AM-GM Inequality for Two Variables

Inequality of Arithmetic and Geometric Means (AM-GM) states that for the positive real numbers $x_1$, $x_2$ it holds that

$$\frac{x_1+x_2}{2}\ge \sqrt{x_1{x}_2}$$

The equality holds if $x_1=x_2$.

Problem (Russia, 1995)

Let $x$ and $y$ be positive real numbers. Show that

$$\frac{x}{x^4+y^2}+\frac{y}{y^4+x^2}\le \frac{1}{xy}$$

 

Solution

First, let us apply the AM-GM inequality to the numbers $x^4$ and $y^2$:

$$x^4+y^2\ge 2\sqrt{x^4{y}^2}=2x^{2}y$$

Now, let us apply the AM-GM inequality to the numbers $y^4$ and $x^2$:

$$y^4+x^2\ge 2\sqrt{y^4{x}^2}=2y^{2}x$$

 

Therefore we have

$$\frac{x}{x^4+y^2}+\frac{y}{y^4+x^2}\le \frac{x}{2x^{2}y}+\frac{y}{2y^{2}x}=\frac{1}{xy}$$

which is what needed to be proven.

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