Use this lemma $a^3+b^3+c^3 \ge sum \frac{c(a^2+b^2)}{2} $ and $\frac{ab}{a^2+b^2}=\frac{1}{c(a^2+b^2)}$

hello, it can also be proved by BW. Sonnhard.

Dear rightways. Show us a full proof. Thank you!

Hope that I didn't make any stupid mistake. This problem is equivalent to the following inequality \[ \frac{2\left(a^3+b^3+c^3\right)}{abc}+\frac{2ab}{a^2+b^2}+\frac{2bc}{b^2+c^2}+\frac{2ca}{c^2+a^2} \ge 9 \] \[ \Leftrightarrow \frac{(a+b+c)(\left(a-b)^2+(b-c)^2+(c-a)^2\right)}{abc} \ge \frac{(a-b)^2}{a^2+b^2}+\frac{(b-c)^2}{b^2+c^2}+\frac{(c-a)^2}{c^2+a^2} \] \[ \Leftrightarrow \sum\limits_{cyc}{(a-b)^2\left(\frac{a+b+c}{abc}-\frac{1}{a^2+b^2}\right)} \] Then notice that we have \[ \frac{a+b+c}{abc} \ge \frac{1}{ab} \ge \frac{2}{a^2+b^2} \] In conclusion, $ \sum\limits_{cyc}{(a-b)^2[\frac{a+b+c}{abc}-\frac{1}{a^2+b^2}]} \ge 0 $, as desired.

For #1 We have \[\frac{a^3+b^3+c^3}{abc}+\sum{\frac{ab}{a^2+b^2}}-\frac{9}{2}=\sum{\frac{(2a+2b+c)(a-b)^2}{4abc}}+\sum{\frac{(a-b)^4}{4ab(a^2+b^2)}}\ge{0}\]

The following inequality is stronger. For positives $a$, $b$ and $c$ prove that: \[\frac{1}{4}\left(\frac{a+b}{c}+\frac{b+c}{a}+\frac{c+a}{b}\right)+\frac{ab}{a^2+b^2}+\frac{bc}{b^2+c^2}+\frac{ca}{c^2+a^2}\geq3\]

By using SOS again, we have the following result. \[\frac{1}{4}\left(\frac{a+b}{c}+\frac{b+c}{a}+\frac{c+a}{b}\right)+\frac{ab}{a^2+b^2}+\frac{bc}{b^2+c^2}+\frac{ca}{c^2+a^2}\geq3\] \[ \Leftrightarrow \sum\limits_{cyc}{(a-b)^2\left(\frac{1}{2ab}-\frac{1}{a^2+b^2}\right)} \ge 0 \] , which is obvious by AM-GM inequality.

Our Solution Anar And Emil Istek Lyceum

For #7 We have \[\frac{1}{4}\sum{\frac{a+b}{c}}+\sum{\frac{ab}{a^2+b^2}}-3=\sum{\frac{(a-b)^4}{4ab(a^2+b^2)}}\ge{0}\]

My Solution: Notice that $\dfrac{ab}{a^2+b^2}=\dfrac{1}{c(a^2+b^2)}$ since $abc=1$ So we have to prove that $a^3+b^3+c^3+\frac{1}{c(a^2+b^2)}+\frac{1}{a(b^2+c^2)}+\frac{1}{b(c^2+a^2)} \geq \frac{9}{2}$ Or $a^3+b^3+c^3+3abc+\frac{1}{c(a^2+b^2)}+\frac{1}{a(b^2+c^2)}+\frac{1}{b(c^2+a^2)} \geq \frac{9}{2}+3$ Applying Schur's inequality We only have to prove $\sum{c(a^2+b^2)}+\frac{1}{c(a^2+b^2)}+\frac{1}{a(b^2+c^2)}+\frac{1}{b(c^2+a^2)} \geq \frac{9}{2}+3$ By AM-GM we have $\dfrac{1}{4}.c(a^2+b^2)+\frac{1}{c(a^2+b^2)}\geq 1$ And since $abc=1$ we have $\sum{c(a^2+b^2)}\geq 6$ Summing all inequalities above, we are done

Apply A.M. G.M Inequality twice in the (a^3+b^3+c^3) portion and the other portion .after that In the 2nd portion use rearrangement, then we are done

