Proving $\frac{a^4}{b^3} +\frac{b^4}{c^3} +\frac{c^4}{a^3} \ge 3$ For each $a,b,c > 0$ and $a^5+b^5+c^5=3 $ .How to prove that :
$$\frac{a^4}{b^3} +\frac{b^4}{c^3} +\frac{c^4}{a^3} \ge 3$$

Since $a, b, c$ are positive, we can set
$$
a = x^2\quad b = y^2 \quad c = z^2
$$
The inequality becomes
$$
\frac {x^8} {y^6} + \frac {y^8} {z^6} + \frac {z^8} {x^6} \geq 3
$$
with the condition $x^{10} + y^{10} + z^ {10} = 3$.
Applying Hölder's inequality we get:
$$
3 = (x^{10} + y^{10} + z^{10})^{9/10} \cdot (1 + 1 + 1)^{1/10} \geq x^9 + y^9 + z^9 
$$
By means of AM - GM inequality we can write
$$
\begin{align}
3 &\geq \frac {(x^9 + y^9 + z^9)^2} {3} \\
 &= \sum_{cyc} x^9 \frac {x^9 + 2y^9} {3} \\
 &\geq \sum_{cyc} x^9 \sqrt[3] {x^9 y^{18}} \\
 &= x^{12} y^6 + y^{12} z^6 + z^{12} x^6
\end{align}
$$
Applying Cauchy-Schwarz inequality we conclude
$$
\left(\frac {x^8} {y^6} + \frac {y^8} {z^6} + \frac {z^8} {x^6} \right) \cdot
(x^{12} y^6 + y^{12} z^6 + z^{12} x^6) \geq (x^{10} + y^{10} + z^{10})^2 = 9
$$
and finally
$$
\frac {x^8} {y^6} + \frac {y^8} {z^6} + \frac {z^8} {x^6} \geq \frac 9 {x^{12} y^6 + y^{12} z^6 + z^{12} x^6} \geq 3
$$
