<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] How $a^2+b^2+c^2=3$, then is equivalent to proof that 
$$\frac{a^3}{3-a^2}+\frac{b^3}{3-b^2}+\frac{c^3}{3-c^2}\geq\frac{3}{2}$$
Now the function $f(x)=\frac{x^3}{3-x^2}$ have $f''(x)>0$ in $(0,\sqrt{3})$ then $f(x)$ is convex in this interval, then:
$$f(a)+f(b)+f(c)\geq3f\left(\frac{a+b+c}{3}\right)$$ 
how $(a+b+c)^2=a^{2}+b^{2}+c^{2}+2ab+2ac+2bc$, for the Rearrangement Inequality,  $2ab+2ac+2bc\leq 2a^2+2b^2+2c^2$, then $(a+b+c)^2=a^{2}+b^{2}+c^{2}+2ab+2ac+2bc\leq 3(a^2+b^2+c^2)=9\Longrightarrow a+b+c\leq3$
A hint for conclude the proof is use the  "smoothing principle", ($a=b=c$), so in this case $f(a)+f(b)+f(c)$ is minimized,  (but I do not understand this topic), and then:
$$f(a)+f(b)+f(c)\geq 3f(1)=\frac{3}{2}\geq3f\left(\frac{a+b+c}{3}\right)$$
God  bless 
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