<s>[INST] An Inequality Problem with not nice conditions How to show that $\dfrac{a^3}{a^2+b^2} + \dfrac{b^3}{b^2+c^2} + \dfrac{c^3}{c^2+a^2} \ge \dfrac32$, where $a^2+b^2+c^2=3$, and $a,b,c > 0$ ?
 [/INST] By weighted AM-HM inequality, $\sum\limits_{cyc}\dfrac{\dfrac{a\cdot a^2}{a^2+b^2}}{3}\ge \dfrac{3}{\sum\limits_{cyc}\dfrac{a^2(a^2+b^2)}{a}}$  
Here we consider the weights to be $a^2$,$b^2$ and $c^2$ respectively.  
From the inequality above, we can have $\sum\limits_{cyc}\dfrac{a\cdot a^2}{a^2+b^2}\ge \dfrac{9}{\sum\limits_{cyc}\dfrac{a^2(a^2+b^2)}{a}}$  
Now, if we can show $a(a^2+b^2)+b(b^2+c^2)+c(c^2+a^2)\le 6$ we will be done.  
We can have, $a(3-c^2)+b(3-a^2)+c(3-b^2)\le 6 \implies 3(a+b+c)-(ac^2+ba^2+cb^2)$  
From Cauchy-Schwarz inequality, we can have, $3(a+b+c)\le 9$  
Now, we are left to prove, $ac^2+ba^2+cb^2\ge 3$  
Again, using weighted AM-HM inequality,$ac^2+ba^2+cb^2\ge \dfrac{9}{\frac{ac}{c}+\frac{ab}{a}+\frac{bc}{b}}$  
Now, $\dfrac{ac}{c}+\dfrac{ab}{a}+\dfrac{bc}{b}= a+b+c\le 3$  
Hence, we are done.  
I still believe there exists a simpler solution by a further application of Chebyshev Inequality.
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