Tsarik wrote: Let $a,b,c$ be positive real numbers such that $abc=1$. Prove the following inequality: $a^3+b^3+c^3+\frac{ab}{a^2+b^2}+\frac{bc}{b^2+c^2}+\frac{ca}{c^2+a^2} \geq \frac{9}{2}$. My solution : We easily see that : $$\dfrac{ab}{a^2+b^2}=\dfrac{(a+b)^2}{2(a^2+b^2)}-\dfrac{1}{2}$$ As a result, $$\dfrac{ab}{a^2+b^2}+\dfrac{bc}{b^2+c^2}+\dfrac{ca}{c^2+a^2}=\dfrac{(a+b)^2}{2(a^2+b^2)}+\dfrac{(b+c)^2}{2(b^2+c^2)}+\dfrac{(c+a)^2}{2(c^2+a^2)}-\dfrac{3}{2}\geq \frac{(a+b+c)^2}{a^2+b^2+c^2}-\dfrac{3}{2}$$ On the other hand, $$a^3+b^3+c^3\geq \dfrac{(a^2+b^2+c^2)^2}{a+b+c}$$ Therefore, we just prove : $$\dfrac{(a^2+b^2+c^2)^2}{a+b+c}+\dfrac{(a+b+c)^2}{a^2+b^2+c^2}\geq 6$$ Which is true because of applying AM-GM, $$\dfrac{(a^2+b^2+c^2)^2}{a+b+c}+\dfrac{(a+b+c)^2}{a^2+b^2+c^2}\geq2\sqrt{(a^2+b^2+c^2)(a+b+c)}\geq 2\sqrt{9abc}=6$$

The problem return to: $ a,b,c>0,abc=1\Rightarrow a^3+b^3+c^3-3abc\ge\sum{\frac{1}{2}-\frac{bc}{b^2+c^2}} $ $ \Leftrightarrow (b-c)^2[a+b+c-\frac{1}{b^2+c^2}]+(c-a)^2[a+b+c-\frac{1}{c^2+a^2}]+(a-b)^2[a+b+c-\frac{1}{a^2+b^2}] $ Because $ a\ge\frac{1}{b^2+c^2}\Leftrightarrow \frac{1}{bc}\ge\frac{1}{b^2+c^2}\Leftrightarrow bc\le b^2+c^2 $ and the other two similar, result it follows that the last inequality is true! _________ Marin Sandu

I am not sure if this is correct though... I'll be leaving out some details It's easy to show that $a^3+b^3 \geq \frac{(a+b)(a^2+b^2)}{2}$. (Muirhead's) Thus $a^3+b^3+\frac{2ab}{a^2+b^2} \geq \frac{(a+b)(a^2+b^2)}{2}+\frac{2ab}{a^2+b^2}$$=2\frac{(a+b)(a^2+b^2)}{4}+\frac{2ab}{a^2+b^2}$. But by AM-GM this is at least $\frac{^3\sqrt{(a+b)^2(a^2+b^2)ab}}{2}$. Again it is easy to show (again by Muirhead's) that this is at least $3ab$. But then we can develop similar expressions corresponding to $(b,c)$ and $(c,a)$. We then add up everything; the LHS of the sum is twice the original LHS. Then we get that the LHS of the original ineq is at least $\frac{3(ab+ac+bc)}{2}$, which is clearly at least $\frac{9}{2}$ by Muirhead's

We have also $ a^3+b^3+c^3 \ge a^2+b^2+c^2 $ and by am-gm for six numbers , so we are done!

