Q:

Inequality problem: with $x,y,z>0$, show that $\frac{x^5}{y^3}+\frac{y^5}{z^3}+\frac{z^5}{x^3}\geq x^2+y^2+z^2$

I am studying AM-GM inequalities in school and have this problem: 
With $x,y,z>0$ show that $\frac{x^5}{y^3}+\frac{y^5}{z^3}+\frac{z^5}{x^3}\geq x^2+y^2+z^2$. 

A:

Idea behind the solution
By AM-GM you have for each $\alpha, \beta, \gamma \in \mathbb N$
$$\frac{\alpha\frac{x^5}{y^3}+\beta\frac{y^5}{z^3}+\gamma\frac{z^5}{x^3}}{\alpha+\beta+\gamma}\geq \sqrt[\alpha+\beta+\gamma]{(\frac{x^5}{y^3})^\alpha(\frac{y^5}{z^3})^\beta(\frac{z^5}{x^3})^\gamma}$$
Then by symmetry you can permute circularly the coefficients and add the inequalities.
Now, set the RHS $=x^2$. This yields
$$5 \alpha -3 \gamma = 2\alpha+2\beta+2\gamma \\
5 \beta- 3 \alpha=0 \\
5 \gamma -3 \beta =0
$$
Since the third equation is the sum of the first two (this comes from the fact that your inequality is homogeneous) your system has infinitely many solutions. Solving you get
$$9 \alpha =15 \beta =25 \gamma$$
In particular the following is a solution:
$$\alpha=25 \\
\beta =15 \\
\gamma=9$$
The solution
By AM-GM you have
$$\frac{25\frac{x^5}{y^3}+15\frac{y^5}{z^3}+9\frac{z^5}{x^3}}{49}\geq \sqrt[49]{(\frac{x^5}{y^3})^{25}(\frac{y^5}{z^3})^{15}(\frac{z^5}{x^3})^{9}}=x^2\\
\frac{9\frac{x^5}{y^3}+25\frac{y^5}{z^3}+15\frac{z^5}{x^3}}{49}\geq \sqrt[49]{(\frac{x^5}{y^3})^{9}(\frac{y^5}{z^3})^{25}(\frac{z^5}{x^3})^{15}}=y^2\\
\frac{15\frac{x^5}{y^3}+9\frac{y^5}{z^3}+25\frac{z^5}{x^3}}{49}\geq \sqrt[49]{(\frac{x^5}{y^3})^{15}(\frac{y^5}{z^3})^{9}(\frac{z^5}{x^3})^{25}}=z^2\\
$$
Add them together.

A:

By AM-GM $$2x^5+3y^5\geq5\sqrt[5]{(x^5)^2(y^5)^3}=5x^2y^3,$$ which gives
$$\frac{x^5}{y^3}\geq\frac{5x^2-3y^2}{2}.$$
Id est, $$\sum_{cyc}\frac{x^5}{y^3}\geq\sum_{cyc}\frac{5x^2-3y^2}{2}=\sum_{cyc}x^2.$$

