \frac {c^5 + a^5}{ca(c + a)}$ $ \geq \frac {1}{4}\{(a + b)^2 + (b + c)^2 + (c + a)^2\}$ $ = \frac {1}{2}(a^2 + b^2 + c^2 + ab + bc + ca)$ $ \geq ab + bc + ca\ \because a^2 + b^2 + c^2\geq ab + bc + ca$ $ \geq 3(ab + bc + ca) - 2$ Then we are to prove $ ab+bc+ca\geq 3(ab+bc+ca)-2$ which can be proved by $ ab+bc+ca\geq 3(ab+bc+ca)-2\Longleftrightarrow 1\geq ab + bc + ca$ $ \Longleftrightarrow a^2 + b^2 + c^2\geq ab + bc + ca\ Q.E.D.$

delegat wrote: IF $ a$, $ b$ and $ c$ are positive reals such that $ a^{2} + b^{2} + c^{2} = 1$ prove the inequality: $ \frac {a^{5} + b^{5}}{ab(a + b)} + \frac {b^{5} + c^{5}}{bc(b + c)} + \frac {c^{5} + a^{5}}{ca(a + c)}\geq 3(ab + bc + ca) - 2$ The following inequality is true too: $ \frac {a^{5} + b^{5}}{ab(a + b)} + \frac {b^{5} + c^{5}}{bc(b + c)} + \frac {c^{5} + a^{5}}{ca(a + c)}\geq6-5(ab + bc + ca).$

arqady wrote: delegat wrote: IF $ a$, $ b$ and $ c$ are positive reals such that $ a^{2} + b^{2} + c^{2} = 1$ prove the inequality: $ \frac {a^5 + b^5}{ab(a + b)} + \frac {b^{5} + c^{5}}{bc(b + c)} + \frac {c^{5} + a^{5}}{ca(a + c)}\geq 3(ab + bc + ca) - 2$ The following inequality is true too: $ \frac {a^{5} + b^{5}}{ab(a + b)} + \frac {b^{5} + c^{5}}{bc(b + c)} + \frac {c^{5} + a^{5}}{ca(a + c)}\geq6 - 5(ab + bc + ca).$ it is equivalent to $ \sum(\frac {a^5 + b^5}}{ab(a + b)} - \frac {1}{2}(a^2 + b^2)} \geq \frac {5}{2}(\sum(a - b)^2)$ $ \sum(a - b)^2(\frac {2a^2 + ab + 2b^2}{2ab} - \frac {5}{2}) \geq 0$ $ \sum\frac {(a - b)^4}{ab} \geq$ 0 which is true i don't know why the latex looks like that

Nice work HTA You have made small mistake in writing Namely you wrote \frac { ...}}

$ a^5 + b^5 \ge \frac{(a+b)^5}{2^4}$ $ a+b \ge 2 \sqrt{ab}$

delegat wrote: IF $ a$, $ b$ and $ c$ are positive reals such that $ a^{2} + b^{2} + c^{2} = 1$ prove the inequality: $ \frac {a^{5} + b^{5}}{ab(a + b)} + \frac {b^{5} + c^{5}}{bc(b + c)} + \frac {c^{5} + a^{5}}{ca(a + b)}\geq 3(ab + bc + ca) - 2$

HTA wrote: arqady wrote: delegat wrote: IF $ a$, $ b$ and $ c$ are positive reals such that $ a^{2} + b^{2} + c^{2} = 1$ prove the inequality: $ \frac {a^5 + b^5}{ab(a + b)} + \frac {b^{5} + c^{5}}{bc(b + c)} + \frac {c^{5} + a^{5}}{ca(a + c)}\geq 3(ab + bc + ca) - 2$ The following inequality is true too: $ \frac {a^{5} + b^{5}}{ab(a + b)} + \frac {b^{5} + c^{5}}{bc(b + c)} + \frac {c^{5} + a^{5}}{ca(a + c)}\geq6 - 5(ab + bc + ca).$ it is equivalent to ${{ \sum(\frac {a^5 + b^5}}{ab(a + b)} - \frac {1}{2}(a^2 + b^2)} \geq \frac {5}{2}(\sum(a - b)^2)$ $ \sum(a - b)^2(\frac {2a^2 + ab + 2b^2}{2ab} - \frac {5}{2}) \geq 0$ $ \sum\frac {(a - b)^4}{ab} \geq$ 0 which is true i don't know why the latex looks like that It can be: it is equivalent to ${ \sum(\frac {a^5 + b^5}{ab(a + b)} - \frac {1}{2}(a^2 + b^2)}) \geq \frac {5}{2}(\sum(a - b)^2)$

By Muirhead inequality we have $a^5+b^5\geq a^4b+b^4a=ab(a^3+b^3)=ab(a+b)(a^2-ab+b^2)\geq a^2b^2(a+b)$ [last by AM-GM], so: $\sum_{cyc}\frac{a^5+b^5}{ab(a+b)}+2\geq \sum_{cyc}\frac{a^2b^2(a+b)}{ab(a+b)}+2=ab+bc+ca+2=ab+bc+ca+2(a^2+b^2+c^2) \geq 3(ab+bc+ca)$ [last by AM-GM].

$a^5+b^5\ge a^4b+b^4a=ab(a+b)(a^2-ab+b^2)$ whriting for all fractions we will get the following inequality which is obvious) $a^2+b^2+c^2\ge ab+ac+bc$

This inequality is equivalent with $ \sum\frac{(a+b)^2}{ab}(a-b)^{2}\geq 0 $ This is SOS form

http://www.artofproblemsolving.com/community/c6h481723p2698355: IF $ a$, $ b$ and $ c$ are positive reals such that $ a^{2}+b^{2}+c^{2}=1$ prove the inequality: \[ \frac{a^{5}+b^{5}}{ab(a+b)}+ \frac {b^{5}+c^{5}}{bc(b+c)}+\frac {c^{5}+a^{5}}{ca(a+b)}\geq 1-\lambda (ab+bc+ca-1).(\lambda\leq5)\]Stronger than B&H 2008

$\frac{c^5+a^5}{ca(a+b)}$ should be corrected as $\frac{c^5+a^5}{ca(c+a)}$

Clearing the denominator, we get $6S(7,3,1)+6S(7,2,2)+6S(6,4,1)+18S(6,3,2) \geq 6S(5,4,2)+6S(5,3,3)+24S(4,4,3)$ This holds by Muirhead inequality

arshakus wrote: $a^5+b^5\ge a^4b+b^4a=ab(a+b)(a^2-ab+b^2)$ whriting for all fractions we will get the following inequality which is obvious) $a^2+b^2+c^2\ge ab+ac+bc$ Then $P(a,b,c)$=LHS(say) $P(a,b,c)\ge 2(a^2+b^2+c^2)–2(ab+bc+ac)\ge 3 (ab+bc+ac)–2$ will suffice $\implies 2(a^2+b^2+c^2)+2\ge 4(ab+bc+ac)$ $2.1=2(a^2+b^2+c^2)$ Above reduces $4(a^2+b^2+c^2)\ge 4(ab+bc+ac)$ obvious true