$a^3+b^3+c^3+\frac{ab}{a^2+b^2}+\frac{bc}{b^2+c^2}+\frac{ca}{c^2+a^2} \geq \frac{1}{2}(a^2+b^2+c^2)+ \frac{ab}{a^2+b^2}+\frac{bc}{b^2+c^2}+\frac{ca}{c^2+a^2} +\frac{3}{2}\geq\frac{9}{2}$. here

Let $t=a^3+b^3+c^3$ , hence $t\geq3\implies (t-3)(2t-3)\geq0\iff t+\frac{9}{2t}\geq \frac{9}{2},$ $a^3+b^3+c^3+\frac{ab}{a^2+b^2}+\frac{bc}{b^2+c^2}+\frac{ca}{c^2+a^2} =t+\frac{1}{c(a^2+b^2)}+\frac{1}{a(b^2+c^2)}+\frac{1}{b(c^2+a^2)}$ $ \geq t+\frac{9}{c(a^2+b^2)+a(b^2+c^2)+b(c^2+a^2)}\geq t+\frac{9}{2t}\geq \frac{9}{2}.$

$a^3+b^3+c^3+\frac{ab}{a^2+b^2}+\frac{bc}{b^2+c^2}+\frac{ca}{c^2+a^2} $ $=a^3+b^3+c^3+\frac{1}{c(a^2+b^2)}+\frac{1}{a(b^2+c^2)}+\frac{1}{b(c^2+a^2)}$ $ \geq a^3+b^3+c^3+\frac{9}{c(a^2+b^2)+a(b^2+c^2)+b(c^2+a^2)}$ $\geq \frac{a^3+b^3+c^3}{2}+\frac{a^3+b^3+c^3}{2}+\frac{9}{2(a^3+b^3+c^3)}$ $\geq 3\sqrt[3]{\frac{a^3+b^3+c^3}{2} \cdot \frac{a^3+b^3+c^3}{2}\cdot \frac{9}{2(a^3+b^3+c^3)}}$ $= \frac{3}{2}\sqrt[3]{9(a^3+b^3+c^3)} \geq \frac{9}{2}.$

Works?

Let $a,b,c$ be positive real numbers such that $abc=1$. Prove the following inequality$$a^3+b^3+c^3+\frac{4ab}{a^2+b^2}+\frac{4bc}{b^2+c^2}+\frac{4ca}{c^2+a^2} \geq 9.$$

Let $a^3+b^3+c^3=t$ hence $t\geq 3$ $ineq= t+\frac{4}{c(a^2+b^2)}+\frac{4}{a(b^2+c^2)}+\frac{4}{b(c^2+a^2)}\geq t+\frac{36}{a^2b+b^2a+c^2b+b^2c+c^2a+a^2c}\geq t+\frac{36}{t+3}\geq 9$ $\iff (t-3)^2\geq 0$

oh, this is my problem!

rightways wrote: Use this lemma $a^3+b^3+c^3 \ge sum \frac{c(a^2+b^2)}{2} $ and $\frac{ab}{a^2+b^2}=\frac{1}{c(a^2+b^2)}$ Rightways, using your idea we have: It's obviously true that $a^{3}+b^{3}+c^{3}\geqslant\sum c(a^{2}+b^{2})$ and that $\sum\frac{ab}{a^{2}+b^{2}}=\sum\frac{1}{c(a^{2}+b^{2})}$, so we have left to prove that $\frac{\sum a(b^{2}+c^{2})}{2}+\sum\frac{1}{c(a^{2}+b^{2})}\geqslant\frac{9}{2}$. From Cauchy inequality we have: $\sum\frac{1}{c(a^{2}+b^{2})}\geqslant\frac{9}{\sum c(a^{2}+b^{2})}$. Now from AM-GM we have that: $\frac{\sum_{sym}a^{2}b}{2} +\frac{9}{\sum_{sym}a^{2}b}\geqslant 3\sqrt{2}$ This means that your method is too weak for this inequality

Easy one! After opening the parentheses and homogenizating (using the fact that $abc=1$) we have to prove the following inequality: $2(\sum_{cyc}a^4b^4c+\sum_{sym}a^4b^3c^2+\sum_{sym}a^7b^2+\sum_{sym}a^5b^4+\sum_{sym}a^3b^4c^2+2\sum_{cyc}a^5b^2c^2) \geq 9(\sum_{sym}a^5b^3c+2a^3b^3c^3)$. Using AM-GM we have that: $\sum_{sym}a^3b^4c^2 \geq 6a^3b^3c^3$ $\sum_{sym}a^5b^4+2\sum_{cyc}a^5b^2c^2 \geq 2(\sum_{sym}a^5b^3c)$ Now our inequality resumes to: $2(\sum_{cyc}a^4b^4c+\sum_{sym}a^4b^3c^2+\sum_{sym}a^7b^2) \geq 6(\sum_{sym}a^5b^3c) \geq 5(\sum_{sym}a^5b^3c)+6a^3b^3c^3$.

