This is not on the shortlist either. Somebody has to explain what is happening.

dangerousliri wrote: This is not on the shortlist either. Somebody has to explain what is happening. Look at the shortlist here: https://pregatirematematicaolimpiadejuniori.wordpress.com/2020/09/12/shl2019/

Okay. Maybe something happened since on the version I have there wasn't these problems. That is why I asked.

Solution Firstly I will arrange this ineq We have $$ \frac{a^2+ac+c^2}{a+b+c} + \frac{(a+c)^2-ac}{a+c} + \frac{b^2}{b+c} + \frac{b^2}{b+a} > a+b+c$$$$ \frac{a^2+ac+c^2}{a+b+c} + \frac{b^2}{b+c} + \frac{b^2}{b+a} > b + \frac{ac}{a+c}$$$$ \frac{a^2+ac+c^2}{a+b+c} + \frac{b^2}{b+c} > \frac{ab}{a+b} + \frac{ac}{a+c}$$$$ \frac{a^2+ac+c^2}{a+b+c} + \frac{b^2}{b+c} + \frac{bc}{b+c} > \sum \frac{ab}{a+b}$$$$ \frac{a^2+b^2+c^2+ab+bc+ac}{a+b+c} > \sum \frac{ab}{a+b}$$$$ a^2+b^2+c^2+ab+bc+ac > \sum ab + abc \sum \frac{1}{a+b}$$$$ a^2+b^2+c^2 > abc \sum \frac{1}{a+b}$$We have $a+b \ge 2\sqrt{ab} \Rightarrow \sum \frac{1}{a+b} \le \sum \frac{1}{2 \sqrt{ab}} = \frac{\sum \sqrt{a}}{2 \sqrt{abc}}$ So it remains to prove that $$2(a^2+b^2+c^2) > \sum a \sqrt{bc}$$We have $a^2+b^2+c^2 = \frac{a^2}{2} + \frac{a^2}{2} +b^2+c^2 \ge \frac{4a \sqrt{bc}}{\sqrt{2}}$ and summing cyclically gives the result.

Steve12345 wrote: Let $a, b, c$ be positive real numbers. Prove the inequality $$(a^2+ac+c^2) \left( \frac{1}{a+b+c}+\frac{1}{a+c} \right)+b^2 \left( \frac{1}{b+c}+\frac{1}{a+b} \right)>a+b+c$$ $$3a^2+4b^2+2c^2+4ab+2ca>2(a+b)(a+b+c)\iff a^2+b^2+c^2+(b-c)^2>0$$$$(a^2+ac+c^2) \left( \frac{1}{a+b+c}+\frac{1}{a+c} \right)+b^2 \left( \frac{1}{b+c}+\frac{1}{a+b} \right)$$$$= \frac{a^2+ac+c^2}{a+b+c}+\frac{a^2}{a+c} +\frac{b^2}{a+b}+\frac{b^2}{b+c}+c\geq\frac{a^2+ac+c^2}{a+b+c}+\frac{(a+2b)^2}{2(a+b+c)} +c$$$$=\frac{3a^2+4b^2+2c^2+4ab+2ca}{2(a+b+c)}+c >a+b+c$$

the inequality is the same as $\frac{a^2}{a+c} +\frac{c^2}{a+c} + \frac{b^2}{b+a}+\frac{b^2}{b+c} + \frac{(a+c)^2}{a+b+c} >a+b+c +ac(negative)$ now $\frac{a^2}{a+c} +\frac{c^2}{a+c} + \frac{b^2}{b+a} ++\frac{b^2}{b+c} + \frac{(a+c)^2}{a+b+c} > \frac{a^2}{a+c} +\frac{c^2}{a+c} + \frac{b^2}{2b+a} +\frac{b^2}{b+c} + \frac{(a+c)^2}{a+b+c} >= a+b+c>a+b+c +ac(negative)$ BY CS

