Framing problem : Show that $\frac{x^2}{yz+1}+\frac{y^2}{zx+1}+\frac{z^2}{xy+1}\leq\frac{x^2+y^2+z^2}{4} +1$

Question

Let $x,y,z\gt 0$ reals and let $x^3+y^3+z^3=4xyz$

Show that : $$\frac{x^2}{yz+1}+\frac{y^2}{zx+1}+\frac{z^2}{xy+1}\leq\frac{x^2+y^2+z^2}{4} +1$$ Can I use Cauchy inequality to solve it ?

Answer 1

Using Cauchy-Inequality:

$$\left(\frac{x^2}{yz}+\frac{x^2}{1}\right)\left(yz+1\right)\geq (x+x)^2\implies \frac{x^2}{yz+1}\leq \frac{x^2}{4yz}+\frac{x^2}{4}$$

Therefore

$$\frac{x^2}{yz+1}+\frac{y^2}{zx+1}+\frac{z^2}{xy+1}\leq\frac{x^2}{4yz}+\frac{y^2}{4zx}+\frac{z^2}{4xy}+ \frac{x^2+y^2+z^2}{4}=1+ \frac{x^2+y^2+z^2}{4}$$