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 $a,b,c > 0$ ?
 A: As has been mentioned in some comments too, this question does not seem to allow a pre-calculus solution. Anyway, it is again a question of the symmetric type, such as listed in:

Why does Group Theory not come in here?

Completed with:


Do symmetric problems have symmetric solutions?

From the latter article comes the following
Theorem (The Purkiss Principle). Let $f$ and $g$ be symmetric functions with continuous second derivatives in the neighborhood of a point $P = (r, \cdots, r)$.
On the set where $g$ equals $g(P)$, the function $f$ will have a local maximum or minimum at $P$ except in degenerate cases.

Our function $f$ in this case is:
$$ f(a,b,c) = \frac{a^3}{a^2+b^2}+\frac{b^3}{b^2+c^2}+\frac{c^3}{c^2+a^2}$$
Applying the Purkiss Principle gives (with $r > 0$):
$$ g(r,r,r)=r^2+r^2+r^2=3 \quad \Longrightarrow \quad r=1 \quad \Longrightarrow \quad f(r,r,r) = \frac{3}{2}$$
The main problem with the Purkiss Principle, most of the time, is to prove that the minimum found is global,
making this part of the answer only a partial answer.


Another possibility would have been to introduce spherical coordinates, thereby eliminating the equation $a^2+b^2+c^2=3$ :
$$
a = \sqrt{3}\cos(\theta)\cos(\phi) \qquad b = \sqrt{3}\cos(\theta)\sin(\phi) \qquad c = \sqrt{3}\sin(\theta)
$$
But this destroys the beautiful symmetry in the first place. And it doesn't help much in the second place, because
the partial derivatives to $\theta$ and $\phi$ are too messy to determine the values where they are zero, which
would be required for establishing a minimum (at least MAPLE - as steered by me - refuses to do the job).
EDIT. Which is a bit too pessimistic view; one can do something with those spherical coordinates:
$$
a = \sqrt{3}\cos(\theta)\cos(\phi)=1 \quad , \quad b = \sqrt{3}\cos(\theta)\sin(\phi)=1 \quad , \quad c = \sqrt{3}\sin(\theta)=1 \\
\Longrightarrow \quad \theta = \arcsin(1/\sqrt{3})=\arctan(1/\sqrt{2}) \quad \Longrightarrow \quad \phi = \arcsin(1/\sqrt{2})=\arctan(1)=\pi/4
$$
Substitution of the spherical coordinates in our function $f$ defines $\;g(\phi,\theta)=f(a,b,c)-3/2\;$.After some tedious calculations (or rather trusting MAPLE :-) for $(\phi_0,\theta_0) = \arctan(1,1/\sqrt{2})$ :
$$
g(\phi_0,\theta_0) = 0 \quad ; \quad
\left[ \begin{array}{c} \frac{\partial g}{\partial \phi} \\ \frac{\partial g}{\partial \theta} \end{array} \right](\phi_0,\theta_0) = 0 \quad ; \quad
\left[ \begin{array}{cc} \frac{\partial^2 g}{\partial \phi^2} & \frac{\partial^2 g}{\partial \phi \, \partial \theta} \\
\frac{\partial^2 g}{\partial \theta \, \partial \phi} & \frac{\partial^2 g}{\partial \theta^2} \end{array} \right](\phi_0,\theta_0) =
\left[ \begin{array}{cc} 3 & 0 \\ 0 & 3 \end{array} \right]
$$
Proving that there is indeed a local extreme at $\;g(\phi_0,\theta_0)\;$ and that this extreme is a minimum.The problem is, again, to prove that the minimum found is global, making this part of the answer a partial answer as well.
A: 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.
A: 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 