https://artofproblemsolving.com/community/c6h1503751p8887849 https://artofproblemsolving.com/community/c6h1503753p8887869: Let $a$,$b$,$c$ be positive real numbers such that $abc=1$. Prove that \[a^{3}+b^{3}+c^{3}+\frac{ab}{a^{2}+b^{2}}+\frac{bc}{b^{2}+c^{2}}+\frac{ac}{c^{2}+a^{2}}\geq \frac{9}{5}(a^2+b^2+c^2-\frac{1}{2})\geq \frac{13}{2}(a+b+c)-15\geq 2(a+b+c)-\frac{3}{2}\geq \frac{9}{2}\]

Jul wrote: Tsarik wrote: Let $a,b,c$ be positive real numbers such that $abc=1$. Prove the following inequality: $a^3+b^3+c^3+\frac{ab}{a^2+b^2}+\frac{bc}{b^2+c^2}+\frac{ca}{c^2+a^2} \geq \frac{9}{2}$. My solution : We easily see that : $$\dfrac{ab}{a^2+b^2}=\dfrac{(a+b)^2}{2(a^2+b^2)}-\dfrac{1}{2}$$As a result, $$\dfrac{ab}{a^2+b^2}+\dfrac{bc}{b^2+c^2}+\dfrac{ca}{c^2+a^2}=\dfrac{(a+b)^2}{2(a^2+b^2)}+\dfrac{(b+c)^2}{2(b^2+c^2)}+\dfrac{(c+a)^2}{2(c^2+a^2)}-\dfrac{3}{2}\geq \frac{(a+b+c)^2}{a^2+b^2+c^2}-\dfrac{3}{2}$$On the other hand, $$a^3+b^3+c^3\geq \dfrac{(a^2+b^2+c^2)^2}{a+b+c}$$Therefore, we just prove : $$\dfrac{(a^2+b^2+c^2)^2}{a+b+c}+\dfrac{(a+b+c)^2}{a^2+b^2+c^2}\geq 6$$Which is true because of applying AM-GM, $$\dfrac{(a^2+b^2+c^2)^2}{a+b+c}+\dfrac{(a+b+c)^2}{a^2+b^2+c^2}\geq2\sqrt{(a^2+b^2+c^2)(a+b+c)}\geq 2\sqrt{9abc}=6$$ ELEGANT!

Tsarik wrote: Let $a,b,c$ be positive real numbers such that $abc=1$. Prove the following inequality: $a^3+b^3+c^3+\frac{ab}{a^2+b^2}+\frac{bc}{b^2+c^2}+\frac{ca}{c^2+a^2} \geq \frac{9}{2}$. Because: $$LHS \geq \frac{a^3+b^3+c^3}{2}+\sum_{cyc}\left(\frac{a^2+b^2}{4}+\frac{ab}{a^2+b^2}\right)\geq \frac{a^3+b^3+c^3}{2}+\sum_{cyc} \sqrt{ab} \geq \frac{3}{2}+3=\frac{9}{2}=RHS$$

Tsarik wrote: Let $a,b,c$ be positive real numbers such that $abc=1$. Prove the following inequality: $a^3+b^3+c^3+\frac{ab}{a^2+b^2}+\frac{bc}{b^2+c^2}+\frac{ca}{c^2+a^2} \geq \frac{9}{2}$. T6/444.

Tsarik wrote: Let $a,b,c$ be positive real numbers such that $abc=1$. Prove the following inequality: $a^3+b^3+c^3+\frac{ab}{a^2+b^2}+\frac{bc}{b^2+c^2}+\frac{ca}{c^2+a^2} \geq \frac{9}{2}$. $LHS+3\geqslant 2\sum a^2+\sum \frac{ab}{a^2+b^2}=\sum \frac{a^2+b^2}{4}+\frac{ab}{a^2+b^2}+\frac{3}{2}\sum a^2\geq \sum \sqrt{ab}+\frac{3}{2}\sum a^2\geqslant 3+\frac{9}{2}=RHS+3$