$xy\leq x^2+y^2$ is a naive form of Cauchy Schwarz? In class, my professor explained that $xy\leq x^2+y^2$ for $x,y \in \mathbb{R}$ is a naive form of the Cauchy-Schwarz inequality. Is this true and how can I see this?
 A: It is not exactly the CS inequality. But there are several ways to see this. Observe for example:
$$
0 \leq (x-y)^2 = x^2+y^2-2xy
$$
Rearrange to get
$$
xy \leq \frac{1}{2}(x^2+y^2).
$$
Or another way to see this: Use that $\log$ is concave.
$$
\log\left(\frac{1}{2}x^2+\frac{1}{2}y^2 \right) \geq \frac{1}{2}\left(\log(x^2) + \log(y^2) \right)= \frac{1}{2}\log(x^2y^2) = \log\left(\sqrt {x^2y^2} \right) = \log(\lvert xy \rvert) 
$$
Take the exponential on both sides to see an even better version of the last inequality.
There are literally dozens of proofs.
This inequality is a special case of the very important Young inequality. It can be used to prove Hölder's inequality a special case of which is the CS inequality.
A: The inequality of Cauchy-Schwarz is :
$$\forall n \in \mathbb{N}^*, \forall x_1, \ldots, x_n, y_1, \ldots, y_n \in \mathbb{R}, \left(x_1 y_1 + \cdots + x_n y_n\right)^2 \leq \left(x_1^2 + \cdots + x_n^2\right) \left(y_1^2 + \cdots + y_n^2\right)$$
When $n = 2$, $x_1 = y_2 = a$ and $x_2 = y_1 = b$ and we have :
$$\left(a b + b a\right)^2 \leq \left(a^2 + b^2\right) \left(b^2 + a^2\right)$$
We deduce that :
$$a b \leq \dfrac{a^2 + b^2}{2} \leq a^2 + b^2$$
