Home » Uncategorized » Inequality 20(Vasile Cirtoaje)

# Inequality 20(Vasile Cirtoaje)

Problem:

Let $\displaystyle x,y,z$ be non-negative numbers, no two of them are zero. Prove that

$\displaystyle \frac{x^2-yz}{x+y}+\frac{y^2-zx}{y+z}+\frac{z^2-xy}{z+x}\geq 0$.

Solution:

Making the Cauchy reverse technique we have that $\displaystyle \frac{x^2-yz}{x+y}=-z+\frac{x^2+zx}{x+y}=\frac{x(x+z)}{x+y}-z$. Doing that cyclic over the $\displaystyle 3$ fractions we need to prove that

$\displaystyle \frac{x(x+z)}{x+y}+\frac{y(y+x)}{y+z}+\frac{z(z+y)}{z+x}\geq x+y+z$.

Let us now multiply each fraction by $\displaystyle x(x+z),y(y+x),z(z+y)$ respectively and apply Cauchy-Schwartz inequality. We get that:

$\displaystyle \frac{x^2(x+z)^2}{x(x+z)(x+y)}+\frac{y^2(y+x)^2}{y(y+x)(y+z)}+\frac{z^2(z+y)^2}{z(z+y)(z+x)}\geq \frac{\left(\sum_{cyc}x^2+\sum_{cyc}xy\right)^{2}}{\sum_{cyc}x(x+y)(x+z)}$, or $\displaystyle \frac{\left(\sum_{cyc}x^2+\sum_{cyc}xy\right)^{2}}{\sum_{cyc}x^3+\sum_{cyc}x\sum_{cyc}xy}$.

So, the current inequality reduces to

$\displaystyle \frac{\left(\sum_{cyc}x^2+\sum_{cyc}xy\right)^{2}}{\sum_{cyc}x^3+\sum_{cyc}x\sum_{cyc}xy}\geq \sum_{cyc}x$.

Doing some manipulations we have that

$\displaystyle \left(\sum_{cyc}x^2+\sum_{cyc}xy\right)^{2}\geq \sum_{cyc}x\sum_{cyc}x^3+\left(\sum_{cyc}x\right)^2\sum_{cyc}xy$.

Now we have to expand the sums. So:

• $\displaystyle \left(\sum_{cyc}x^2+\sum_{cyc}xy\right)^{2}=\sum_{cyc}x^4+2\sum_{cyc}x^2y^2+\left(\sum_{cyc}xy\right)^2+2\sum_{cyc}x^2\sum_{cyc}xy$.
• $\displaystyle \sum_{cyc}x\sum_{cyc}x^3=\sum_{cyc}x^4+\sum_{cyc}xy(x^2+y^2)$.
• $\displaystyle \left(\sum_{cyc}x\right)^{2}\sum_{cyc}xy=\sum_{cyc}x^2\sum_{cyc}xy+2\left(\sum_{cyc}xy\right)^{2}$
• $\displaystyle \sum_{cyc}x^2\sum_{cyc}xy=\sum_{cyc}xy(x^2+y^2)+xyz\sum_{cyc}x$

Back to our inequality now, if we substitute those sums we reach to the obvious conclusion

$\displaystyle \sum_{cyc}x^2y^2\geq xyz\sum_{cyc}x$, Q.E.D.

Advertisements