Proving $\frac{a^4}{b^3} +\frac{b^4}{c^3} +\frac{c^4}{a^3} \ge 3$

Question

For each $a,b,c > 0$ and $a^5+b^5+c^5=3 $ .How to prove that :

$$\frac{a^4}{b^3} +\frac{b^4}{c^3} +\frac{c^4}{a^3} \ge 3$$

Answer 1

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 $$ \frac {x^8} {y^6} + \frac {y^8} {z^6} + \frac {z^8} {x^6} \geq \frac 9 {x^{12} y^6 + y^{12} z^6 + z^{12} x^6} \geq 3 $$