electrovector wrote: Solution Firstly I will arrange this ineq We have $$ \frac{a^2+ac+c^2}{a+b+c} + \frac{(a+c)^2-ac}{a+c} + \frac{b^2}{b+c} + \frac{b^2}{b+a} > a+b+c$$$$ \frac{a^2+ac+c^2}{a+b+c} + \frac{b^2}{b+c} + \frac{b^2}{b+a} > b + \frac{ac}{a+c}$$$$ \frac{a^2+ac+c^2}{a+b+c} + \frac{b^2}{b+c} > \frac{ab}{a+b} + \frac{ac}{a+c}$$$$ \frac{a^2+ac+c^2}{a+b+c} + \frac{b^2}{b+c} + \frac{bc}{b+c} > \sum \frac{ab}{a+b}$$$$ \frac{a^2+b^2+c^2+ab+bc+ac}{a+b+c} > \sum \frac{ab}{a+b}$$$$ a^2+b^2+c^2+ab+bc+ac > \sum ab + abc \sum \frac{1}{a+b}$$$$ a^2+b^2+c^2 > abc \sum \frac{1}{a+b}$$We have $a+b \ge 2\sqrt{ab} \Rightarrow \sum \frac{1}{a+b} \le \sum \frac{1}{2 \sqrt{ab}} = \frac{\sum \sqrt{a}}{2 \sqrt{abc}}$ So it remains to prove that $$2(a^2+b^2+c^2) > \sum a \sqrt{bc}$$We have $a^2+b^2+c^2 = \frac{a^2}{2} + \frac{a^2}{2} +b^2+c^2 \ge \frac{4a \sqrt{bc}}{\sqrt{2}}$ and summing cyclically gives the result. I've got a bit easier solution. The original inequality can be easily rewritten as $\frac{a^{2}+ac+c^{2}}{a+b+c}+b > \frac{ab}{a+b}+\frac{bc}{b+c}+\frac{ac}{a+c}$. Note that by AM-HM we have $\frac{a+b}{4}>\frac{ab}{a+b}$. Thus RHS of inequality is not greater than $\frac{a+b+c}{2}$. We are left to prove that $\frac{a^{2}+ac+c^{2}}{a+b+c}>\frac{a+c-b}{2}$. After some routine calculations we can see that it is the same as $a^{2}+b^{2}+c^{2}>0$ which is true. We are done:)

sqing wrote: Steve12345 wrote: Let $a, b, c$ be positive real numbers. Prove the inequality $$(a^2+ac+c^2) \left( \frac{1}{a+b+c}+\frac{1}{a+c} \right)+b^2 \left( \frac{1}{b+c}+\frac{1}{a+b} \right)>a+b+c$$ $$3a^2+4b^2+2c^2+4ab+2ca>2(a+b)(a+b+c)\iff a^2+b^2+c^2+(b-c)^2>0$$$$(a^2+ac+c^2) \left( \frac{1}{a+b+c}+\frac{1}{a+c} \right)+b^2 \left( \frac{1}{b+c}+\frac{1}{a+b} \right)$$$$= \frac{a^2+ac+c^2}{a+b+c}+\frac{a^2}{a+c} +\frac{b^2}{a+b}+\frac{b^2}{b+c}+c\geq\frac{a^2+ac+c^2}{a+b+c}+\frac{(a+2b)^2}{2(a+b+c)} +c$$$$=\frac{3a^2+4b^2+2c^2+4ab+2ca}{2(a+b+c)}+c >a+b+c$$ How did you get this $$=\frac{3a^2+4b^2+2c^2+4ab+2ca}{2(a+b+c)}+c >a+b+c$$

Can someone explain to me why is this true $$=\frac{3a^2+4b^2+2c^2+4ab+2ca}{2(a+b+c)}+c >a+b+c$$

besnikhaziri wrote: Can someone explain to me why is this true $$=\frac{3a^2+4b^2+2c^2+4ab+2ca}{2(a+b+c)}+c >a+b+c$$ Just multiply both sides by $2(a+b+c)$ to get $$3a^2+4b^2+2c^2+4ab+2ca+2(c)(a+b+c)>2(a+b+c)^2$$which is equivalent to $3a^2+4b^2+4c^2+4ab+2bc+4ca>2a^2+2b^2+2c^2+4ab+4bc+4ca$ or $$a^2+2b^2+2c^2>2bc$$but it's obviously true because $a^2+2b^2+c^2>b^2+c^2\geq 2bc$.

letting $b=1$ and $s=a+c$, $p=ac$ the inequality can be written as: \[\frac{(s^2-p)(2s+1)}{s(s+1)}+\frac{s+2}{s+p+1}>s+1\]after expanding and cancelling we need to show \[s^4+s^3+s^2+s+s^3p\ge 3ps^2+2p^2s+4ps+p^2+p\]Using the AM-GM inequality $s^2\ge 4p$ we prove that using: $s^4\ge 4ps^2\ge 3ps^2+4p^2$, $s^3\ge 4ps$, $s^2\ge 4p$, $s^3p\ge 4p^2s$ I have explained this problem in my inequalities tutorial playlist on my youtube channel little fermat. Here is the link