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

Question

I am studying AM-GM inequalities in school and have this problem:

With $x,y,z>0$ show that $\frac{x^5}{y^3}+\frac{y^5}{z^3}+\frac{z^5}{x^3}\geq x^2+y^2+z^2$.

Answer 1

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.

Answer 2

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, $$\sum_{cyc}\frac{x^5}{y^3}\geq\sum_{cyc}\frac{5x^2-3y^2}{2}=\sum_{cyc}x^2.$$