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

$$ \frac{ x_1+x_2 }{2} \geq \sqrt{ x_1x_2 } $$

The equality holds if $x_1=x_2$.

Problem (Russia, 1995)

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

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

 

Solution

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

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

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

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

 

Therefore we have

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

which is what needed to be proven.

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